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Post by edder villegas on March 2, 2014

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Post by Tiffany Huang on December 11, 2013

GOOD

Solving Systems of Equations Algebraically

  • You can solve a system of equations by substitution or by elimination.
  • Sometimes each equation must be multiplied before elimination can be used.
  • If you obtain an equation that is always true, the system has an infinite number of solutions.
  • If you obtain an equation that is never true, the system has no solution.

Solving Systems of Equations Algebraically

Solve the system of equations algebraically
4x − y = − 4
− 4x + y = 0
  • Whenever you are solving a system of equations algebraically you have two options, elimination method
  • or substitution method. For elimination method, look out for variables with equal, opposite sign coefficients.
  • With this method, one variable will drop out. For substitution method, look out for positive variables that have
  • a leading coefficient of one.
  • In this problem, the best method is elimination because the x's have equal, opposite sign coefficients.
  • Step 1: Add the system of equations so x's drop out.
  • (4x − y = − 4) + (− 4x + y = 0)=
  • 0 + 0 = − 4
  • 0 = − 4
This is never true, therefore, you have an inconsistent system. No solution.
Solve the system of equations algebraically
(a) 4x + y = 1
(b) − 3x − 3y = − 3
  • Notice that y on equation (a) is positive with a coefficient of 1. Therefore solve by substitution.
  • Step 1: solve (a) for y
  • 4x + y = 1
  • 4x+y− 4x = 1− 4x
  • y = − 4x − 1
  • Step 2:substitute y = 4x + 1 into equation (b) and solve for x.
  • − 3x − 3y = − 3
  • − 3x − 3( − 4x + 1) = − 3
  • Simplify
  • − 3x + 12x − 3 = − 3
  • 9x − 3 = − 3
  • 9x−3 + 3 = −3 + 3
  • 9x = 0
  • x = 0
  • Step 3: Substitute x = 0 into y = − 4x + 1
  • y = − 4x + 1 = − 4(0) + 1 = 1
Solution : (0,1)
Solve the system of equations algebraically
(a) 2x + 2y = 8
(b) x − 3y = 0
  • Notice that x on equation (b) is positive with a coefficient of 1. Therefore solve by substitution.
  • Step 1: solve (b) for x
  • x − 3y = 0
  • x−3y+3y=0+3y
  • x = 3y
  • Step 2:substitute x = 3y into equation (a) and solve for y.
  • 2x + 2y = 8
  • 2(3y) + 2y = 8
  • Simplify
  • 6y + 2y = 8
  • 8y = 8
  • y = 1
  • Step 3: Substitute y = 1 into x = 3y
  • x = 3y = 3(1) = 3
Solution : (3,1)
Solve the system of equations algebraically
(a) 8x − 6y = − 16
(b) − 8x + 6y = 18
  • Notice that both the x and the y have equal, opposite sign coefficients. That means that both variables will drop out.
  • Add the system
  • (8x − 6y = − 16)+(−8x+6y=18)
  • 0 + 0 = − 2
  • 0 = − 2
Since 0 is never going to equal − 2, you have an Inconsistent System of equations. No Solution.
Solve the system of equations algebraically
(a) − 9x − 3y = 15
(b) 9x + 5y = − 25
  • Notice that the variables x has equal, opposite sign coefficients. That means if we add the system of equations the x's will drop out.
  • Step 1. Add the system
  • (− 9x − 3y = 15)+(9x + 5y = − 25)
  • 0 + 2y = − 10
  • Step 2: Solve for y
  • 2y = − 10
  • y = − 5
  • Step 3: Solve for x. You may use equation (a) or equation (b). Substitute y = − 5
  • − 9x − 3y = 15
  • − 9x − 3( − 5) = 15
  • − 9x + 15 = 15
  • −9x + 15 − 15 = 15 − 15
  • − 9x = 0
  • x = 0
Solution(0, − 5)
Solve the system of equations algebraically
(a) 6x − 10y = − 16
(b) − 2x + 10y = 12
  • Notice that the variables y has equal, opposite sign coefficients. That means if we add the system of equations the y's will drop out.
  • Step 1. Add the system
  • (6x − 10y = − 16)+(−2x + 10y = 12)
  • 4x + 0 = − 4
  • Step 2: Solve for x
  • 4x = − 4
  • x = − 1
  • Step 3: Solve for y. You may use equation (a) or equation (b). Substitute x = − 1
  • − 2x + 10y = 12
  • − 2( − 1) + 10y = 12
  • 2 + 10y = 12
  • 2+10y− 2=12 − 2
