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Lecture Comments (2)

1 answer

Last reply by: Professor Selhorst-Jones
Thu Jul 11, 2013 12:15 PM

Post by Thalia Carver on July 3, 2013

How do I fast forward to one part of the lecture? If I move the orange bar and then hit play it jumps back to my previous location and continues playing from that point.

Systems of Linear Equations

  • A linear equation in two variables is of the form
    Ax + By = C,
    where A, B, and C are all constant numbers. Notice that each of the variables has just a power of 1, similar to a linear polynomial only having a degree of 1.
  • A system of equations is a group of multiple equations that are all true at the same time. The solution to a system is some set of variables that satisfies all the equations in the system at the same time.
  • When we first introduced graphs in this course, we talked about how a graph can be viewed as the location of all points that make the system true. With this idea in mind, we can graph each equation in a system of equations. Wherever they intersect is a solution to the system because all of the equations are simultaneously true at an intersection.
  • There are three possibilities when solving a system of linear equations:
    • 1 Solution-Independent: the equations only agree at one location;
    • 0 Solutions-Inconsistent: it is impossible for the equations to agree;
    • ∞ Solutions-Dependent: the equations agree at infinitely many locations (they are overlapping);
  • A system of linear equations can be solved by substitution: solving for one variable in terms of the other(s), then "plugging in". This gives an equation we can solve normally, then go back and find the other variable(s).
  • A system of linear equations can also be solved by elimination: adding a multiple of one equation to the other equation(s) to eliminate variables and give an equation that can be be solved normally.
  • When working with either of the two above methods, it's possible for the system to be any of the three types mentioned above. Here's how to tell which type you're working with:
    • Independent (1 solution): The system solves "normally" and you get values for the variables.
    • Inconsistent (0 solutions): While solving, you get an impossible equation (like 5=8).
    • Dependent (∞ solutions): While solving, you get an always true equation (like 7=7).
  • If you have access to a graphing calculator, you can also find the solutions to a system of equations by graphing all the equations, then find any intersection points using the calculator.
  • If a system of linear equations has more than two variables, the above methods (substitution and elimination) still work great. They become more time-consuming the more variables we have to solve for, but there is no issue in using them for any arbitrary number of variables.

Systems of Linear Equations

Solve the system of equations by substitution (if possible).
3x+y = 22               2x−3y = −11
  • To solve by substitution, we need to substitute ("plug in") one variable for an expression using only the other variable. To do this, we first must solve for one of the variables in one of the equations in the system.
  • It's quite easy to solve for y in terms of x from the first equation, so we do that:
    3x+y = 22     ⇒     y = 22−3x
  • We now substitute this in for y in the second equation:
    2x −3(22−3x) = −11
  • From here, we can simplify and solve for x:
    2x − 66 +9x = −11     ⇒     11x −66 = −11     ⇒     11x = 55     ⇒     x = 5
  • Now that we know x=5, we can plug that in and solve for y. Notice that we could plug it in to either of the two equations we started with and be able to use algebra to find y. However, the easiest way of all is to use the equation where we found y in terms of x:    y = 22−3x.  Plugging in to this equation will take the least work, since all the algebra has already been done for us. All that's left is simplifying!
    y = 22−3(5)     =     22 −15     =     7
x=5,    y=7
Solve the system of equations by elimination (if possible).
3a+2b = −1               a−2b = −11
  • To solve by elimination, we can add multiples of each equation to eliminate variables. For this problem, we see that the b variable will be eliminated if we add each equation together.
  • Add the two equations together:
    3a + 2b
    a − 2b
  • Now that one of the variables has been eliminated, we can use standard algebra to solve for the other variable in the resulting equation:
    4a = −12     ⇒     a = −3
  • Now that we know a=−3, we can plug it into either of the two starting equations from the system and solve for b. Either equation would work fine, so we will default to the first one: 3a+2b = −1 becomes
    3 (−3) + 2b = −1     ⇒     −9 + 2b = −1    ⇒     2b = 8    ⇒     b = 4
a=−3,    b=4
Graph each of the two equations below on the same set of axes, then use the resulting graph to solve the system of equations.
2x+4y = 6              5x+y = −3
  • In general, the easiest form to graph a linear equation in is slope-intercept form: y=mx+b (where m=slope and b=y-intercept). Begin by converting each of the above equations to the form.
    2x+4y = 6     ⇒     4y = −2x + 6    ⇒     y = − 1

    x + 3


    5x+y = −3     ⇒     y = −5x − 3
  • Now that each equation is in the appropriate form, it is relatively easy to graph. Place the first point for each equation where the y-intercept is, then plot the next point based on one "step" forward or backward using the corresponding slope and starting from the y-intercept:
  • Now that we have points down for each line, use a straightedge to draw in the rest of the line by connecting the two dots. Wherever the two lines intersect is the solution to the system of equations.
  • When solving by graphing, it is always a good idea to check your solution. Because we're reading a graph, the solution we find is necessarily less precise than if we had found it by another method. However, if we check it, we can be sure it is correct. From the graph, it looks like the solution is at the point (−1, 2)  ⇒  x=−1,  y=2. Plug this into each of the starting equations and check:
    2x+4y = 6              5x+y = −3

    2(−1)+4(2) = 6               5(−1) + (2) = −3

    6 = 6               −3 = −3
x=−1,    y=2
Solve the system of equations by substitution (if possible).
−4u + 20v = 1               2u − 10v = −2
  • To solve by substitution, we need to substitute ("plug in") one variable for an expression using only the other variable. To do this, we first must solve for one of the variables in one of the equations in the system. Looking at the two equations, we might see that the easiest thing to solve for will be u in the second equation (this is because when we divide to cancel out u's coefficient, it will not create any fractions).
  • Solving for u in the second equation:
    2u − 10v = −2     ⇒     2u = 10v − 2    ⇒     u = 5v−1
  • Now that we have u=5v−1, we can plug it in to the first equation:
    −4(5v−1) + 20v = 1
    Simplifying things out, we get
    −20v + 4 + 20v = 1     ⇒     4 = 1     IMPOSSIBLE!
  • Because we arrive at an impossible equation (4=1), we see that it is impossible for the system of equations to be true. That is, it has no solution.
No solution exists to the system of equations.
Solve the system of equations by elimination (if possible).
−2x+4y = −12               3x−6y=18
  • To solve by elimination, we can add multiples of each equation to eliminate variables. For this problem, we see that we can use the first equation to eliminate the x variable of the second equation by multiplying the first equation by [3/2]:

    −2x + 4y = −12
        ⇒     −3x + 6y = −18
  • Now we can add our multiple of the first equation to the second equation:
    −3x + 6y
  • Over the course of solving, we've ended up at an equation that is always true (0=0), thus the system has infinitely many solutions. We can check this by showing that the starting equations are equivalent to each other. One way to do this is to put each in slope-intercept form (y=mx+b):
    −2x+4y = −12     ⇒     4y = 2x −12     ⇒     y = 1

