Chapter 6.2, Problem 6P

Solve each system by substitution. Check your solution.

$2x-3y=13$

$y=\frac{1}{2}x-\frac{7}{2}$

To determine

To solve: the system by substitution and Check solution.

  $\begin{array}{l}2x-3y=13\\ y=\frac{1}{2}x-\frac{7}{2}\end{array}$

Answer to Problem 6P

Therefore the solution is

  $\left(x,y\right)=(5,-1)$

Explanation of Solution

Given information:

  $\begin{array}{l}2x-3y=13\\ y=\frac{1}{2}x-\frac{7}{2}\end{array}$

Calculation:

In order to solve two system by substitution, we solve one of the equations for one of the

Variables, and plug this into other equation.

  $\begin{array}{l}2x-3y=13\\ y=\frac{1}{2}x-\frac{7}{2}\end{array}$

As it is given that

  $y=\frac{1}{2}x-\frac{7}{2}$

Now we substitute $y=\frac{1}{2}x-\frac{7}{2}$ into $2x-3y=13$ , we get,

  $\begin{array}{l}2x-3y=13\\ \Rightarrow 2x-3\left(\frac{1}{2}x-\frac{7}{2}\right)=13\\ \Rightarrow 2x-\frac{3x}{2}+\frac{21}{2}=13\\ \Rightarrow \frac{4x-3x}{2}=13-\frac{21}{2}\\ \Rightarrow \frac{x}{2}=\frac{26-21}{2}\\ \Rightarrow x=5\end{array}$

By plugging $x=5$ in equation $y=\frac{1}{2}x-\frac{7}{2}$ we get

  $\begin{array}{l}y=\frac{1}{2}x-\frac{7}{2}\\ \Rightarrow y=\frac{1}{2}\times 5-\frac{7}{2}\\ \Rightarrow y=\frac{5-7}{2}\\ \Rightarrow y=-1\end{array}$

Therefore the solution is

  $\left(x,y\right)=(5,-1)$

In order to check it, we replace $\left(x,y\right)=(5,-1)$ in any one system, we get

  $\begin{array}{l}2x-3y=13\\ \Rightarrow 2\times 5-3\times (-1)=13\\ \Rightarrow 10+3=13\\ \Rightarrow 13=13\end{array}$