Chapter 6.2, Problem 22P

Solve each system by substitution. Tell whether the system has one solution, infinitely many solutions, or no solution.

$7x+2y=-13$

$-3x-8y=-23$

To determine

To solve: the system by substitution find if the system has one solution, finitely many solutions, or no solution. .

  $\begin{array}{l}7x+2y=-13\\ -3x-8y=-23\end{array}$

Answer to Problem 22P

The solution is

  $(x=-3,y=4)$

The system has one solution.

Explanation of Solution

Given information:

  $\begin{array}{l}7x+2y=-13\\ -3x-8y=-23\end{array}$

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.

  $\begin{array}{l}7x+2y=-13\\ -3x-8y=-23\end{array}$

First we solve equation $-3x-8y=-23$ as follows:

  $\begin{array}{l}-3x-8y=-23\hfill \\ \Rightarrow 3x=23-8y\hfill \\ \Rightarrow x=\frac{23}{3}-\frac{8}{3}y\hfill \end{array}$

Now we substitute $x=\frac{23}{3}-\frac{8}{3}y$ into $7x+2y=-13$ , we get,

By pluggingin equation $x=\frac{23}{3}-\frac{8}{3}y$ we get

  $\begin{array}{l}7x+2y=-13\\ \Rightarrow 7\left(\frac{23}{3}-\frac{8}{3}y\right)+2y=-13\\ \Rightarrow \frac{161}{3}-\frac{56}{3}y+2y=-13\\ \Rightarrow \frac{-56+6}{3}y=\frac{-161-39}{3}\\ \Rightarrow \frac{-50}{3}y=\frac{-200}{3}\\ \Rightarrow y=4\end{array}$

  $\begin{array}{l}x=\frac{23}{3}-\frac{8}{3}y\hfill \\ \Rightarrow x=\frac{23}{3}-\frac{8}{3}\times 4\hfill \\ \Rightarrow x=-3\hfill \end{array}$

Therefore the solution is

  $(x=-3,y=4)$

The system has one solution.