Chapter 13.6, Problem 27WE

To determine

To solve: the equation algebraically then draw the graph and also label the intersection point.

Answer to Problem 27WE

The value of $x=4$ and $y=3$ and the intersection point is $\left(4,3\right)$ .

Explanation of Solution

Given information: The equation is:

  $\begin{array}{l}x+2y=10\\ 3x-2y=6\end{array}$

Calculation:

  $\begin{array}{l}x+2y=10\mathrm{......}\left(1\right)\\ 3x-2y=6\mathrm{......}\left(2\right)\\ x=10-2y\mathrm{......}\left(3\right)\end{array}$

Put value of $x$ in equation $\left(2\right)$

  $\begin{array}{l}3\left(10-2y\right)-2y=6\\ 30-6y-2y=6\\ -8y=6-30\\ -8y=-24\\ y=\frac{-24}{-8}\\ y=3\end{array}$

Put value of $y$ in equation $\left(3\right)$

  $\begin{array}{l}x=10-2y\\ x=10-2\left(3\right)\\ x=10-6\\ x=4\end{array}$

The graph will be:

  

The intersection point is $\left(4,3\right)$