Chapter 6.4, Problem 13P

Solve each system. Explain why you chose the method you used.

$2x-3y=13$

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

To determine

To find: the solution and identify which method is used to solve it.

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

Answer to Problem 13P

In order to solve two systems we used substitution method, as in second equation y has 1as coefficient. The required solution is $(x,y)=(5,-1)$ .

Explanation of Solution

Given information:

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

Calculation:

In order to solve two systems we used substitution method, as in second equation y has 1

Coefficient as:

  $y=\frac{1}{2}x-\frac{7}{2}$

Replace y in the first equation we get,

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

Replace x =5 in the second equation we get,

  $\begin{array}{l}y=\frac{1}{2}x-\frac{7}{2}\hfill \\ \Rightarrow y=\frac{1}{2}\times 5-3.5\hfill \\ \Rightarrow y=-1\hfill \end{array}$

We have one solution, $(x,y)=(5,-1)$ which are consistent and independent.