Chapter 9.4, Problem 14WE

To determine

To solve: The given equations by the addition-or-subtraction method.

Answer to Problem 14WE

The solution is $\left(\frac{3}{4},\frac{1}{5}\right)$ .

Explanation of Solution

Given:

The equations are $-28g+15h=-18$ and $-28g-40h=-29$ .

Calculation:

Let $-28g+15h=-18$ be equation (1) and $-28g-40h=-29$ be equation (2).

Observe that equation (1) has a term $-28g$ and equation (2) has a term $-28g$ . So, subtraction of the two equations will be easier to obtain the solution of the set.

  $\begin{array}{l}-\cancel{28g}+15h=-18\\ \underset{\_}{\begin{array}{l}-\cancel{28g}-40h=-29\\ (+)(+)(+)\end{array}}\\ \text{ 55}h=11\\ h=\frac{1}{5}\left[\text{Dividing both sides by 5}\right]\end{array}$

Substitute $\frac{1}{5}$ for h in equation (1) and solve for g .

  $\begin{array}{c}-28g+15\left(\frac{1}{5}\right)=-18\\ -28g+3=-18\\ -28g=-21\left[\text{Subtracting 3 on both sides}\right]\\ g=\frac{3}{4}\left[\text{Dividing both sides by 28}\right]\end{array}$

So, the solution of the given equations is $\left(\frac{3}{4},\frac{1}{5}\right)$ .