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Solving Systems of Two Equations - Addition Method

Reference > Mathematics > Algebra > Systems of Equations
 

In the previous section of this unit on systems of equations, we talked about the following system of equations, and found a solution using a trial-and-error method:

x + y = 12
x - y = 2

Remember that we need an x value and a y value that make both equations true. We made a list of solutions that made the first equation true, and then tried them out in the second equation, hoping that we would find a pair of numbers that made that one true also. We found that x = 7; y = 5 worked. But now let's see if we can solve this using an algebraic method. What we're going to do is called the "addition method." Before I show it to you, I need to warn you that I picked the most simple example possible to use this method on; things will get more difficult as we go along. So don't think it will always be this easy.

Example #1

In order to solve this equation, I notice that if we simply added these two equations together, since we have +y in one of them and -y in the other, the y variables would cancel us out leaving us with one equation and one unknown, which is not indeterminate!

x + y = 12
x - y = 2
__________
2x    = 14

Now we can divide both sides by 2, and we get x = 7. How do we find y? Simple! Now that we know what x is, we can just plug that number into either equation, and solve for y:

 x + y = 12
 5 + y = 12
-5      -5
_________
     y = 7 

So now we have the answer! x = 7; y = 5. And we didn't have to use trial and error to solve!

We should check our answer, though, to make sure our answer works in both equations. Since we used the first equation to solve for x, we should plug both of our variables into the other equation to solve for y: 

x - y = 2
7 - 5 = 2

This is true, so we can feel confident we didn't make any mistakes along the way.

Why did this work so nicely? It worked nicely because when we added the two equations, one of the variables canceled out. Why did the y variable cancel? It canceled because the coefficients of y were equal and opposite; one coefficient was +1 and the other was -1. Will that always happen? No! So we have to be paying attention very carefully. Take a look at the system below.

Example #2

x + y = 10
x - 2y = 4

What will happen if we add these two equations together? Let's give it a try and see...

 x +  y = 10
 x - 2y =  4
____________
2x -  y = 14

Yeah, that didn't work so well, did it? We still have an indeterminate equation!  It would have been nice if the second equation had a negative x; then we could have added them and the x variables would cancel, right? So let's fix that! Let's turn the x into negative x! How will we do that? We'll multiply both sides of the second equation by -1:

-1(x - 2y) = -1(4)
-x + 2y = -4

Now we can add them, and watch the magic happen:

 x +  y = 10
-x + 2y = -4
____________
     3y = 6

This leads to y = 2. Now we can plug that value back into whichever equation we want to find x:

x + y = 10
x + 2 = 10
x = 8.

Our solution is: x = 8; y = 2. You should take the time to check your solution; I'll leave that as an exercise for you. Notice that this time we made the x variable cancel instead y. You'll choose which variable to make cancel depending on what you think is going to be easier to do. You'll see what I mean in a few minutes.

Example #3

x + 2y = 17
2x - y = 9

Sadly, in this problem, adding the equations won't work, and multiplying one of them by -1 won't work either. Why not? Because the coefficients of x don't match, and the coefficients of y don't match, so no matter what we do, nothing will cancel.

But...can we make the coefficients match? Sure! If we multiplied the second equation by two, we'd have +2y in the first equation and -2y in the second equation!

2(2x - y) = 2(9)
4x - 2y = 18

Now we can add them and watch the magic happen:

 x + 2y = 17
4x - 2y = 18
____________
5x      = 35
 x      =  7

Now we can plug that value back into the first equation and solve for y:

x + 2y = 17
7 + 2y = 17
    2y = 10
     y =  5

Then you can check your solution.

Example #4

2x + 7y = 49
3x + 9y = 69

Okay, this one looks a little more gross. If we wanted to make the x coefficients match, we'd need to either multiply the first equation by -
3
2
or the second equation by -
2
3
. Who wants to do that? And making the y coefficients match is even worse; we'd have to multiply the first equation by -
9
7
or the second equation by -
7
9
. Yuck! Surely there must be a better way! 

There is.

Instead of dealing with fractions, we're going to find the least common multiple of the coefficients of x. The coefficients of x are 2 and 3. The least common multiple of 2 and 3 is 6. So we're going to ask ourselves, "What do I have to multiply the first equation by to have 6x as one of the terms?" And the answer is, we multiply the first equation by 3. Similarly, we ask ourselves, "What do I have to multiply the second equation by to have -6x as one of the terms?" And the answer is, we multiply the second equation by -2.

3(2x + 7y) = 3(49) 
-2(3x + 9y) = -2(69)

 6x + 21y =  147
-6x - 18y = -138
________________
      3y =     9
       y =     3

Now we can plug this back into the first equation, and solve for x: 

2x + 7y   = 49
2x + 7(3) = 49
2x + 21   = 49
2x        = 28
x         = 14

And, of course, we can check our answer. At this point, I should point out that we could have multiplied the first equation by 9 and the second equation by -7 in order to make the y variable cancel. But really...who wants to do that? That just gives us a whole bunch of really big numbers to deal with. So it's a good idea to find the least common multiple of the x coefficients and  the y coefficients before you decide which one you're going to use!

I should also point out that the equations won't always be nicely formatted with the x first, the y second, and the constant on the other side. You might need to rearrange things a bit. Always make sure you pay close attention to where your like terms are, so you don't end up combining the wrong things. We'll deal with that some more in the next section.

Questions

1.
In example #2, show how you would you check your answer.
2.
In example #3, show how you would check your answer.
3.
Solve the system of equations: x + y = 0; x - y = 2
4.
Solve the system of equations: x + 2y = 14; -x + 4y = 28
5.
Solve the system of equations: x + 2y = 8; 2x - y = 22
6.
Solve the system of equations: 2x + 3y = 12; 3x + 2y = 13
7.
Solve the system of equations: 3x + 2y = 11; 4x + 5y = 10
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Introducing Systems of EquationsIntroducing Systems of Equations
Solving Systems of Equations - Rearranging TermsSolving Systems of Equations - Rearranging Terms
 

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