Chapter 2.1, Problem 29E

To determine

To solve: equations algebraically.

Answer to Problem 29E

  $\begin{array}{l}x=-\frac{6}{43}\\ y=-\frac{64}{43}\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}4x+5y=-8\mathrm{...}\text{(1)}\\ 3x-7y=10\mathrm{...}\text{(2)}\end{array}$

Calculation:

Multiply equation (1) by 3 and equation (2) by 4 and rewrite the equation as follows:

  $\begin{array}{l}\left(4x+5y=-8\right)\left(3\right)\\ 12x+15y=-24\\ \\ \left(3x-7y=10\right)\left(4\right)\\ 12x-28y=40\end{array}$

Use elimination method to solve the equations as follows:

  $\begin{array}{l}12x+15y=-24\\ 12x-28y=40\\ -\text{ + }-\\ \overline{\begin{array}{l}0+43y=-64\\ \\ 43y=-64\\ y=-\frac{64}{43}\end{array}}\end{array}$

Substitute $y=-\frac{64}{13}$ in equation (2) to solve x as follows:

  $\begin{array}{l}3x-7\left(-\frac{64}{43}\right)=10\\ 3x+\frac{448}{43}=10\\ 3x=10-\frac{448}{43}\\ 3x=\frac{430-448}{43}\\ 3x=\frac{-18}{43}\\ x=\left(\frac{-18}{43}\right)\left(\frac{1}{3}\right)\\ =-\frac{6}{43}\end{array}$

The values of x and y are $-\frac{6}{43}$ and $-\frac{64}{43}$ respectively.