    x − 3

    3x−6y=18     ⇒     −6y = −3x +18     ⇒     y = 1

    x −3
    Thus, we see that the two equations are equivalent to each other. So for any (x, y) point that is true for one equation, it must also be true for the other. Therefore we have an infinite number of solutions described by the line y = [1/2] x −3.
Infinitely many solutions [The infinite set of solutions is described by the line y = [1/2] x −3, since both equations in the system are equivalent to that.]
Solve the system of equations (if possible) in whatever manner you choose.
13t+6s = 4               −7t + 2s = 24
  • We could solve this either by substitution or elimination. When given the choice, you should choose whichever method you feel more comfortable with. If you feel equally comfortable with both methods, choose the method that will make it slightly easier/faster to do the problem.
  • In this case, it looks like elimination will be a little bit easier/faster because we can easily multiply the second equation by −3 to later cancel out the s variables.
    −3 ·
    −7t + 2s = 24
        ⇒     21 t − 6s = −72
    [Note: If you decided to do it by substitution, that method is shown after the method of elimination if you want to see it.]
  • We now add this resultant equation to the first equation:
    21 t − 6s
    34 t       
    We can now use a little algebra to solve and find t = −2.
  • Once we know t=−2, we can plug it in to either of the two starting equations and solve for s. Let's choose the second equation, just because the coefficients on the variables are slightly smaller:
    −7 (−2) + 2s = 24     ⇒     14 + 2s = 24     ⇒     2s = 10     ⇒     s = 5
    Now that we have found t=−2 and s=5, we are done with the problem.
  • Alternatively, if you had decided to do it by substitution, you would begin by solving for one of the variables. Let's choose s from the second equation:
    −7t + 2s = 24     ⇒     2s = 7t + 24     ⇒     s = 7

    t + 12
  • Plug that in to the first equation and solve for t:
    13t + 6

    t + 12
    = 4     ⇒     13t + 21 t + 72 = 4     ⇒     34t = −68     ⇒     t = −2
    Once you know t=−2, plug that in to the equation we had for s:
    s = 7

    (−2) + 12     =     −7 + 12     =     5
t=−2,    s=5
Solve the system of equations (if possible).
5x + 2y = 6               3x − z = 7               −4x + 4y + z = 5
  • For this problem, we're dealing with three variables instead of just two. This isn't an issue! Substitution and elimination work fine for any number of variables. Focus on the methods as you did before: Substitution-Solve for each variable in terms of the others, then plug it in to another equation; Elimination-Add multiples of equations together to cancel out variables.
  • Since the problem does not specify which method you should use, choose the method you are more comfortable with. For this problem, elimination will work well because y and z can easily be canceled (we see how in a moment). That said, if you don't like elimination or find it difficult to follow, just use substitution instead!
  • Doing this by elimination, focus on how to cancel out each variable. x is not easy because we would have to multiply two equations as opposed to just one. The appropriate multiple of the first equation will cancel out the y's in the third equation:
    −2 ·
    5x + 2y = 6
        ⇒     −10x − 4y = −12
    The second equation is already perfectly set up to cancel the z in the third equation.
  • Add the three equations together (the multiple of the first, the second, and the third):
    −10x − 4y     
    3x        − z
    −4x + 4y + z
    Thus, we see x=0.
  • Now that we know x=0, we can plug that in to the starting equations:
    5(0) + 2y = 6     ⇒     2y = 6     ⇒     y = 3

    3(0) − z = 7     ⇒     −z = 7     ⇒     z = −7
x=0,    y = 3,    z=−7
Solve the system of equations (if possible).
2a−4b−2c = 10               −3a+6b+3c = 17               3b + c = 4
  • For this problem, we're dealing with three variables instead of just two. This isn't an issue! Substitution and elimination work fine for any number of variables. Focus on the methods as you did before: Substitution-Solve for each variable in terms of the others, then plug it in to another equation; Elimination-Add multiples of equations together to cancel out variables.
  • Since the problem does not specify which method you should use, choose the method you are more comfortable with. For this problem, let's approach it using substitution simply because the steps in the previous problem we're done with elimination. That said, elimination is a fine method, and you should feel free to use it if you're more comfortable with it.
  • In doing substitution, we want to begin by figuring out which equation we will solve for which variable. Looking at the first equation, we see that it would be reasonably easy to solve for a or c (because when we divide by the 2 coefficient, it will divide into everything evenly). Looking at the third equation, we see it would be very easy to solve for c. Thus, we will solve the first for a and the third for c, then plug those in to the second equation.
    2a−4b−2c = 10     ⇒     2a = 4b + 2c + 10     ⇒     a = 2b + c + 5

    3b + c = 4     ⇒     c = −3b + 4
  • However, before we plug these in, notice that our equation for a has two variables on the right: a = 2b + c + 5. We will have to eventually get rid of one of those, so let's do it now by plugging in c = −3b + 4:
    a = 2b + (−3b+4) + 5     ⇒     a = −b + 9
  • Now we have a and c both in terms of b (a = −b + 9 and c = −3b + 4), so we're ready to plug into the second starting equation:
    −3a+6b+3c = 17    ⇒     −3(−b+9)+6b+3(−3b+4) = 17
    Expand and simplify:
    3b − 27 + 6b −9b + 12 = 17     ⇒     −15 = 17     IMPOSSIBLE!
    Because we arrive at an impossible equation (−15=17), we see that it is impossible for the system of equations to be true. That is, it has no solution.
No solution exists to the system of equations.
The distance by airplane from Los Angeles to New York City is 2500 miles. Two passenger jets, each having the same airspeed (the speed of the plane relative to the air it is in) take off simultaneously from Los Angeles and New York City, each headed to the other city. However, because of the wind, they do not arrive simultaneously. This is because, depending on the direction of flight, the wind either increases or decreases the speed of the plane relative to the ground. If the wind is pushing the plane from behind, it is called a tailwind, and it adds the wind's speed to the plane's airspeed. If the wind is pushing against the plane, it is called a headwind, and it subtracts the wind's speed from the plane's airspeed. The wind is blowing from the direction of Los Angeles towards New York City. The plane leaving from Los Angeles takes 4 hours, 10 minutes to complete its trip, while the one leaving from New York City takes 5 hours. What is the airspeed of the planes and the speed of the wind? (Give your answers in mph.)
  • This problem is about speed, time, and distance. We already know the distance and the time, so now we need to figure out more about the speeds involved. Let's find a way to express the speed of each plane relative to the ground. Begin by naming the two types of speed the problem talked about: a=airspeed of each plane,  w=windspeed.
  • With these variables in place, we can now express the speed of each plane relative to the ground. The LA→NYC plane is being sped up by the wind, so its speed is a+w. The NYC→LA plane is being slowed down by the wind, so its speed is a−w. Furthermore, we know the speed of the plane multiplied by the time of its travel will give the distance. Since we need to use the unit of hours for time, convert 4 hours, 10 minutes to the quantity [25/6] hours. Thus, we have the following two equations:


    = 2500               5
    = 2500
  • We now have a system of equations. Looking at them, we see that it is probably easiest to solve by substitution. Solve for a in the second equation:
    = 2500    ⇒     a−w = 500     ⇒     a = w+500
    We can now plug this into our first equation:


    = 2500
  • Solve for w:
    2w + 500 = 600     ⇒     2w = 100     ⇒     w = 50
    Then use w=50 to find a:
    a = w+500     =     50 + 500     =     550
The airspeed of each plane is 550 mph, while the wind speed is 50 mph.
You need to make 700 mL of 3% NaOH solution for an experiment. The lab you are working in only has 1% and 10% solutions available, though. How much of each solution should you use to arrive at the desired result without wasting anything?
  • Begin by naming the variables. In this problem, we're trying to figure out how much of each solution type we will be using. Let's call them l and h for low and high percentages.
    l = volume (in mL) of 1% solution               h = volume (in mL) of 10% solution
  • From the problem, we know that we're not supposed to waste any of the solutions. Thus, since our desired result is a total of 700 mL, it must be that the two solutions put together make that much volume:
    l+ h = 700
  • The more difficult part to figure out is how they contribute to making a percentage for the newly mixed solution. It can be thought through as follows: each solution type adds a certain number of "points" per mL. For example, if we use 7 mL of the 10% solution, we would get 7·10 = 70 points. However, we need to keep in mind that this is a percentage. Thus, after we find how many total points the mixture has, we must divide it by its volume to find the resulting percentage. Thus, for our mixture we will have a total of 1 ·l+ 10 ·h points. Our goal is a percentage, so we divide that by the total volume of the final solution (700), and we will have arrived at the desired percentage (3%):
    l+ 10h

    = 3
  • We now have two equations to work as a system of equations. Let us solve by substitution. Solve for l in the first equation:
    l+ h = 700     ⇒     l = −h + 700
    Then plug in to the second equation:
    l+ 10h

    = 3     ⇒     (−h+700) + 10h

    = 3
  • Solve by algebra:
    9h + 700 = 2100     ⇒     9h = 1400     ⇒     h ≈ 155.56
    Now that we know h, plug in to find l:
    l = −h + 700     =     − 155.56 + 700     =     544.44
1% solution: 544.44 mL,    10% solution: 155.56 mL

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.


