For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
Systems of Linear Equations
 A linear equation in two variables is of the form
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.Ax + By = C,  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 SolutionIndependent: the equations only agree at one location;
 0 SolutionsInconsistent: it is impossible for the equations to agree;
 ∞ SolutionsDependent: 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 timeconsuming 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

 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

 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 = −1 a − 2b = −11 4a = −12  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

 In general, the easiest form to graph a linear equation in is slopeintercept form: y=mx+b (where m=slope and b=yintercept). Begin by converting each of the above equations to the form.
2x+4y = 6 ⇒ 4y = −2x + 6 ⇒ y = − 1 2x + 3 25x+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 yintercept is, then plot the next point based on one "step" forward or backward using the corresponding slope and starting from the yintercept:
 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

 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:
Simplifying things out, we get−4(5v−1) + 20v = 1 −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.

 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]:
3 2· ⎡
⎣−2x + 4y = −12 ⎤
⎦⇒ −3x + 6y = −18  Now we can add our multiple of the first equation to the second equation:
−3x + 6y = −18 3x−6y = 18 0 = 0  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 slopeintercept form (y=mx+b):
−2x+4y = −12 ⇒ 4y = 2x −12 ⇒ y = 1 2x − 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 −3.3x−6y=18 ⇒ −6y = −3x +18 ⇒ y = 1 2x −3

 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.
[Note: If you decided to do it by substitution, that method is shown after the method of elimination if you want to see it.]−3 · ⎡
⎣−7t + 2s = 24 ⎤
⎦⇒ 21 t − 6s = −72  We now add this resultant equation to the first equation:
We can now use a little algebra to solve and find t = −2.21 t − 6s = −72 13t+6s = 4 34 t = −68  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:
Now that we have found t=−2 and s=5, we are done with the problem.−7 (−2) + 2s = 24 ⇒ 14 + 2s = 24 ⇒ 2s = 10 ⇒ s = 5  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 2t + 12  Plug that in to the first equation and solve for t:
Once you know t=−2, plug that in to the equation we had for s:13t + 6 ⎛
⎝7 2t + 12 ⎞
⎠= 4 ⇒ 13t + 21 t + 72 = 4 ⇒ 34t = −68 ⇒ t = −2 s = 7 2(−2) + 12 = −7 + 12 = 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: SubstitutionSolve for each variable in terms of the others, then plug it in to another equation; EliminationAdd 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:
The second equation is already perfectly set up to cancel the z in the third equation.−2 · ⎡
⎣5x + 2y = 6 ⎤
⎦⇒ −10x − 4y = −12  Add the three equations together (the multiple of the first, the second, and the third):
Thus, we see x=0.−10x − 4y = −12 3x − z = 7 −4x + 4y + z = 5 −11x = 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