  • 10y = 10
  • y = 1
Solution ( − 1,1)
Solve the system of equations algebraically
(a) 9x − 4y = 10 (b) − 9x + 2y = 4
  • Notice that the variables x has equal, opposite sign coefficients. That means if we add the system of equations
  • the x's will drop out.
  • Step 1. Add the system
  • (9x − 4y = 10)+(− 9x + 2y = 4)
  • 0 − 2y = 14
  • Step 2: Solve for y
  • − 2y = 14
  • y = − 7
  • Step 3: Solve for x. You may use equation (a) or equation (b). Substitute y = − 7
  • − 9x + 2y = 4
  • − 9x + 2( − 7) = 4
  • − 9x − 14 = 4
  • −9x−14 + 14=4 + 14
  • − 9x = 18
  • x = − 2
Solution ( − 2, − 7)
Solve the system of equations algebraically
(a) 5x − 2y = 16
(b) − 7x − 7y = 7
  • Notice that we've run into a problem: none of the variables have coefficient = 1, and neither x's or y's have equal, different sign coefficients.
  • That means we need to do extra work in order to create an equivalent system that can be solved by elimination.
  • Since the x's already have opposite signs, now it's time to look for the Least Common Multiple of 5 and 7.
  • The Least Common Multiple of 5 and 7 is 35. That means we'll multiply (a) by 7 and (b) by 5
  • Step 1. Multiply (a) by 7, (b) by 5 to create an equivalent system.
  • (a) 35x − 14y = 112
  • (b) − 35x − 35y = 35
  • Step 2: Add the System
  • (35x − 14y = 112)+(− 35x − 35y = 35)
  • 0 − 49y = 147
  • − 49y = 147
  • Step 3: Solve for y
  • − 49y = 147
  • y = − 3
  • Step 4: Solve for x. You may use equation (a) or equation (b). Substitute y = − 3
  • In this step you may use the original equations, or equations from the Equivalent System.
  • 5x − 2y = 16
  • 5x − 2( − 3) = 16
  • 5x + 6 = 16
  • 5x +6 − 6 = 16 − 6
  • 5x = 10
  • x=2
Solution (2, − 3)
Solve the system of equations algebraically
(a) − 15x − 15y = − 15 (b) 6x + 6y = 6
  • Notice that we've run into a problem: none of the variables have coefficient = 1, and neither x's or y's have equal, different sign coefficients.
  • That means we need to do extra work in order to create an equivalent system that can be solved by elimination.
  • Since the x's and y's have opposite signs, now it's time to look for the Least Common Multiple of 6 and 16.
  • The Least Common Multiple of 6 and 15 is 30. That means we'll multiply (a) by 2 and (b) by 5
  • Step 1. Multiply (a) by 2, (b) by 5 to create an equivalent system.
  • (a)− 30x − 30y = − 30
  • (b)30x + 30y = 30
  • Step 2: Add the System
  • (− 30x − 30y = − 30)+(30x + 30y = 30)
  • 0+ 0 = 0
  • 0 = 0
In this case we have a Dependent System because 0 is always going to equal 0. Therefore, you will have infinite number of solutions.
Solve the system of equations algebraically
(a) − 5x + 6y = 19
(b) − 2x − 7y = 17
  • Notice that we've run into a problem: none of the variables have coefficient = 1, and neither x's or y's have equal, different sign coefficients.
  • That means we need to do extra work in order to create an equivalent system that can be solved by elimination.
  • Since the easiest variable to find the Least Common Multiple is x, we'll Drop Out x even if that means (a) or (b) needs to be multiplied by a negative
  • in order to have the required ( + )( − ) coefficients.
  • The Least Common Multiple of 5 and 2 is 10. That means we'll multiply (a) by 2 and (b) by − 5 or (a) by − 2 and (b) 5
  • Step 1. Multiply (a) by 2, (b) by − 5 to create an equivalent system.
  • (a) − 10x + 12y = 38
  • (b) 10x + 35y = − 85
  • Step 2: Add the System
  • (− 10x + 12y = 38) + (10x + 35y = − 85)=
  • 0 + 47y = − 47
  • Step 3: Solve for y
  • 47y = − 47
  • y = − 1
  • Step 4: Solve for x. You may use equation (a) or equation (b). Substitute y = − 1.
  • In this step you may use the original equations, or equations from the Equivalent System.
  • − 2x − 7y = 17
  • − 2x − 7( − 1) = 17
  • − 2x + 7 = 17
  • − 7 − 7
  • − 2x = 10
  • x = − 5
Solution ( − 5, − 1)