Systems of Linear Equations

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Introduction 0:04
  • Graphs as Location of 'True' 1:49
    • All Locations that Make the Function True
    • Understand the Relationship Between Solutions and the Graph
  • Systems as Graphs 4:07
    • Equations as Lines
    • Intersection Point
  • Three Possibilities for Solutions 6:17
    • Independent
    • Inconsistent
    • Dependent
  • Solving by Substitution 8:37
    • Solve for One Variable
    • Substitute into the Second Equation
    • Solve for Both Variables
    • What If a System is Inconsistent or Dependent?
    • No Solutions
    • Infinite Solutions
  • Solving by Elimination 13:56
    • Example
    • Determining the Number of Solutions
  • Why Elimination Makes Sense 17:25
  • Solving by Graphing Calculator 19:59
  • Systems with More than Two Variables 23:22
  • Example 1 25:49
  • Example 2 30:22
  • Example 3 34:11
  • Example 4 38:55
  • Example 5 46:01
  • (Non-) Example 6 53:37

Transcription: Systems of Linear Equations

Hi--welcome back to

Today, we are going to talk about systems of linear equations.0002

A linear equation in two variables is something of the form Ax + By = C, where A, B, and C are all constant numbers.0005

So, all of the capital letters that we will be seeing in this represent constant values.0015

Notice that each of the variables is just a power of 1: we have x1, effectively, y1, effectively...0019

x without an exponent is the same thing as just saying x to the power of 1.0026

So, since each of these variables has a power of 1, that is similar to a linear polynomial, which has a degree of 1.0031

And so, that is where we get this idea of calling it a linear equation--because everything has a degree of 1.0038

It is not really a degree, technically, because it is not a polynomial; but there is sort of this parallel idea going on.0044

All right, here are some examples: 3x + 2y = 4 and x - 5y = 7.0049

Both of these would be linear equations, because they are just some constant times a variable,0056

plus some other constant times a variable, until we equal, finally, some constant at the end.0061

A system of equations is a group of multiple equations that are all true at the same time.0066

So, we have this equation here and this equation here, and we know that they are both going to be true at the same time.0072

For example, the two equations above make up a system of linear equations.0077

We put these two things together with each other, and we have a system of linear equations.0082

The solution is going to be some pair of x and y, two numbers where one is going to be x, and the other one is going to be y.0089

And we plug them both in, and it will satisfy both equations at the same time.0095

It is some x and y that is going to be true here and here--in both of our equations--at the same time.0099

That is why we are fulfilling a system of equations, as opposed to just one equation.0105

Let's talk about graphs to really get into this.0111

First, long ago, when we first discussed the concept of a graph, we saw that there were two ways to look at it.0114

We can talk about how input x is transformed into output y.0120

We can think that, if we plug in x at 1, then y is going to come out at 1, as well.0125

If we plug in 2, y is going to come out as 4 (22).0132

If we plug in 3, y is going to come out as 9, because 32 is 9, and so on and so on.0136

We could also go backwards: 0, -1...all of that sort of thing.0141

But alternatively, we can also think of it as the location of all of the points that make the equation true--where is this going to be true?0144

The reason why (0,0) is a point on our graph is because, if we plug that in, 0 = 02--that checks out.0153

If we try plugging in the point (1,1), this point here, if we plug that in, we have 1 = 12; that checks out.0163

If we try plugging in the point (2,4), then we have 4 = 22, and that checks out.0172

All of the points on this parabola--the reason why these points are on the parabola,0183

the reason why this graph is these points, is because those points make the equation that makes that graph true.0188

Any other point--if we were to choose some other point on it, like (2,1)--if we plugged that into our equation, we would have 1 = 22.0196

That is not true; this is not a true statement, so this point here does not exist; it is not on our graph.0207

So, the location of all of the points that make the equation true is a great way of thinking about this.0216

When working with systems of equations, it is useful to think of graphs in this second way.0225

It will help us understand the connection between graphs and solutions.0230

This way is still useful; it is a great way, in general, whenever we have to graph something.0234

But for what we are doing right here with the systems of equations, this second way of thinking of the points that are true0238

is going to really help us understand what is going on.0245

We first want to put all of the equations from the system into a form that we can easily graph when we are going to graph this.0248

So, if we want to graph a system of equations, we first want to change these things that are kind of hard, like 3x + 2y = 4.0253

That is not very easy to graph immediately; so we want y = stuff, where that "stuff" is going to involve our x inside of it.0260

y =...with our x over there: we can graph that easily--we are used to graphing lines like that all the time.0267

So, we have 3x + 2y = 4; we can convert 3x + 2y = 4; this converts into this, -3/2x + 2.0274

You can work that out; we can get this as 2y = -3x + 4 by subtracting 3x from both sides,0286

and then we divide both sides by 2, and we get -3/2x + 4.0294

And that ends up giving us that line right there; that is what we get from this equation, so our red one is this one.0298

Similarly, we can solve x - 5y = 7 for y to make it more easy to graph; and that ends up giving us this.0307

We have both of these graphed together.0314

Now, what does this tell us about a solution to the system?0317

Where the graphs intersect, we have a solution to the system; where they intersect, right here, we have a solution.0320

Why is that a solution? Well, what we are looking for is...we are looking for a point, some (x,y), that is going to be true here and true here.0328

And since this and this are just equivalent to these two equations up here,0339

if we can find some (x,y) that is true here and true here,0344

then we have found something that is going to be the answer to our system of equations.0349

Well, any point that is on the graph of an equation is going to make that equation true.0353

So, what we have found here is that this point here is a point that is on both graphs.0360

Since it is on both graphs, it must be true for both equations.0366

So, since it is on both graphs, it is true for both equations; so we have just found the solution at that point of intersection.0369

With this idea in mind, we can see that there are three possibilities for solutions to a system of linear equations.0378

The first one is independent: we call it independent if we have just one solution--they intersect in just one place.0384

Notice that they have to have different slopes for them to intersect in just one place.0392

The next one is inconsistent: we say that a system of equations is inconsistent if they never intersect each other.0397

Notice that they have the same slope, but different intercepts.0404

They are parallel lines, but they don't intersect; they are parallel, and they never touch each other in any way.0406

They are not on top of each other; they are just parallel.0412

The system is inconsistent because, if one equation is true, the other equation can't be true.0415

There is no point that will be on both of the graphs; thus, there is no point that is going to fulfill both equations at the same time.0420

And finally, a dependent system is a system where the two lines are just completely on top of each other.0427

They are collinear--they are the same line.0433

So, in that case, they have the same slope (they are parallel), but they also have the same intercept; so they are actually just plain the same line.0435

So, in the case where it is dependent, any point on either of them is going to be a solution,0443

because if it is a point on one of them, it is a point on the other one, as well.0448

So, that gives us a total of three possibilities for how many solutions a linear system can have:0452

one solution in the case where it is independent; zero in the case where it is inconsistent0456