 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: SubstitutionSolve for each variable in terms of the others, then plug it in to another equation; EliminationAdd 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:
Expand and simplify:−3a+6b+3c = 17 ⇒ −3(−b+9)+6b+3(−3b+4) = 17
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.3b − 27 + 6b −9b + 12 = 17 ⇒ −15 = 17 IMPOSSIBLE!
 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:
25 6⎛
⎝a+w ⎞
⎠= 2500 5 ⎛
⎝a−w ⎞
⎠= 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:
We can now plug this into our first equation:5 ⎛
⎝a−w ⎞
⎠= 2500 ⇒ a−w = 500 ⇒ a = w+500 25 6⎛
⎝(w+500)+w ⎞
⎠= 2500  Solve for w:
Then use w=50 to find a:2w + 500 = 600 ⇒ 2w = 100 ⇒ w = 50 a = w+500 = 50 + 500 = 550
 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 700= 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:
Then plug in to the second equation:l+ h = 700 ⇒ l = −h + 700 l+ 10h 700= 3 ⇒ (−h+700) + 10h 700= 3  Solve by algebra:
Now that we know h, plug in to find l:9h + 700 = 2100 ⇒ 9h = 1400 ⇒ h ≈ 155.56 l = −h + 700 = − 155.56 + 700 = 544.44
*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.
Answer
Systems of Linear Equations
Lecture Slides are screencaptured 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
 Introduction
 Graphs as Location of 'True'
 All Locations that Make the Function True
 Understand the Relationship Between Solutions and the Graph
 Systems as Graphs
 Three Possibilities for Solutions
 Solving by Substitution
 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
 Why Elimination Makes Sense
 Solving by Graphing Calculator
 Systems with More than Two Variables
 Example 1
 Example 2
 Example 3
 Example 4
 Example 5
 (Non) Example 6
 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
Precalculus with Limits Online Course
Transcription: Systems of Linear Equations
Hiwelcome back to Educator.com.0000
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 x^{1}, effectively, y^{1}, 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 equationbecause 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 herein both of our equationsat 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 (2^{2}).0132
If we plug in 3, y is going to come out as 9, because 3^{2} 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 truewhere 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 = 0^{2}that checks out.0153
If we try plugging in the point (1,1), this point here, if we plug that in, we have 1 = 1^{2}; that checks out.0163
If we try plugging in the point (2,4), then we have 4 = 2^{2}, and that checks out.0172
All of the points on this parabolathe 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 pointif we were to choose some other point on it, like (2,1)if we plugged that into our equation, we would have 1 = 2^{2}.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 easilywe 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 solutionthey 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 collinearthey 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 parallelthey 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 dependentjust because there are infinitely many solutionsdoesn'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 pointwhere 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 quantitybecause 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 parentheses0587
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 thingwe just substitute our new yvalue 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 systemthey 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 falseit can't be true0697
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 5we 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 trueit 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 falsesomething 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 true0762
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 outExample 2 and Example 3.0829
We will explore thatwe 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 sidewe 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 15ythey 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, oldfashioned 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 equationsomething 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 alwaystrue equationsomething that is just automatically true, like 7 = 7.1022
At this point, we understand how to use eliminationwe 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 workswhat gives us the ability to add whole equations together.1038
Why are we allowed to do thiswhy 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 = yjust 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 lefthand side and the righthand side are the same thingthey 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 + 3distribution 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 intersectwherever we have intersectionwe 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 graphingif we wanted to use graphing1232
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 suchandsuch 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 thisif this sounds interestingif 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 showdefinitely provethat 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 math1366
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 ljust 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 Dall of these capital lettersbe constants;1419
and then we could have a threevariable 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 Asubsomenumber is going to be some constant,1440
and xsubsome number is going to be each of our named variables.1447
So, it would be A_{1} times x_{1} + A_{2} times x_{2} +...1451
up until we get to A_{n} times x_{n} = some constant number, A_{0}.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 variablesif you have a 3variable system,1476
and you have 3 equationsyou are good to solve it.1481
If you have a 4variable system, and you have 4 equations, you are good to solve it.1482
If you have a 57variable 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 twodimensional.1509
But we can visualize it, because we are used to living in a 3D world.1513
But once you talk about a fourdimensional or fivedimensional 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 threedimensional 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 threevariable 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 4everything 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 right1813
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 = 6let'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 impossiblewe 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 lefthand side, because we have nothing left over there; 0 = 45that is impossible.1954
We can never have that be the case, so we have no solutionsthis 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 lefthand side is going to be the same thing as the lefthand side to our other equation.1985
You have 3a + 9b = 18; so notice, 3a + 9b matches up to 3a + 9b.1998
So, the lefthand side is the same, but the righthand 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 touchwe 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 lefthand sides equal,2035
but the righthand sides won't be equal, even when the lefthand sides are equal.2040
Therefore, the lefthand sides of the equations are the same thing; the righthand 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 equationthe two equations in our systemare 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 = 6since 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 solutionsnot 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 acidthe 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 dealingthe amount of the acid stuff in there,2452
divided by the volumethe number of millilitersis 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 as2594
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 now22 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 threevariable 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 work2787
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 "numbercrunching"; 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 thingat 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 1that 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 xvalue, our yvalue, and our zvalue.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 thisyuck!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 Educator.com latergoodbye!3338
1 answer
Last reply by: Professor SelhorstJones
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.