*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

Solving Systems of Equations Algebraically

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
  • Solving by Substitution 0:08
    • Example: System of Equations
  • Solving by Multiplication 7:22
    • Extra Step of Multiplying
    • Example: System of Equations
  • Inconsistent and Dependent Systems 11:14
    • Variables Drop Out
    • Inconsistent System (Never True)
    • Constant Equals Constant
    • Dependent System (Always True)
  • Example 1: Solve Algebraically 13:58
  • Example 2: Solve Algebraically 15:52
  • Example 3: Solve Algebraically 17:54
  • Example 4: Solve Algebraically 21:40

Transcription: Solving Systems of Equations Algebraically

Welcome to Educator.com.0000

In today's lesson, we will be talking about solving systems of equations algebraically.0002

In the previous lesson, we talked about solving systems of equations by graphing.0008

However, that method has some limitations; therefore, we are going to talk about a few other methods of solving.0013

And this is some review from Algebra I; so again, if you need more detail--more review on these concepts--check out our Educator.com Algebra I series.0020

First, we are going to review solving by substitution, first jotting down a system of equations:0030

2x + 3y = 12, and the second equation in the system is x + 4y = 1.0036

In this method, you are going to solve one equation for one variable in terms of the other variable.0047

Then, substitute the expression for the variable in the other equation.0053

Let's look at what this means: first step: solve one equation for one variable in terms of the other one.0059

The easiest thing to do is to find a situation where you have a variable that has a coefficient of 1; and I have that right here.0065

So then, I can solve for x in terms of y pretty easily.0071

So, x + 4y = 1, so I am going to solve for x in terms of y; this gives me x = -4y + 1.0075

The second step is to substitute this expression for the variable in the other equation.0088

So, I am going to substitute -4y + 1 for x in the first equation; and that is going to create0094

an equation that has only one variable, and then I will be able to solve that.0101

It is important that you go back into the other equation to substitute in.0105

This is going to give me 2(-4y + 1) + 3y = 12; then, I can just solve for y.0111

2 times -4y; that is -8y, plus 2 times 1--that is just 2--plus 3y, equals 12.0124

Combine like terms: -8y and 3y gives me -5y, plus 2 equals 12; subtract 2 from both sides to get -5y = 10.0135

Dividing both sides by -5, I get y = -2.0150

Once I have one variable solved, I can easily solve for the other.0155

So, I just go back to either one of these (and this one is easier to work with), and I am going to substitute in...I am going to let y equal -2.0159

x + 4(-2) = 1; this is going to give me x + -8 = 1, or x - 8 = 1; adding 8 to both sides, I get x = 9.0170

Here, y equals -2 and x equals 9.0192

This method of substitution works really well when the coefficient of one of the variables is 1.0197

So, use substitution when a variable in one of the equations has a coefficient of 1.0203