(they are parallel--they never touch); or infinitely many in the case where it is a dependent system,0463

where it is just the same line on top of itself, where anything that is going to be on it is going to be an answer to both equations.0470

Now, notice: I want to point out that, just because it is dependent--just because there are infinitely many solutions--doesn't mean any point is a solution.0477

If we consider that point there, it is not on either of our lines, so it is not going to be one of them.0485

So, a dependent system doesn't mean that every point is a solution.0490

What it means is that all of the points on the line are going to be solutions, because they are both just the same line.0493

So, if you find a point on one of the lines, you know it is going to work for the entire system, because it is on both of the lines.0500

But that doesn't mean that any point whatsoever is going to be true for both of your equations, for all of your entire system.0507

All right, the first method to talk about finding that point--where is that point of intersection located?"--is through substitution.0517

Substitution works by plugging in one variable for another; then we solve the resulting equation.0526

For example, consider if we have 3x + 2y = 4 and x - 5y = 7.0532

We want to begin by solving one equation for one variable in terms of the other.0537

We want to be able to plug in y for x, or x for y.0541

So, we have to get either y on its own or x on its own.0544

Now, either equation and either variable will work; we are happy plugging in x, just as we are happy plugging in y.0548

So, just choose whichever one looks easiest.0553

Now, to me, it seems really easy to get x on its own, because all we have to do is add 5y to both sides.0555

So, we have x - 5y = 7; we add 5y to both sides, and we get x = 5y + 7.0560

At this point, we are in a great position, because we have x here; we have x here.0566

We can swap out those x's, and we can get an equation that uses just y.0571

So, we substitute that into the other equation; we have x = 5y + 7 here; we have 3x + 2y = 4; so we swap out the x;0575

and now we have 3 times the quantity--because remember, we have to plug it in as a quantity,0583

because it is not just 3 times 5y; it is not just 3 times 7; it is 3 times all of 5y + 7, so we put it in parentheses--0587

3 times the quantity (5y + 7), plus 2y, equals 4.0593

We simplify that a bit: 3 times 5y is 15y, plus 2y gets us 17y; and 3 times 7 gets us 21.0597

So, we have 17y + 21 = 4; subtract 21 on both sides and divide by 17, and finally we get to the answer y = -1.0603

Great; how do we get the x?--the same thing--we just substitute our new y-value in for y here or y here; either one will work.0611

And look, we actually have this really great thing to use: we have x = 5y + 7.0620

So, we have already figured out something where, as soon as you plug in y, you will get x as soon as you just do a little bit of arithmetic.0626

So, let's plug it into this one, because we made this equation from other equations we already knew.0632

And it is going to be the easiest way to get x, because we won't really have to do any algebra; we will just have to do some basic arithmetic.0636

So, we will plug into this equation here.0641

We swap out the y that we know here for -1; we have 5(-1) + 7.0644

We simplify that: x = -5 + 7; so we get x = 2, and at this point, we have found the point.0650

We have found the pair (x,y) that is going to solve both of these equations.0656

We know that the point (2,-1) solves both of the equations above; great.0661

Of course, the example on the previous slide was an independent system.0669

What we just looked at was an independent system of linear equations, because it only had one solution.0672

In the end, we only got (2,-1); we didn't get other solutions; we didn't get 0 solutions; so it was an independent system--they had different slopes.0677

How do you tell if a system is inconsistent (no solutions) or dependent (infinitely many solutions)?0685

Well, in the case where it is inconsistent (that is, it has no solutions), if a system has no solutions,0691

you eventually get to a nonsense equation, some sort of equation that has to be clearly false--it can't be true--0697

something like 0 = 5 or -8 = 42 or √2 = 2; all of these things are completely ridiculous.0704

They are absurd: you can't have 0 equal to 5--we just know that that is not true.0712

So, what that tells us is that the only way the system can be solved is if something impossible is true.0716

But you can't have something impossible be true--it is impossible.0723

So therefore, the only other possibility is that the system has no solutions.0727

There are no solutions if we end up getting some statement that is clearly false--something ridiculous like 0 = 1 or 5 = 20.0731

At that point, we say, "Oh, it must be impossible for this thing to be true, because we are getting impossible statements out of it."0739

So, we know that we have an inconsistent system, a system that has no solutions.0745

On the other hand, we could also have something that is a dependent system, something that has infinitely many solutions.0751

In that case, if we have a dependent system, one with infinitely many solutions,0758

you are eventually going to arrive at an equation that will always be true--0762

things that are going to be like 0 = 0, -47 = -47, π = π.0766

No matter what we plug in there...there is nothing to plug in: -47 = -47...they are both constants; it is always true.0772

It is just a true statement; so since these statements are always true, just as the statement is always true, the system is always true.0779

So, we see that the system is always true, meaning that we have infinitely many solutions.0787

And remember: I want to point out what I talked about before.0791

That doesn't mean that any point whatsoever will solve the system.0794

It just means that any point on one of the equations is going to be true for the other equation, as well.0797

So, if you find a point on one of the lines, since the other line is just that same line, it is going to be true for the second equation, as well.0803

We have infinitely many solutions; we have that entire line of solutions.0810

You can also show this by showing that the two equations are equivalent.0814

And we will see an example for inconsistent; that will be Example 2 in this lesson.0818

And we will see a dependent system in Example 3.0825

So, if you are confused about that, you want to watch those specifically; check those out--Example 2 and Example 3.0829

We will explore that--we will see how that is the case.0834

All right, there is another way to solve, though: we don't have to use substitution.0836

We can also use a method called elimination, where we eliminate one of the variables from the equation.0842

Doing this, we add multiples of one equation to the other, which allows us to eliminate variables.0848

So, we are able to multiply one of the equations by a number, any constant we want.0853

And then, we can add that result to the other equation, and that will allow us to eliminate variables.0858

For example, if we have 3x + 2y = 4 and x - 5y = 7, well, we see that I have an x here; I have a 3x here;0863

well, if we could get a -3x to show up and then add that, it would cancel it out.0873

So, we multiply x - 5y = 7 by -3; we multiply both the left side and the right side--we multiply the whole equation.0877

That will come out to be -3x + 15y = -21; and we know that that is equal, because x - 5y = 7 is equal;0884

so if we multiply both sides by -3, by algebra, we know that we still have an equal thing.0891

So, -3x + 15y = -21; now, we bring these down: 3x + 2y = 4...bring it down here;0895

and we add -3x + 15y = -21 to that; we add those together.0902

3x and -3x cancel each other out; we have 2y and 15y--they come out to be 17y; 4 + -21 comes out to be -17.0908

We divide both sides by 17, and we have y = -1; great.0918

And remember: that is the same thing that we got when we were working through substitution.0922

So that is good, because both of our methods should give us the same answer.0925

At this point, since we have y, we can solve for x by substitution with the other equations.0928

We can plug this y = -1 in here or here, and we would be able to figure out what the answer is, just by good, old-fashioned substitution.0934

Alternatively, we could do it by further elimination.0941

If we know y = -1, then we see over here that there is 2y.0944

Well, we could multiply this by -2, and we get -2y = +2, because it is -1 times -2.0948

So, we have -2y = +2; we can add that over here, and we have 3x + 2y, what we started with, equals 4.0956

And we will add the thing that we just created to that, the multiple we just created: -2y = +2.0964

Add that on both; these cancel each other out; we have 3x = 6; divide by 3 on both sides; x = 2, so we end up getting (2,-1).0972

All right, both of our values are there; that is the exact same answer that we got through substitution, so it looks like elimination works.0983

Cool; just like with substitution, you can tell how many solutions this system has by the same things.0990

Independent (that means it has one solution): the system will solve in a normal fashion.0995

