Chapter 6.3, Problem 19PPE

To determine

To solve: The system of linear equations using elimination.

Answer to Problem 19PPE

x = 2 and y = −1

Explanation of Solution

Given information:

The equations $6x-3y=15$ and $7x+4y=10$

Calculation:

Multiply equation $6x-3y=15$ by 4 to obtain

  $\begin{array}{l}6x-3y=15\\ \Rightarrow 4\left(6x-3y\right)=4\times 15\\ \Rightarrow 24x-12y=60\mathrm{...}\left(1\right)\end{array}$

Multiply equation $7x+4y=10$ by 3 to obtain

  $\begin{array}{l}7x+4y=10\\ \Rightarrow 3\left(7x+4y\right)=3\times 10\\ \Rightarrow 21x+12y=30\mathrm{...}\left(2\right)\end{array}$

Now add equation (1) and (2)to obtain

  $\begin{array}{l}24x-12y=60\\ \underset{\_}{21x+12y=30}\\ 45x+0=90\end{array}$

Now simplify above equation.

  $\begin{array}{l}45x+0=90\\ \Rightarrow x=2\end{array}$

Use above value ofx in equation $7x+4y=10$

  $\begin{array}{c}7\left(2\right)+4y=10\\ 4y=10-14\\ 4y=-4\\ y=-1\end{array}$

Therefore the solution is x = 2 and y = −1.