So again, we looked at this system of equations, saw that this variable had a coefficient of 1,0232

then solved for this variable x in terms of the other variable y to get x = -4y + 1.0240

And then, I went ahead and substituted this 4x in the other equation; that gave me one equation with one variable, and I can solve for y.0247

Once you have y, you can substitute that value into either equation and then solve for x.0260

The second method is solving by elimination: in elimination, you add or subtract the two equations to eliminate one of the variables from the resulting equation.0267

And this system works well when you have variables, either the two x's or the two y's, that have either the same coefficient or opposite coefficients.0277

By "opposite coefficients," I mean the same number with opposite signs, such as 2 and -2 or 3 and -3.0286

For example, if you were asked to solve this system, you look and see that there is no variable with a coefficient of 1.0293

Therefore, substitution is not the ideal--it is not the easiest way to go.0306

But what you see is that you have one variable--you have y--that has the same coefficient.0312

So, adding is certainly not going to help me; if I add these together, I will get 6x + 10y = 8.0318

What I need to do is subtract, because my goal is to get one variable to drop out.0324

So, I am going to subtract the second equation from the first.0328

And just to help me keep my signs straight, I rewrite that as adding the opposite; so I am adding -4x, -5y, and -3.0333

OK, this is going to give me 2x - 4x is -2x; the y's are going to drop out: 5y - 5y gives me 0, so I can write + 0, but they just drop out.0352

And then, 5 minus 3 equals 2; this is -2x = 2; I can easily solve for x, and x equals -1.0365

Once I have one variable, I can go to either original equation, and then substitute in order to solve for the other variable.0376

I know that x equals -1, so I am going to substitute that up here.0385

Add 2 to both sides (that is going to give me 5y = 7), and divide both sides by 5 to get y = 7/5.0395

So, solving by elimination, I was able to come up with the solution that x equals -1 and y = 7/5, which will satisfy both of these equations.0403

Again, this works well; use elimination when the same variable (meaning both x's or both y's)0412

have the same or opposite coefficients (the same coefficient, but opposite signs).0426

Sometimes, you look, and you see that there is no variable with a coefficient of 1; and then you say, "OK, I will use elimination."0442

But then, you realize that none of the variables have the same or opposite coefficients.0449

In that case, you can still use elimination, but you are going to have to take an extra step before you do the elimination.0453

And the extra step is to multiply one or both equations so that one of the variables has the same or opposite coefficient in the two new equations.0459

So, your goal is to create a situation where one of the variables has the same or opposite coefficients.0469

Then, you just use elimination, as we did previously.0476

If you had a system of equations, 5x + 4y = 1, and 6x + 3y = 3, you see that neither of these0480

has a variable with a coefficient of 1, and neither set of variables has the same or opposite coefficients.0489

But if I look at these two, 4y and 3y, the least common multiple is 12; so what I want to do0497

is multiply each of these by something, in order to end up with a coefficient of 12 in front of the y.0505

Sometimes you will be lucky, and all you will have to do is multiply one of the equations by a number to get the opposite or same coefficient.0511

Other times, like this, you are going to have to multiply both.0519

So, for the first equation, I am going to multiply by 3; and this is going to give me 15x + 12y = 3.0522

The second equation I am going to multiply by 4: this is going to give me...4 times 6x is 24x + 12 y = 12.0542

OK, I now see that I have the same coefficient for y; so rewriting this over here, I need to subtract.0559

So, I am going to subtract the second equation from the first.0576

And to keep everything straight, I like to go about this by adding the opposite to make sure that I keep my signs straight.0581

So, I am going to add -24x - 12y, and then that is going to be a -12.0589

OK, -24x - 12y - 12: add these together: I get 15 - 24x--that is going to give me -9x; the y's drop out, and then I have 3 - 12; that is -9.0603

Divide both sides by -9 to get x = 1.0618

From here, I substitute back; I will choose this top equation, 5x + 4y = 1, and I know that x equals 1.0622