You will end up just getting values for the variables, and there will be only one solution.1000

If it is inconsistent (it has no solutions whatsoever), then when you are solving,1004

you will get an impossible equation--something like 5 = 8 that just can never be true.1008

And then finally, dependent (infinite solutions, where anything on one of the lines is going to be on the other line,1013

so the entire line is going to be answers): while solving, 1019

you get an always-true equation--something that is just automatically true, like 7 = 7.1022

At this point, we understand how to use elimination--we can multiply one of the equations by some constant number,1029

and then add that to the other one and eliminate variables.1035

But we might wonder why it works--what gives us the ability to add whole equations together.1038

Why are we allowed to do this--why does this work?1043

OK, imagine you had a simple equation, like something as simple as x = y; we just started out with that.1046

Now, if we wanted to, we could add 2 to each side: x + 2 = y + 2, if x = y--just basic algebra.1052

But we could also say, "2 is equal to 1 + 1, and I feel like adding 2 on one side and 1 + 1 on the other."1059

Now, 2 = 1 + 1; they are the same thing; what algebra says is that you have to do the same thing to both sides.1067

If we add the same thing on each side, we still have equality.1078

So, if we add 2 on one side and 1 + 1 on the other side, we are still adding the same thing on both sides, so there is nothing wrong with doing that.1082

We can add this equation: we can add 2 = 1 + 1, because the important thing is that 2 and 1 + 1 are the same thing.1089

Since 1 + 1 becomes 2, that is just another way of expressing 2, and that is what an equation really is.1097

We can express it this way, or we can express it this way; both the left-hand side and the right-hand side are the same thing--they are equivalent.1102

So, we can take x = y here, and we can add 2 = 1 + 1 to that, and we have x + 2 = y + 1 + 1; it makes sense.1111

Cool; if we wanted to, we could also start by multiplying 2 = 1 + 1 by some number on both sides.1123

We start, and we have some constant multiple; so we multiply by 3 on both sides, and we get 6 = 3 + 3--distribution on the right side.1129

By algebra, we know that this equation is equal, too, since we just multiplied the left side and the right side by the same number, by this 3.1137

So, we know that 6 does equal 3 + 3, because we just did algebra on it by multiplying both sides by the same number.1144

So, we can add it to x = y by the same logic: we have x = y as what we started with, and we add 6 = 3 + 3,1150

because we just figured out that that is true, as well; and we will get x + 6 = y + 3 + 3.1157

Great; the method of elimination is working by the exact same reasoning.1163

The way we add an equation to both sides is because, since it is an equation, it has equality.1167

We know that the left side and the right side are the same thing, so that means that we are adding the same thing to both sides.1174

What we are adding on the left side is the same thing as what we are adding to the right side.1182

They might look different, but we know, because of that equality, that the left side and the right side are the same thing.1186

They are equivalent, so we can add them to both sides.1192

And this allows us to have that nice process of elimination, where we can just knock out variables.1195

All right, solving by a graphing calculator is the final thing we will talk about for ways to solve systems of linear equations.1200

Earlier, we talked about how where the graphs of a system intersect--wherever we have intersection--we have solutions.1207

Now, normally that isn't very useful in practice.1213

It is a great theory for helping us understand how this works.1216

We can either cross once, cross never, or just be on top of each other (one solution/zero solutions/infinite solutions).1219

And so, that is really useful as a theory thing, but in practice, it is a pain to graph in a really accurate way.1226

If we wanted to find solutions from graphing--if we wanted to use graphing--1232

we would have to spend so much time making accurate graphs1237

that we would be better off just doing the system directly.1240

We have substitution; we have elimination; those are just direct methods that will get us the answer.1244

So, we can just use those methods if we want to figure out what the answer is, exactly.1248

If we need accuracy, it takes a long time to make a really accurate graph.1252

So, we would be better off solving it directly if we are doing it by hand.1256

However, it is possible to get around this if you have a graphing calculator.1259

If you have a graphing calculator, a graphing calculator allows you to graph both equations very quickly and very accurately.1265

Then, there is a way on the graphing calculator to find accurate intersections.1272

You can analyze the graph, and it will tell you that these two lines, these two curves, intersect at such-and-such a point.1276

That allows you to immediately find the solution; you graph one of them; you graph the other one.1283

You tell your graphing calculator to look for the intersection point, and it gives out some number, and there is your point that solves the system of equations.1287

If you want more information on this--if this sounds interesting--if you have a graphing calculator,1297

or if you are interested in buying a graphing calculator, there is an entire appendix on graphing calculators.1300

It has lots of useful information; and even if you don't own an actual graphing calculator,1304

a physical calculator, there are lots of programs that will work on computers, like the one you are probably using right now,1308

or tablets or smartphones, which you might be...whatever you are watching this on right now,1313

there is a way to be able to use graphing programs on that, as well.1318

And so, you can go and look those up; we have talked about that in the appendix on graphing calculators, as well.1322

There are lots of great things out there; you can probably be able to do this, even without having to buy a graphing calculator.1327

Now, of course, I want to caution you: you have to show your work in most math classes.1332

So, if you are taking a math class, you still have to learn these other methods of solving,1338

because it is not enough to say, "The calculator said so"--you have to be able to show--definitely prove--that this is the case.1342

We can't just say, "This piece of technology said it."1348

And also, you don't want to become reliant on your calculator, even if you are not taking any tests.1351

Even if you are not doing this and taking homework or anything else, if it is just for you, you want to understand what is going on.1355

And while the graphing calculator is a useful tool, you get a lot out of being able to do this,1362

and understand what is going on through substitution and elimination, because that will show up in other things in math--1366

whereas this graphing calculator is really useful, but it is just a trick for finding answers to this one thing.1370

So, you don't want to become reliant on your calculator, and you are probably going to have to be able to do this yourself for math class.1375

So, you have to know the other methods, as well.1380

Nonetheless, it is great for checking your answers, and when you don't need to show work.1382

So, if you don't need to show work (like if you are taking a multiple choice exam,1387

and you are allowed to use your graphing calculator), or if you just want to check your answers1391

after you have written the whole thing out by hand and shown your work,1395

it is a really great way to be able to do it quickly and accurately and move on.1398

All right, we can also expand this idea of a linear equation to more than two variables.1402

So far, we have just talked about systems that are x and y, or maybe a and b, or k and l--just two different variables.1407

But we could also have more; we could have more variables, like three variables.1415

For example, we could have A, B, C, and D--all of these capital letters--be constants;1419

and then we could have a three-variable linear equation that had A times x, B times y, C times z, and then finally all equal to some constant.1424

We could take it even further to four variables, where the capital letters are still constants:1431

A times x, B times y, C times z, D times w, equals some constant E;1435

or any arbitrary number, n, of variables, where each of our A-sub-some-number is going to be some constant,1440

and x-sub-some number is going to be each of our named variables.1447

So, it would be A1 times x1 + A2 times x2 +...1451

up until we get to An times xn = some constant number, A0.1455

So, the idea is that we can just have as many variables as we want;1459

and as long as they are just being added together and multiplied by some constant,1462

and equal some constant at the end, in this nice linear form, we still have a linear equation.1465

We can solve a system of such equations if we have at least the same number of equations as variables.1471

If you have the same number of equations as you do variables--if you have a 3-variable system,1476

and you have 3 equations--you are good to solve it.1481

If you have a 4-variable system, and you have 4 equations, you are good to solve it.1482

If you have a 57-variable system, and you have 57 equations, you are good to solve it.1488

So, as long as you have that, you can work through it.1493

We solve these systems in pretty much the exact same manner as when we did two variables.1496

