Chapter 6.3, Problem 15PPE

To determine

To solve: The system of linear equations using elimination.

Answer to Problem 15PPE

x = 3 and y = 1

Explanation of Solution

Given information:

The equations $2x+3y=9$ and $x+5y=8$

Calculation:

Multiply equation $x+5y=8$ by 2 to obtain

  $\begin{array}{l}x+5y=8\\ \Rightarrow 2\left(x+5y\right)=2\times \left(8\right)\\ \Rightarrow 2x+10y=16\end{array}$

Now subtract equation $2x+3y=9$ from $2x+10y=16$ to obtain

  $\begin{array}{l}2x+10y=16\\ \underset{\_}{2x+3y=9}\\ 0+7y=7\end{array}$

Now simplify above equation.

  $\begin{array}{l}0+7y=7\\ \Rightarrow y=1\end{array}$

Use above value of y in equation $2x+3y=9$

  $\begin{array}{c}2x+3\left(1\right)=9\\ 2x=9-3\\ 2x=6\\ x=3\end{array}$

Therefore the solution is x = 3 and y = 1.