This is 5(1) + 4y = 1; so that is 5 + 4y = 1, or 4y = -4, so y = -1.0631

So, x equals 1; y equals -1.0645

Again, this is just the first step in elimination; and you use it in a situation like this,0648

where you want to use elimination, but you don't already have a set of variables with the same or opposite coefficients.0653

So, you figure out what the least common multiple is, and multiply one or both equations in order to achieve that.0660

From there, you just proceed as we did before, solving by elimination.0667

There are several different possibilities that can occur when you are solving systems of equations algebraically.0675

Usually, in the problems that you will see, what will happen is what just happened previously,0683

that I showed you, where you will end up with a value for x and a value for y that satisfy both equations.0688

However, there are times when that doesn't occur.0695

You can be going along, doing substitution, doing elimination, and things are going fine;0699

and then you end up with something like this: c = d.0705

Your variables drop out, and you end up with an equation that is saying that a constant is equal to a different constant--for example, 4 = 5.0710

Well, that is not true; so when you see this, this is an equation that is never true.0719

And what this tells me is that the system of equations is inconsistent.0727

So, this is a situation where there is no solution.0742

If you start seeing a situation where the constants drop out, and then you see something like 4 = 50752

or 9 = 10--something that is not true--then you know you have an inconsistent system.0758

The other possibility is that you could have a system that is dependent, or always true.0764

In that case, you are going along; you are doing elimination; you are doing substitution;0773

and then you see that you end up with variables dropping out, and a constant that equals a constant--the same constant--for example, 3 = 3.0778

Well, that is always true; so if you end up with an equation--the sum or difference of the two equations--0788

the system of equations--that is always true, this system is dependent, and it has an infinite number of solutions.0795

Recall, when we were solving systems of equations by graphing: this is analogous to the situation where you would end up with the same line.0812

So, if you have two equations (a system of equations), you graph both equations,0819

and you find out that they are the same line, well, then, there is an infinite number of solutions--all points along those lines.0823

Here, no solution would be like parallel lines, where they never intersect; there is no solution to that system.0829

OK, the first example is: Solve algebraically for the system of equations.0839

As soon as I see that I have variables (or even one variable) with a coefficient of 1, I recognize that substitution would be a really good method to use.0844

So, use substitution when a variable has a coefficient of 1.0853

So, I am going to solve for x in terms of y: x + y = 5, so x = 5 - y.0856

Now, I have solved for x in terms of y; and now I am going to substitute this value into the other equation.0867

2x + 3y = 13: I am going to substitute this expression for x, so 2 times (5 - y), plus 3y, equals 13.0875

Or, 10 - 2y + 3y = 13; this is 10 + y = 13; then subtract 10 from both sides to get y = 3.0889

Once I have this, I can plug y into this simple equation up here: x + y = 5.0906

Substitute in; let y equal 3; and I get x = 2.0916

And this is very easy to check; you can see that, if x is 2 and y is 3, this equation holds true.0924

And you could do the same--plug it back into that first equation, put these values in for x and y, and just verify that these solutions are correct.0929

OK, so again, we are solving by substitution because you had a situation where you had a variable (actually, two variables) with a coefficient of 1.0942

That is the easiest method to use.0950

Here, I am solving this system of equations algebraically; and I don't have any variables with a coefficient of 1, so I am not going to use substitution.0954

But if I look; I have variables (y) that have the opposite coefficient, -3 and 3.0963

What that means is that, if I add these equations, the y's are going to drop out; then, I can just solve for x.0970

Rewrite this right here and add: I am going to add these two.0977

2x + 4x: that gives me 6x; the y's drop out: -3y + 3y is 0; and then 0 + 0 is 9.0988

I am going to divide both sides by 6; this is going to give me 9/6, and that simplifies to 3/2.1004

I have x = 3/2; I am going to pick either one--I will go ahead and pick the top one: 2x -3y = 0.1011

And let x equal 3/2; substitute that for x.1018

The 2's cancel out; this gives me 3 - 3y = 0, or -3y = -3; if I divide both sides of the equation by -3, I will get y = 1.1028

