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Section 6.2 Solving a System of Equations: Substitution

In this section, we will learn one technique for finding a solution to a system of equations.

When we look at a system of equations, remember that we are looking for one \(x\) and one \(y\) that work for both equations. That means that the \(y\) in the first equation is going to be the same \(y\) that's in the second equation. Same thing for the \(x\)'s. So, we can substitute from one equation to the next.

Example 6.8.

Let's say we want to solve the following system of equations using substitution:

\begin{equation*} \begin{cases}y \amp = 2x-1\\ 3 \amp= y+x\end{cases} \end{equation*}

From the first equation, we can see that \(y\) is equal to (so it's the same as) \(2x-1\text{.}\) Since that's the same as the \(y\) that's in the second equation, we can plug in \(2x-1\) in for that \(y\) and then solve the equation we have left:

\begin{align*} 3 \amp= y+x\\ 3 \amp = (2x-1) + x\\ 3 \amp=3x-1\\ 4 \amp= 3x\\ \frac{4}{3} \amp= x \end{align*}

Now that we figured out what \(x\) is, don't forget that a solution is a point, so we still need to figure out what \(y\) is. To do that, we plug our \(x=\frac{4}{3}\) into either of the starting equations and solve for \(y\text{:}\)

\begin{equation*} y = 2x-1 = 2\left(\frac{4}{3}\right)-1 = \frac{8}{3} - 1 = \frac{8}{3} - \frac{3}{3} = \frac{5}{3} \end{equation*}

Now, in the example above, we chose the first equation to plug \(x\) into in order to find \(y\text{.}\) However, you should be able to plug into either equation and you'll get the same value. In fact, that's a great way to check your work: plug the \(x\) you got into both equations and make sure you get the same \(y\text{.}\) If you don't, then you've probably made an algebra mistake somewhere, and you should go back and check your work.

Another way to check your work is to use what you know about solutions to a system: plug in the point to both equations and make sure it works for both.

Checkpoint 6.9.

Solve the following system of equations using substitution:

\begin{equation*} \begin{cases}1 \amp= x-y \\ 5 \amp= 2x+y\end{cases} \end{equation*}
Answer.

\((2,1)\)

Solution.

We need to start by solving one of the equations for one of the variables. It doesn't matter which equation we pick, and it doesn't matter which variable we pick. Since we solved for \(y\) in the example above, let's solve for \(x\) in this case. The first equation does not have a number multiplied by \(x\text{,}\) so let's use that one:

\begin{align*} 1 \amp = x-y \\ 1+y \amp= x \end{align*}

So, we are going to plug \(1+y\) in for \(x\) in the second equation:

\begin{align*} 5 \amp = 2x+y \\ 5 \amp= 2(1+y)+y \end{align*}

And now we can solve for \(y\text{:}\)

\begin{align*} 5 \amp= 2(1+y)+y\\ 5 \amp= 2+2y+y\\ 5 \amp= 2+3y\\ 3 \amp= 3y\\ 1 \amp= y \end{align*}

Now that we've figured out that \(y=1\text{,}\) we need to plug back into an equation to find \(x\text{.}\) We could use either of the starting equations, but in this case, it will be easiest to use the equation we already solved for \(x\text{:}\)

\begin{equation*} x = 1+y = 1+1 = 2 \end{equation*}

Therefore, our answer is \((2,1)\text{.}\)

Let's check our work using the strategies in the paragraph above. First, let's try plugging \(y=1\) into the other starting equation, and make sure we get the same value for \(x\text{:}\)

\begin{align*} 5 \amp= 2x+y\\ 5 \amp= 2x+1\\ 4 \amp= 2x\\ 2 \amp= x \end{align*}

That worked! So, we can be pretty confident that we have the right answer. If we want to be really sure, let's plug our point into both starting equations and make sure they both work:

\begin{align*} 5 \amp= 2x+y\\ 5 \amp= 2(2)+1\\ 5 \amp 5 \end{align*}
\begin{align*} 1 \amp=x-y\\ 1 \amp=2-1\\ 1 \amp= 1 \end{align*}

It works for both equations! Therefore, we can be very, very confident that we did indeed compute the right answer.