We can use substitution; we can use elimination.1500

Graphing gets really difficult, because it is difficult to visualize this stuff in higher than two dimensions.1503

We can visualize three dimensions to some extent; it is hard to write it on paper, since paper is two-dimensional.1509

But we can visualize it, because we are used to living in a 3-D world.1513

But once you talk about a four-dimensional or five-dimensional system, we can't really visualize four dimensions or five dimensions.1516

How do you add an extra dimension to space?1523

So, living in three-dimensional space, we are not used to thinking in more than three dimensions, so we can't really graph in it.1525

Graphing can be hard or just simply impossible; but substitution and elimination still work great.1532

So, if you encounter a system of equations that has more than two variables, you can still use substitution; you can still use elimination.1537

And we will see that when we get to Example 5: we will see a three-variable system that we will work through.1545

All right, we are ready for some examples.1551

The first example: Solve the system of equations, if it is possible: 3x + 4y = 6, 2x + y = -1.1553

Let's try both substitution and elimination; we will work through it with substitution first.1560

If we are working through it with substitution, to me it looks like 2x + y = -1 is easy to get just the y alone.1566

So, we subtract by 2 on both sides; we have y = -2x - 1.1572

At this point, we can plug that in here; we have 3x (switch to a new color); we bring that down; 3x + 4,1576

times what we substitute in, -2x - 1 = 6; work that out: 3x - 8x - 4 = 6; 3x - 8x...we simplify that to -5x;1584

we add 4 to both sides: -5x = 10; divide by -5 on both sides; x = -2.1600

10 divided by -5 becomes -2; great.1607

We have what our x is equal to; and then, to figure out what our y is equal to, we already built this nice equation, y = -2x - 1.1611

We will switch colors again at this point.1620

y = -2x - 1; we can take this and swap it out here; y = -2 times...what was x?...-2, minus 1.1622

That is -2(-2) is positive 4, minus 1 is 3; so y = 3.1634

So, the point for this comes out to be (-2,3); great.1640

Alternatively, we could do this through the method of elimination.1647

At this point, we are looking for how we can get these things to mesh nicely so that we can knock out some things.1654

So, 2x and 3x...we are going to have to multiply both of the equations to do it.1659

But y...we can easily multiply y by -4, and we will have brought it up to the +4 here.1663

So, let's start with 3x + 4y = 6; and then we are going to take our 2x + y = -1,1669

and we want to knock out that +4, so we are going to multiply everything by -4.1679

We multiply the whole thing by -4--everything in that equation; that comes out to be -8x - 4y =...-1 times -4 is +4.1685

At this point, we bring this down; we will add that together; we have -8x - 4y = positive 4.1698

The positive 4y and the -4y cancel each other out; 3x - 8x becomes -5x; that equals 10; x = -2.1706

At this point, we can either plug x = -2 into 2x + y = -1, or 3x + 4y = 6.1716

We could use substitution, or we could just continue on our way with more elimination.1723

So, let's bring down 2x + y = -1; and notice: if we multiply x = -2 by -2 on both sides, we will have -2x =...-2 times -2 is positive 4.1727

We add this whole thing together, 2x - 2x; they cancel; we have y = positive 3, and there are our two answers.1744

And so, we get the same thing, (-2,3).1751

If you wanted to check this, you could plug it into both equations.1754

You would have to plug it into both equations, because you can't be sure that you solved both of them, unless you know it works in both of them.1757

But you can also, if you are running low on time, check just one of them; and that will probably help you figure out whether or not you got it right.1763

So, 2x + y = -1; we can plug in 2(-2) + 3 = -1; -4 + 3 = -1; and sure enough, that checks out: -4 + 3 = -1.1770

We can do the same thing with 3x + 4y; 3(-2) + 4(3) = 6; that gets us -6 + 12 = 6; sure enough, that checks out, as well.1787

So, we have performed our check; we know for sure that (-2,3); x = -2, y = 3; is the solution to this system.1800

So, checking your work is great if you have the time.1808

If you are on an exam, it is really nice to be able to check your work.1810

So, I recommend checking your work whenever you have the time, if it is an exam and it really matters that you get this right--1813

or if it is an important problem, and you are engineering something, and you can't let it fail.1819

All right, the second example: Solve the system of equations, if possible: -a + 3b = 6; -3a + 9b = -27.1823

Once again, let's see it both through substitution and through elimination.1832

Over here, we will do substitution; -a + 3b = 6--let's add a to both sides: 3b = 6 + a, a + 6; and subtract 6 on both sides, so 3b - 6 = a.1835

That is the same thing; we can plug this into -3a + 9b.1849

We have -3 times a, which is the same thing as 3b - 6, plus 9b, equals -27.1854

-9b...-3(-6) gets us + 18...+ 9b = -27; our 9b and our -9b cancel out, and we get 18 = -27.1865

That is impossible--we can never have this be true.1878

So, since 18 cannot be equal to -27, we know that there are no solutions to this.1881

Similarly, we can work through this with elimination.1890

Working through this with the process of elimination, we see -a + 3b, so we can multiply -a + 3b by -3 on both sides.1895

So, -a + 3b = 6; we can multiply that by -3 on both sides, and that will end up giving us positive 3a - 9b = -18.1903

-3a + 9b = -27; and then, we are going to add what we just figured out is a multiple of -a + 3b = 6.1921

We have that; we multiply it by -3 on both sides.1931

And we get positive 3a - 9b = -18; we add those two together (method of elimination): -3a + 3a knock each other out;1935

9b + -9b knock each other out; -27 + -18 gets us -45.1945

So, we have 0 on the left-hand side, because we have nothing left over there; 0 = -45--that is impossible.1954

We can never have that be the case, so we have no solutions--this can never be true; great.1960

Alternatively, one other way to be able to see what is going on is: we can show that these two things are very similar.1968

Notice: -a + 3b = 6, and -3a + 9b = -27.1976

Well, notice that if we just multiply this one by 3, then the left-hand side is going to be the same thing as the left-hand side to our other equation.1985

You have -3a + 9b = 18; so notice, -3a + 9b matches up to -3a + 9b.1998

So, the left-hand side is the same, but the right-hand sides are totally different.2008

In one world, in one equation, -3a + 9b = 18; in the other world, our other equation, -3a + 9b, the same thing, is equal to -27.2015

So, in one world, we have a totally different meaning than the other world.2024

Thus, the worlds can never match up; it means that our lines never touch--we have no solutions, based on this.2028

So, one other way to be able to see what is going on is the fact that we can make the left-hand sides equal,2035

but the right-hand sides won't be equal, even when the left-hand sides are equal.2040

Therefore, the left-hand sides of the equations are the same thing; the right-hand sides will be different.2044

Therefore, it must be never possible for the two things to meet.2049

Solve the system of equations, if possible: -3/7p + 2/7q = -6/7 and 9p - 6q = 18.2053

The first thing I would do is say, "Wow, 7...divided by 7...divided by 7...divided by 7; I don't like fractions."2061

So, what I am going to do is just multiply the whole thing by 7.2068

At that point, we have -3p + 2q = -6; it is easier to work from this point on, I think.2073

So now, let's do substitution first: -3p + 2q = -6, so we can work out...let's go with q.2082

2q = 3p - 6; divide both sides by 2; we have 3/2p; 6 divided by 2 becomes 3, so 3/2p - 3.2093

At this point, we have 9p - 6q = 18; we can substitute in our q there: 9p - 6 times what we swap out for q, (3/2p - 3), equals 18.2105

Simplify this: 9p - 6(3/2)...half of our 6 goes away, so we can break this into 3 times 2: 6 breaks into 3 times 2;2127

so that knocks out the 2 down here, and we have 3 times 3, or 9p, here; and -6 times -3 got a little confusing,2139

because I was distributing, but it didn't cancel out both of them...sorry about that;2147

