Chapter 3.4, Problem 4E

To determine

To find: Solve system of equations $\frac{1}{2}x+\frac{1}{3}y=7$ and $\frac{1}{5}x-\frac{2}{3}y=-2$ to find values of x and y .

Answer to Problem 4E

  $x=10;y=6$

Explanation of Solution

Given information: There are two linear equations to solve $\frac{1}{2}x+\frac{1}{3}y=7$ and $\frac{1}{5}x-\frac{2}{3}y=-2$ .

Calculation:

Solve the system of equations using elimination method.

Decide which variable going to eliminate first. Let’s eliminate the y terms first.

The first equation has $\frac{1}{3}y$ and second equation has $\frac{-2}{3}y$

Make the coefficients of y terms as same. So, multiply the first equation by 2.

  $2(\frac{1}{2}x+\frac{1}{3}y=7)\to 1x+\frac{2}{3}y=14$

Add this equation to the second equation to get rid the y terms

  $\begin{array}{l}1x+\frac{2}{3}y=14\\ \frac{1}{5}x-\frac{2}{3}y=-2\\ ----------\\ \frac{6}{5}x=12\end{array}$

Multiply both sides by $\frac{5}{6}$ to get rid the fraction with x .

  $\frac{5}{6}*\frac{6}{5}x=12*\frac{5}{6}$

Simplify it,

  $x=10$

Plug in x as 10 into any of the given equations to find the corresponding y value.

  $\frac{1}{2}(10)+\frac{1}{3}y=7$

Simplify it,

  $5+\frac{1}{3}y=7$

Subtract both sides 5.

  $\frac{1}{3}y=2$

Multiply both sides by 3.

  $y=6$

Hence we get $x=10;y=6$