Chapter 6.2, Problem 30P

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

$5y=7x+22$

$x=-6y+17$

To determine

To solve: The system by substitution and suggest whether the system has one solution, infinitely many solutions, or no solution.

  $\begin{array}{l}5y=7x+22\\ x=-6y+17\end{array}$

Answer to Problem 30P

Therefore the solution is

  $\left(x,y\right)=(-1,3)$

The system has one solution.

Explanation of Solution

Given information:

  $\begin{array}{l}5y=7x+22\\ x=-6y+17\end{array}$

Calculation:

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

the variables, and plug this into other equation.

  $\begin{array}{l}5y=7x+22\\ x=-6y+17\end{array}$

First we solve equation $5y=7x+22$ as follows:

  $\begin{array}{l}5y=7x+22\\ \Rightarrow y=\frac{7}{5}x+\frac{22}{5}\end{array}$

Now we substitute $y=\frac{22}{5}+\frac{7}{5}x$ into $x=-6y+17$ , we get,

  $\begin{array}{l}x=-6y+17\\ \Rightarrow x=-6\left(\frac{22}{5}+\frac{7}{5}x\right)+17\\ \Rightarrow x+\frac{42}{5}x=\frac{-132}{5}+17\\ \Rightarrow \left(\frac{5+42}{5}\right)x=\left(\frac{-132+85}{5}\right)\\ \Rightarrow x=-1\end{array}$

By plugging $x=-1$ in equation $y=\frac{22}{5}+\frac{7}{5}x$ we get

  $\begin{array}{l}y=\frac{22}{5}+\frac{7}{5}x\\ \Rightarrow y=\frac{22}{5}+\frac{7}{5}(-1)\\ \Rightarrow y=3\end{array}$

Therefore the solution is

  $\left(x,y\right)=(-1,3)$