I hope that it wasn't too confusing...-6 times -3 gets us positive 18 equals 18.2150

9p - 9p means that we have 18 = 18; and that is always true, so we have infinite solutions.2157

We have a dependent system: infinite solutions, as long as one of these equations is true, so let's go with 9p - 6q = 18.2165

So, as long as one of our equations is true, we know that the entire system is going to work,2178

because we know, from what we just figured out, that they are actually just the same line.2182

They make the same linear thing.2186

All right, we can also do this through elimination.2188

I think this way is actually easier; once we get to this point, the -3p + 2q = -6, we see that -3p + 2q...we can eliminate the p's pretty easily.2192

So, we multiply this one, -3p + 2q = 6, by 3 on both sides: 9p - 6q = 18.2201

And then, we multiply -3p + 2q = -6 by 3 on both, so we get -9p; 2 times -3 becomes -6q; -6 times -3 becomes -18.2212

We add those together, and we get that these cancel; these cancel; these cancel; 0 = 0.2224

And that always ends up working out; so we know that we have infinite solutions over here, as well.2231

It always ends up being the case; great.2238

One last way to see what is going on here: we can actually show that these two things are just the same equation.2241

If you have -3p + 2q = -6, well, if we want, we can just multiply both sides by 3, at which point we have -9p + 6q = -18.2249

At this point, we multiply both sides by -1; we have 9p - 6q (because we multiplied by -1) = +18.2263

Oh, that is what we started with here; the equations are equivalent.2275

They are just the same equation, put in different ways of phrasing it.2284

This equation here is the exact same thing as this equation here.2289

The left equation and the right equation--the two equations in our system--are just the same equation.2293

So, if the equations are equivalent, that means we have infinitely many solutions,2298

because anything that fulfills one of the equations will fulfill the other.2302

Now once again, remember: that doesn't mean that all points are true.2305

But anything on the line created by 9p - 6q = 18, which is the same thing as the line -3/7p + 2/7q = -6/7,2309

which is the same thing as the line -3p + 2q = -6--since they are all equivalent equations, they make the same line.2318

So, anything that fulfills one of the equations fulfills both of the equations.2325

We have infinitely many solutions; we have an entire line of solutions--not the whole plane, but a line of solutions.2330

All right, let's try a word problem here.2337

You need to make 100 milliliters of 22% acid solution for chemistry.2340

However, the lab only has 10% and 50% solutions available.2344

How much of each should you use to arrive at the desired acid solution, without wasting any?2349

We are going to have to mix 10% solution acid and 50% solution acid together to finally make a 22% acid solution with a quantity of 100 milliliters.2354

So, the first thing: we have to mix some quantity of the low acid and some quantity of the high acid, so let's name those things.2364

We have l, which will be the name for the amount, which will be the number of milliliters that we use of low acid, which is the 10% acid.2371

And we will use h to describe the amount of the high acid, the 50%.2383

So, l is the milliliters of 10% acid; h is the milliliters of 50% acid (for low acid and high acid).2392

All right, so if we are not going to waste any low acid or high acid, that means that the amount that we mix of low acid2401

and the amount that we mix of high acid--the amount that we are using of both of those must come up to be together exactly 100 milliliters.2407

Otherwise, we have gone over the amount that we are going to use eventually, so we are being wasteful.2413

We are using too much in creating our thing.2417

So, our first piece of knowledge is that h + l = 100.2420

The number of milliliters of our high acid, plus the number of milliliters of our low acid, must come out to a total of 100 milliliters.2426

Our other piece of information is that we want to make a 22% acid solution.2433

So, at this point, we have to start thinking out: what does it mean to be a percent acid solution?2438

Well, that means that there is some quantity of acid there; and when divided by the total volume2443

of whatever we are dealing with, that is going to give us a percentage number.2447

So, amount, divided by volume that we are dealing--the amount of the acid stuff in there,2452

divided by the volume--the number of milliliters--is going to be equal to some ratio, 0.something,2463

which we can then convert up into a percent, remember.2470

Percent is the same thing as the decimal shifted over to the side; so, 10% is the same thing as .10;2473

50% is the same thing as .50; and 22% is the same thing as .22.2480

So, for every milliliter of high acid that we put in, we will put in .5, times the milliliters of high acid, of acid stuff going in.2485

With that idea in mind, we can start creating a formula here.2497

.50, 50% acid here, times the number of milliliters, plus .10, the low acid,2501

is the total amount of acid stuff that we have put into our mixture.2513

And then, we are going to divide it by...we know, at the end, that we are going to end up with 100 milliliters, so that will be our final volume.2517

And we want that to come out to be .22.2523

So, we know that, at the end, it is going to come out being that.2527

These are our two equations.2530

Now, alternatively, I am going to tell you a trick that I think is a lot easier way to think about this.2531

Alternatively, we can think of this in terms of points.2538

Every milliliter of 10% brings 10 points to the table; every milliliter of 50% brings 50 points to the table.2541

We know that, in the end, if we have 100 milliliters of 22%, then it is going to be 100 times 22 points.2551

We can think of each of the percentages, times the amount of it, as "it is going to be that many points on the table."2559

It is a sort of abstract way of thinking about it; but I think it makes it a lot easier to make these equations.2565

We know that, if it is 10% for the low and 50% for the high, then we have 50 times h2570

(the number of high points that come in), plus 10 times the low (the number of low points that come in,2575

the points mixed together)...we know that, in the end, we want to have 22 times 100 points.2582

That is going to be our final solution, 22 times 100, which is 2200.2588

50h + 10l = 2200; so we can either think in terms of the percentage that our final solution comes out as--2594

this .50h + .10l, divided by 100, equals .22; or we can think of the number of points that are being put in to make our final number of points.2601

It is kind of abstract; it doesn't really make as much sense; but I think it is a whole lot easier to understand.2611

So, that is why I am telling it to you.2615

50h + 10l = 2200; I also want you to notice that what we have here and what we have here are actually equivalent equations.2617

If you multiply the top one, the red one, by 10000 (by 100 and then by 100 again),2627

the first 100 would cancel out the fraction on the left and bring us to 22 on the right.2636

And then, the next 100 would bring the .50 to 50, the .10 to 10, and the now-22 to 2200.2640

We see that we have the same thing; they are actually equivalent.2648

It is just this multiplication; so this point method and the percentage method are going to end up giving us the same thing.2650

But I think this point method is a lot easier to understand, in terms of the mechanics of creating it mathematically.2656

And it is an easy, fast way to create it on a test, when you have a low amount of time on your hands.2662

I recommend thinking about it; but if you don't like it, don't use it.2667

All right, at this point, we are ready to solve it.2670

Solving it actually won't end up being that hard.2672

I am in the mood for using elimination, because we have these nice h's and l's that are ready to go.2674

We can multiply them by -10 pretty easily, so we know that -10h - 10l =...multiplied by -10, so -1000.2679

We add that to 50h + 10l = 2200; so -10h - 10l = -1000 gets added together here.2691

These cancel out entirely; we are left with 40h = 2200 - 1000, or 1200.2702

40h...divide by 40 on both sides; we have 1200/40, or 120/4, which is 30.2710

So, we know that h = 30, and then from h + l = 100, if h + l = 100, then we know that 30 + l = 100, so l = 70.2718

Great; so in the end, these two pieces of knowledge here and here mean that we want to add 30 milliliters of 50% solution and 70 milliliters of 10% solution.2733

And that is how we will build our 22% acid solution with 100 milliliters in the end.2754

Great; Example 5: Solve the system of equations, if possible: 3x + y - 2z = -6, 4x + 1/2y + 2z = -5, and -3x + 2y - z - 9.2760