So, the solution is x = 3/2, y = 1; again, I could always check these solutions1042

by substituting the values in here and making sure that this equation holds true.1051

And this was a perfect setup to use elimination, because I already had variables that had the opposite coefficients.1056

I simply added these equations together; the y dropped out, allowing me to solve for x, and then substitute 3/2 in for x.1063

OK, Solve algebraically: again, there are no variables with a coefficient of 1, so I am going to go to elimination.1075

But these do not have the same or opposite coefficients, and the y's do not have the same or opposite coefficients.1083

So, this time I am going to need to use that extra step; I am going to need to use multiplication1090

in order to create a situation where I have variables with the same or opposite coefficients.1095

And I see that, if I multiply the top equation by 3, I will get 6x; and if I multiply the bottom equation by 2, I will get 6x.1102

So, I am going to go ahead and do that--multiply the top equation by 3: that is going to give me 3 times 2x - 3y = -16.1112

And this comes out to 3(2x) is 6x; 3 times -3...that is -9y; and 3 times -16 is -48.1129

The second equation I am going to multiply by 2.1144

So, 2 times 3x, plus 5y, equals 14; OK, 2 times 3x gives me 6x; 2 times 5y is 10y; and 2 times 14 is 28.1147

I have the same coefficient here; so what I need to do is subtract.1165

I am going to subtract, and again, I am going to convert this so that I am adding the opposite.1171

Rewrite the top equation the same, and the bottom as adding the opposite (adding -6x - 10y - 28).1185

OK, when I do that, the 6x's drop out, and that is going to give me -19y equals...-48 and -28 comes out to -76.1196

Divide both sides by -19, and -76 divided by -19 actually equals 4, so it divided very nicely and evenly; so I ended up with y = 4.1215

Now that I have that, I am going to substitute y = 4 into either equation; I am going to choose the top one.1227

So, 2x - 3y = 16; let y equal 4; so, 2x - 3(4) = -16, or 2x - 12 = 16.1233

Add 12 to both sides to get 2x = -4, and divide both sides by 2 to get x = -2.1251

So, the solution is that x equals -2 and y equals 4.1259

So, when I looked at this, I saw that I could use elimination, if I got the x's to have the opposite coefficient1264

by multiplying the top equation by 3 to give me 6x, and the bottom equation by 2 to give me 6x here, as well.1274

So, I did that; and then, these have the same coefficient, so I subtracted and solved for y.1283

Once I had a value for y, I substituted that value into one of the equations and then solved for x to get the solution.1291

Here, I have 2x + 3y = 8, and -4x - 6y = -16.1303

Since there is no variable with a coefficient of 1, I am going to use elimination.1311

But again, I am going to have to do a little work to get the opposite coefficients.1314

And I see here that all I have to do is multiply the top by 2, and that will give me 4x; that is opposite coefficients.1319

I don't have to do anything to the bottom equation.1329

So, that is 2 times 2x, plus 3y, equals 8, and this is going to give me 4x + 6y = 16.1330

That is the opposite coefficient, so I am going to add this new equation and this equation: + -4x - 6y - 16.1343

And you may have already realized that each one of these terms is opposite.1354

And so, here is what is going to happen: 4x - 4x is 0; 6y - 6y is 0; 16 - 16 is 0; so I end up with 0 = 0.1360

And I didn't make any mistakes--I did everything correctly--but then all my variables went away.1373

And what this tells me is that I have a dependent system, and it has an infinite number of solutions.1377

If I were to graph this, I would see that I have an infinite number of solutions--that these are intersecting at every point along the line.1386

So, we call this a dependent system.1399

If I had been going along, and all of the variables dropped out, and then I got something that wasn't true,1405

like 4 = 2 or 4 = 0, then I would have a situation where it is an inconsistent system and there are just no solutions.1410

But here, this is always true; so I have an infinite number of solutions.1418

So today, we covered solving systems of equations algebraically.1425

And that concludes today's lesson for Educator.com; I will see you again.1430