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

### Example 2

We're asked to solve this system of equations:
\begin{aligned} 7x+10y &= 36\\\\ -2x+y&=9 \end{aligned}
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{aligned} -2x+y&=9 \\\\ y&=2x+9 \end{aligned}
Now we can substitute the expression 2, x, plus, 9 in for y in the first equation of our system:
\begin{aligned} 7x+10\blueD{y} &= 36\\\\ 7x+10\blueD{(2x+9)}&=36\\\\ 7x+20x+90&=36\\\\ 27x+90&=36\\\\ 3x+10&=4\\\\ 3x&=-6\\\\ x&=-2 \end{aligned}
Plugging this value back into one of our original equations, say y, equals, 2, x, plus, 9, we solve for the other variable:
\begin{aligned} y&=2\blueD{x}+9\\\\ y&=2\blueD{(-2)}+9\\\\ y&=-4+9 \\\\ y&=5 \end{aligned}
The solution to the system of equations is x, equals, minus, 2, y, equals, 5.

## Practice

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

Want more practice? Check out this exercise.

## Want to join the conversation?

• 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?
• In example 2, when it says we have to solve for x or y, how do I get -2x+y=9 into slope intercept form? And also, how did we get rid of the negative once it was in slope intercept form?
Thanks
• You cannot get rid of negatives if they are there, so let the signs work themselves out. To isolate the y, all you need to do is add 2x on both sides to get y=2x+9.
• 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 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.
• how about a problem that has a fraction?
• The formatting of the problem never changes. If the fraction is a coefficient, you can multiply it by its own reciprocal (negative flip of the fraction) or you can add or subtract based on whether the fraction is positive or negative. All in all, the way you solve the system stays the same, whether you have fractions, integers or variables.

Hope this helps:)
• If i may ask,could anyone help me in this equation:

3y+2x=7
y-3x=6

• I'll get you started.
Take the 2nd equation and add 3x to both sides to isolate "y".
y = 3x+3
Now, substitute this value of "y" into the first equation.
3(3x+6) + 2x = 7
You can now solve for "x".
Once you have the value of "x", use it to calculate the value of "y".

Hope this helps. Comment back if you have questions.
• Any tips for turning a wordy problem solving question into simultaneous equations? I can find the answers with working on google and it seems so easy, but I can never get started without being given the equations - I can't find a pattern/rule for determining the equations.
• There isn’t a concrete way to do this. Practicing is the best way to get better at this. I would suggest learning what certain phrases translate to in math. For exmaple, “26 is 3 less than 2k” would translate to 26 = 2k - 3.
• how would i do y=2(x+3)^2 -5 y=14x+17
(1 vote)
• To solve using the substitution method, you find what y is, and plug it in to the other equation.
To do this one: y=14x+17. That means you just plug 14x+17 into the other equation.

y=2(x+3)^2-5 ---> Now substitute