Chapter ISG, Problem 6.3.6P

To determine

To solve the system of equations

Answer to Problem 6.3.6P

  $q=12$ and $r=-2$ are the solutions

Explanation of Solution

Given information:

  $\begin{array}{l}\frac{1}{2}q+4r=-2\\ \frac{1}{6}q-4r=10\end{array}$

Calculation:

Let’s assume equations to be (1) and (2) respectively.

To solve equations by elimination, the given equations are either added or subtracted to get an equation in only one variable.

Let’s eliminate variable x here. To eliminate a variable, the coefficients of the variable must be same in both the equations.

The coefficients of r is same. Now, add the equations

  $\frac{1}{2}q+4r+\frac{1}{6}q-4r=-2+10$

Solving,

  $\frac{1}{2}q+\frac{1}{6}q=-2+10$

Taking LCM,

  $\frac{3q+q}{6}=8$

Cross multiplying the sides,

  $q=\frac{8\times 6}{4}$

So,

  $q=12$

Put value of q in (1),

  $\frac{1}{2}(12)+4r=-2$

Solving the equation,

  $\begin{array}{l}6+4r=-2\\ 4r=-8\end{array}$

So,

  $r=-2$

Hence, $q=12$ and $r=-2$ are the solutions