Chapter 3, Problem 8MCQ

To determine

To calculate: The solution set for the system of equations with the help either substitution or elimination.

  $\begin{array}{c}\frac{1}{3}x-\frac{3}{8}y=28\\ \frac{1}{7}x+\frac{5}{7}y=-37\end{array}$

Answer to Problem 8MCQ

The solution set for the system of equations is $\left(21,-56\right)$ .

Explanation of Solution

Given information:

The system of equations

  $\begin{array}{c}\frac{1}{3}x-\frac{3}{8}y=28\\ \frac{1}{7}x+\frac{5}{7}y=-37\end{array}$

Calculation:

Consider the system of equations

  $\begin{array}{c}\frac{1}{3}x-\frac{3}{8}y=28\\ \frac{1}{7}x+\frac{5}{7}y=-37\end{array}$

Use the method of elimination to solve the system above.

The lowest common multiple of first equation is $24$ ; simplify it as,

  $\begin{array}{c}\frac{1}{3}x-\frac{3}{8}y=28\\ \left(\frac{1}{3}x\cdot \frac{8}{8}\right)-\left(\frac{3}{8}y\cdot \frac{3}{3}\right)=28\\ 8x-9y=28\cdot 24\\ 8x-9y=672\end{array}$

Simplify second equation as,

  $\begin{array}{c}\frac{1}{7}x+\frac{5}{7}y=-37\\ x+5y=-37\cdot 7\\ x+5y=-259\end{array}$

Now, multiply $x+5y=-259$ by $-8$ and then add into $8x-9y=672$ .

  $\begin{array}{c}\left(-8\right)\left(x+5y\right)=\left(-8\right)\left(-259\right)\\ -8x-40y=2072\end{array}$

Add both the equations,

  $\begin{array}{ccccc}8x& -& 9y& =& 672\\ -8x& -& 40y& =& 2072\\ +& & & & \\ 0& -& 49y& =& 2744\end{array}$

Therefore,

  $\begin{array}{c}-49y=2744\\ y=-56\end{array}$

Now substitute $y=-56$ in the equation $x+5y=-259$ ,

  $\begin{array}{c}x+5\left(-56\right)=-259\\ x-280=-259\\ x=280-259\\ x=21\end{array}$

Therefore, solution is the coordinate pair, $\left(21,-56\right)$ .

Thus, the solution set for the system of equations is $\left(21,-56\right)$ .