Chapter 2.1, Problem 28E

To determine

To solve: equations algebraically.

Answer to Problem 28E

  $\begin{array}{l}x=5\\ y=12\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}\frac{3}{5}x-\frac{1}{6}y=1\mathrm{...}\text{(1)}\\ \frac{1}{5}x+\frac{5}{6}y=11\mathrm{...}\text{(2)}\end{array}$

Calculation:

Multiply equation (1) by 5 and rewrite the equation as follows:

  $\begin{array}{l}\left(\frac{3}{5}x-\frac{1}{6}y=1\right)\left(5\right)\\ \frac{3}{5}\left(5\right)x-\frac{1}{6}\left(5\right)y=1\left(5\right)\\ 3x-\frac{5}{6}y=5\mathrm{...}\left(3\right)\end{array}$

Use elimination method to solve the equations (3) and (2) as follows:

  $\begin{array}{l}3x-\frac{5}{6}y=5\\ \frac{1}{5}x+\frac{5}{6}y=11\\ \overline{\begin{array}{l}\frac{16}{5}x+0=16\\ \\ \frac{16}{5}x=16\\ x=16\left(\frac{5}{16}\right)\\ =5\end{array}}\end{array}$

Substitute $x=5$ in equation (2) to solve y as follows:

  $\begin{array}{l}\frac{1}{5}\left(5\right)+\frac{5}{6}y=11\\ 1+\frac{5}{6}y=11\\ \frac{5}{6}y=11-1\\ \frac{5}{6}y=10\\ y=10\left(\frac{6}{5}\right)\\ =12\end{array}$

The values of x and y are $5$ and $12$ respectively.