What is the Substitution Method?

Hi, and welcome to this video on solving systems of equations using the Substitution Method.

Solving a system of equations is considering two or more equations and finding the point, or points, at which they intersect. If you are considering linear equations, then there will only be one point of intersection, unless the lines are parallel, in which case there would be zero points of intersection. If you are considering equations of a higher degree, there may be more points of intersection.

There are four methods of solving systems of equations: graphing, substitution, elimination, and matrix. Today we will focus on the substitution method, specifically in regards to two linear equations.

When using the substitution method, we are going to substitute one of our variables with an expression using the other variable. What do I mean exactly? Well, you want to solve one of your equations so that either x or y is by itself on one side, and then plug that expression into your other equation and solve from there. Let’s look at a few examples to see what I’m talking about.

Solve the following system of equations.

\(x=2y+7\)
\(2x+3y=14\)
 
Our first step is to make one of our equations have either an x or a y by itself on one side. For this example, we have already been given an equation like that, \(x=2y+7\), so we won’t have to do any extra manipulating.

Now we want to think for a minute about what the equals sign means. The equals sign is telling us that the values on either side of the equation equals the other side; they’re the same! So, in our example, x is the same as \(2y+7\). Knowing that, we can then substitute \(2y+7\) in for x in our second equation, like this:

\(2(2y+7)+3y=14\)
 
Now we are left with only one variable in our equation, so we can follow the normal steps of solving an equation to find out what our y-value is.

\(4y+14+3y=14\)
 
So \(4y+3y=7y.\)
 
\(7y+14=14\)
 
If we subtract 14 from both sides, we get:

\(7y=0\)
 
And when we divide by 7 on both sides, we’re left with:

\(y=0\)
 
Remember, solving a system of equations means finding the point where these two equations intersect. That means that we only have half of our answer. Our y-coordinate for this point is 0, but we still need to find our x-coordinate. We do this by substituting our newly found y-value in for y in either of our original equations and solving for x. I’m going to use our first equation because it will take less manipulating to get our x.

\(x=2(0)+7=0+7=7\)
 
So our x-coordinate for this point is 7. That means our final answer, the point of intersection for these two lines is \((7, 0)\).

Let’s look at another example so we can get some more practice using this method.

\(9x-3y=12\)
\(-6x+8y=4\)
 
Neither one of our equations is set up where x or y is on one side by itself, so we are going to have to manipulate one of them to make it that way. I am going to solve our first equation for y.

I’m gonna subtract 9x from both sides. Gives me:

\(-3y=12-9x\)
 
And then divide by -3 on both sides. Which gives us:

\(y=3x-4\)
 
Now we can substitute this expression in for y in our other equation.

\(-6x+8(3x-4)=4\)
\(-6x+24x-32=4\)
 
And when we add these two together, we get:

\(18x-32=4\)
 
Add 32 to both sides. You get:

\(18x=36\)
 
Divide by 18 on both sides. And we get:

\(x=2\)
 
Now that we have our x-value, we can plug this number into the equation we found for y earlier to get our y-value. So:

\(y=3(2)-4=6-4=2\)
 
So our final point is \((2, 2)\).

Before I ask you to try one more on your own, let’s do one more example together.

\(8x-3y=14\)
\(2x+y=7\)
 
First, solve one of the equations so there is an independent variable on one side. I’m going to solve the second equation for y.

\(y=7-2x\)
 
Now, we plug in this expression for y in our first equation and solve for x.

\(8x-3(7-2x)=14\)
\(8x-21+6x=14\)
\(14x-21=14\)
\(14x=35\)
\(x=\frac{35}{14}=\frac{5}{2}\)
 
Finally, we plug in our x-value to our equation for y, and solve for y.

\(y=7-2(\frac{5}{2})=7-5=2\)
 
So our final point is \((\frac{5}{2}, 2)\).

Now that we’ve gone through three examples together, I want you to try one on your own. After I give you the equations, pause the video and try to solve it on your own. Then, once you have your answer, press play and see if it matches up with mine.

\(x+5y=7\)
\(7x-2y=12\)
 
Ready to check your answer?

First, we want to solve one of our equations for an independent variable. I’m going to solve the first equation for x.

\(x=7-5y\)
 
Now, substitute this expression into the second equation.

\(7(7-5y)-2y=12\)
\(49-35y-2y=12\)
\(49-37y=12\)
\(-37y=-37\)
\(y=1\)
 
Then, plug the y-value into the equation for x and solve for the x-coordinate.

\(x=7-5(1)=7-5=2\)
 
So our final answer for the system of equations is \((2, 1)\).

We only looked at examples with two linear equations, but this method can be used for more than two equations and for equations with higher degrees. Simply follow the same steps covered in this video. Solve equations for independent variables and then substitute them into the other equations until you get a singular variable to solve for. Once you have that variable known, work backward to find all the other variables.
I hope this review of the substitution method was helpful. Thanks for watching and happy studying!

Substitution Method Practice Questions