Chapter 0.8, Problem 15E

To determine

To solve the given system of equation.

Answer to Problem 15E

  $x=-6$ and $y=4$

Explanation of Solution

Given information:

The system of equations are,

  $\begin{array}{l}x+3y=6\\ 4x-2y=-32\end{array}$

Formula used:

Substitution

Calculation:

Method used to solve the system of equation is substitution.

In equation 1, writing x in terms of y.

Shifting left hand side terms to right-hand side,

  $x=6-3y$ _____________(1)

Substituting (1) in the 2^nd equation,

  $4(6-3y)-2y=-32$

Opening the brackets,

  $24-12y-2y=-32$

Shifting left- hand side values to right- hand side,

  $-14y=-32-24$

On solving,

  $-14y=-56$

Cross-multiplying,

  $y=\frac{-56}{-14}$

On solving,

  $y=4$

Substituting in (1),

  $x=6-3(4)$

On solving,

  $x=-6$

Hence, the solution to the system of equation is,

  $x=-6$ and $y=4$