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## Algebra 1

### Course: Algebra 1>Unit 6

Lesson 2: Solving systems of equations with substitution

# Substitution method review (systems of equations)

The substitution method is a technique for solving a system of equations. This article reviews the technique with multiple examples and some practice problems for you to try on your own.

## What is the substitution method?

The substitution method is a technique for solving systems of linear equations. Let's walk through a couple of examples.

### Example 1

We're asked to solve this system of equations:
$\begin{array}{rl}3x+y& =-3\\ \\ x& =-y+3\end{array}$
The second equation is solved for $x$, so we can substitute the expression $-y+3$ in for $x$ in the first equation:
$\begin{array}{rl}3x+y& =-3\\ \\ 3\left(-y+3\right)+y& =-3\\ \\ -3y+9+y& =-3\\ \\ -2y& =-12\\ \\ y& =6\end{array}$
Plugging this value back into one of our original equations, say $x=-y+3$, we solve for the other variable:
$\begin{array}{rl}x& =-y+3\\ \\ x& =-\left(6\right)+3\\ \\ x& =-3\end{array}$
The solution to the system of equations is $x=-3$, $y=6$.
We can check our work by plugging these numbers back into the original equations. Let's try $3x+y=-3$.
$\begin{array}{rl}3x+y& =-3\\ \\ 3\left(-3\right)+6& \stackrel{?}{=}-3\\ \\ -9+6& \stackrel{?}{=}-3\\ \\ -3& =-3\end{array}$
Yes, our solution checks out.

### Example 2

We're asked to solve this system of equations:
$\begin{array}{rl}7x+10y& =36\\ \\ -2x+y& =9\end{array}$
In order to use the substitution method, we'll need to solve for either $x$ or $y$ in one of the equations. Let's solve for $y$ in the second equation:
$\begin{array}{rl}-2x+y& =9\\ \\ y& =2x+9\end{array}$
Now we can substitute the expression $2x+9$ in for $y$ in the first equation of our system:
$\begin{array}{rl}7x+10y& =36\\ \\ 7x+10\left(2x+9\right)& =36\\ \\ 7x+20x+90& =36\\ \\ 27x+90& =36\\ \\ 3x+10& =4\\ \\ 3x& =-6\\ \\ x& =-2\end{array}$
Plugging this value back into one of our original equations, say $y=2x+9$, we solve for the other variable:
$\begin{array}{rl}y& =2x+9\\ \\ y& =2\left(-2\right)+9\\ \\ y& =-4+9\\ \\ y& =5\end{array}$
The solution to the system of equations is $x=-2$, $y=5$.

## Practice

Problem 1
Solve the following system of equations.
$\begin{array}{rl}-5x+4y& =3\\ \\ x& =2y-15\end{array}$
$x=$
$y=$

Want more practice? Check out this exercise.

## Want to join the conversation?

• I am curious if there are times when either the elimination method or the substitution method would be more appropriate, and or if there would be times when only one way or the other would work. Thank you for the advice in advance!
• Yes, both equations will always work, but yes, at times it is more logical to use one over the other. for instance:
x-4y=6
x+4y=12

we see here that elimination would best fit:
x+x+4y-4y=6+12

or:
x=2y-3
7x+3y-81=23

substitute (2y-3) for x.

Also, once you have a single placeholder, put it ito quadratic form (ax^2+bx+c=0)
x=(-b±sqrt(b^2-4ac))/2a
• This is pretty challenging not gonna lie
• If you practice a lot and have a good teacher, it'll get less challenging.
• how do I solve y=2x-1 and y=3x+2 using the substitution method
• You would set them equal to each other because they both equal "y": 2x-1=3x+2
• are there any easy tips or tricks i can use to remember this?
• This is so hard I can't.
• This is the review. Start at the beginning of the lesson. Go thru each video step by step and make sure you understand before moving to the next one. Ask specific questions when there is some part of the video that you don't understand.

As you do the practice problems, if you get one wrong, use the hints to learn from your mistakes.

Then, come back and try the review again.
• Lol this is actually kind of simple, though it takes a different mindset. I like to think of the first equation as the problem I'm trying to solve and the second as a hint. See the second one says what x equals even if it doesn't necessarily give you the answer. Because it doesn't give the answer for y, we have to replace the x with the second equation to be able to solve for it. After solving for y, you would plug y into one of the two original equations and solve for x the way you did for y. After you solve for x, you would have the two answers you needed, x and y :) (I hope this was helpful, sorry if it wasn't :[ ...)
• Practice Solutions:

First system of equations:

1st equation: -5x + 4y = 3
2nd equation: x = 2y - 15

Substitute for x:
-5(2y - 15) + 4y = 3

Solve for y:
-10y + 75 + 4y = 3
-6y = -72
y = 12

Solve for x by substituting in one of the equations:
x = 2y - 15
x = 2(12) - 15
x = 9

Check your solution by substituting for x and y in the first equation:

-5x + 4y = 3
-5(9) + 4(12) = 3
-45 + 48 = 3
3 = 3 √

Second system of equations:

5x - 7y = 58

y = -x + 2

Subsitute for y:
5x - 7(-x + 2) = 58

Solve for x:
5x + 7x - 14 = 58
12x - 14 = 58
12x = 58 + 14
x = 72/12
x = 6

Solve for y by substituting in one of the equations:
y = -x + 2
y -(6) + 2
y = -4

Check your solution by substituting for x and y in the first equation:

5x - 7y = 58
5(6) - 7(-4) = 58
30 + 28 = 58
58 = 58 √

I hope this could help someone. It seems like a lot, but once you understand the steps and get the hang of it, it will become pretty quick!