All right, this is a three-variable system of equations; but ultimately, we are just going to use substitution and elimination.2772

We can either stick just to substitution, stick just to elimination, or use a combination of them.2779

I am going to use a combination of them, because I feel like it.2783

But I want you to know that you can really use any method, as long as you are careful with your work--2787

you are making sure that you don't make mistakes.2791

Either method will work, or a combination of them.2793

You can switch over; sometimes the easiest thing won't be obvious,2795

and you will just have to try playing with it for a while, and then you will figure out what it is.2798

Just play with it, even if you are not quite sure what will be easiest; just get started on it, and things will work out in the end.2802

All right, at this point, I notice that we have 3x here and -3x here; so I am going to add those two equations, and I will get cancellation.2807

3x + y - 2z = -6 and -3x + 2y - z = 9: we add those two together (I'll put a line under it):2815

3x and -3x cancel out, so we have 3y - 3z = +3.2828

Let's divide everything by 3 to make it a little bit simpler: y - z =...sorry, not 3; but we divided by 3, so y - z = 1.2835

OK, now we will come back to that in a few moments.2845

And now, we also might see that there is a -2z here; there is a positive 2z here; we can add those ones together, as well.2848

3x + y - 2z = -6, and 4x + 1/2 + 2z = -5; now, if you wanted to, you could have previously just gone to substitution once you had y - z = 1.2854

You could have gone to substitution from the beginning.2870

But I decided to start with elimination, because I saw these things that could cancel out easily.2872

3x + 4x is 7x, plus 1/2 + 1 becomes 3/2y; -6 + -5 becomes equal to -11.2876

Great; so at this point, we have stuff involving x and y and stuff involving y and z.2885

So, if that is the case, we want to overlap for the y; we want to be able to figure out what y is, so let's get y as the one that is not going to be substituted out.2890

So, y - z = 1; we will change that into z = y - 1; we subtract y, and then multiply by -1 on both sides.2901

z = y - 1; and over here, we can get what x is equal to: 7x = -3/2y - 11, or x = -3...we divide that by 7,2911

so that would become 14, minus 11, over 7; great.2926

Now, that is not that friendly to substitute; but that is what we have gotten to at this point.2931

It might have been easier if we had gone with a slightly different method, but we have something; we can work it out.2935

It might be a little bit more math than we were hoping to have to do.2940

It might be a little bit more "number-crunching"; but it is not going to be that hard to work through.2943

At this point, let's figure out which one we want to plug it into.2948

Let's choose 3x + y - 2z = -6; that will be fine: 3x + y - 2z = -6.2951

Now, we know that x is equal to this stuff over here, so we have 3 times -3...2961

Oh, oops, that was one big mistake here; I forgot to put that y down.2966

I am sorry about that; so -3 over 14 times y minus 11 over 7, plus...2970

we don't have anything substituting for y, because we want y, because we are going to solve for y at the end of this...2981

minus 2 times...and over here, z is equal to y - 1, equals -6.2985

We start working this thing out: 3 times -3/14 will become -9/14y; minus 33/7, plus y, minus 2y, plus 2, equals -6.2992

Let's compact some stuff together: we will compact our y's to what we can...3006

-9/14 times y...and we have the y here and the -2y here, so that would become -y.3011

And we can't really combine 33/7 and 2 very easily, without bringing in more fractions.3020

So, let's just subtract them to the other side, because we know that eventually we are going to have to subtract on the other side.3025

So, we subtract by 2 on both sides; -6 - 2 becomes -8; we add 33/7 on both sides, so we get + 33/7.3029

OK, now, at this point, we can...I don't like dealing with the fractions here;3037

so at this point, I am going to just say, "Let's get rid of the fractions."3042

As opposed to trying to put things over common denominators, we are going to get rid of those denominators eventually.3045

So, we will just multiply the whole thing by 14, 14 times this whole thing.3049

That will get us -9y - 14y =...-8 times 14 becomes -112, and 33/7 times 14...3056

well, that will cancel the 14 to just 2 times 33; the 14/7 becomes 2, so 14 times 33/7 is the same thing as 2 times 33, so that is + 66.3068

That is a negative; and so, -9y - 14y becomes -23y; -112 + 66 becomes -46.3080

Divide by -23 on both sides, and we get y = +2.3092

Great; so there is our first thing--at this point, we have these two equations.3096

We have our z equation and our x equation, so it is just a matter of plugging into those.3100

z =...plug in our 2 for y...2 - 1; so z is equal to positive 1--that one is pretty easy.3104

This one is a little bit more difficult; we have x = -3/14 times 2 minus 11/7.3112

So, that gets us to -6/14, minus 11/7; actually, it is easier to just cancel out.3122

2/14 becomes 7, so that will cancel the denominator to just a 7; minus 11/7.3129

That gets us -6 - 11...3138

Oops, let's go back a few moments to before I made that mistake.3140

-3/14, times 2...well, we can break this into 7 times 2, so the 2's cancel out, and we have -3/7 - 11/7.3147

-3/7 - 11/7 gets us -14/7, because they combine (they have the same denominator); -14/7 simplifies to x = -2.3157

Great; at this point, we have found each of our values.3170

We have found our x-value, our y-value, and our z-value.3172

So, in the end, we know that the point that works in all three of these equations is (-2,2,1).3175

Great; now, once again, I am just going to say: you could have gone through and done this in other ways.3184

You could have used just substitution; you could have used just elimination.3191

You could have combined them in a different way than I did here.3194

There might have been an easier way; there probably was definitely going to be a harder way, as well.3197

But you end up just choosing one way, and you work through it, and eventually it will end up working out.3201

As you do more of these problems, you will figure out what works better, what is easier for you,3206

a general sense of how to get these things done fastest.3211

But really, it is just a matter of practice and just slogging through it sometimes.3213

All right, the final example: what if we wanted to solve this system of equations? u + 2v + 7w - 3x + 4y + 2z = 41,3218

and then another one, and then another one, and then another one, and then another one, and then another one?3227

Well, notice: we have 1, 2, 3, 4, 5, 6 variables total, and we have 1, 2, 3, 4, 5, 6 equations total.3230

So we know that this is possible; so we could solve this system of equations.3243

Or at least, we could figure out if it has no solutions, if it has one solutions, or if it has infinitely many solutions.3247

This is horrifying; I don't want to do this--yuck!3253

Solving this problem with the methods we know now would be possible.3256

We could work through this with elimination; we could work through this with substitution.3260

But it would take us forever to be able to work through this thing.3265

It would be awful; it would be this huge task; it would just take us so much paper.3267

We could get it done, but I don't want to; and it turns out that there are some great ways to do this.3271

We have some tricks ready; later on, when we understand vectors and matrices,3277

we will be able to see how to solve a monster like this, in the lesson using matrices, to solve systems of linear equations.3283

You will be able to use what you know about matrices, matrix multiplication, matrix inverses...things that we will all learn about in the future; don't worry.3291

You will have to learn all of this stuff, and you will be able to see that there is a really, really easy way to be able to just solve this stuff.3298

You have access to a calculator, so you can compute the inverses easily.3304

You will be able to just finish this thing really, really quickly.3308

Solving this thing will be really easy once we talk through this stuff.3311

So, we will actually come back to this many lessons in the future, when we get to using matrices to solve systems of linear equations.3314

We will take this thing, and we will be able to solve it, just like that.3321

It will be really, really easy to work through, which is pretty cool.3323

All right, I hope you have a good idea of how systems of linear equations work.3327

It is all about figuring out where the thing is intersecting in terms of that.3331

That is really the idea of when these two things are going to be true.3335

All right, we will see you at later--goodbye!3338