Chapter 6.3, Problem 27PPS

To determine

To solve the system of equations using elimination method

Answer to Problem 27PPS

  $x=\frac{-5}{6}$

  $y=3$

Explanation of Solution

Given:

Equations:

  $\begin{array}{l}6x-7y=-26\\ 6x+5y=10\end{array}$

Calculation:

Let the equations be numbered as

  $6x-7y=-26\to \text{(1)}$

  $6x+5y=10\to \text{(2)}$

Subtract (1) from (2), so that x is eliminated.

  $\begin{array}{l}\text{ 6}x+5y=10\\ \frac{-\text{6}x+7y=26}{12y=36}\text{ [When subtracting the sign of (1) changes]}\end{array}$

  $\begin{array}{c}12y=36\text{ [Divide 12 on both sides]}\\ y=\frac{36}{12}\\ y=3\end{array}$

Substitute the value of $y=3$ in any one of the equations.

Say, equation (1)

  $\begin{array}{c}6x-7(3)=-26\\ 6x-21=-26\\ 6x=-26+21\\ 6x=-5\\ x=\frac{-5}{6}\end{array}$

Conclusion:

Therefore, the required solution for the given system of equation is $x=\frac{-5}{6}$ and $y=3$