Chapter EP, Problem 3.2.6EP

To determine

Solve equation using elimination method.

Answer to Problem 3.2.6EP

  $\boxed{\begin{array}{l}n=-2\\ q=4\end{array}}$

Explanation of Solution

Given information:

The given equations are:

  $\begin{array}{l}6n+8q=20\\ 5n-4q=-26\end{array}$

  $6n+8q=20$ ….1

  $5n-4q=-26$ ….2

In the elimination method you either add or subtract the equations to get an equation in one variable. When the coefficients of one variable are opposites you add the equations to eliminate a variable and when the coefficients of one variable are equal you subtract the equations to eliminate a variable.

Now multiplying equation 2 by 2 to make the coefficients of $q$ in both the equation same:

  $2\left(5n\right)-2\left(4q\right)=2\left(-26\right)$

  $10n-8q=-52$ …...3

Now as we can see coefficient of $q$ in equation 1 and 3 are same with opposite sign, therefore adding equation 1 from equation 3 to find the value of $n$ .

  $\begin{array}{l}6n+8q=20\\ 10n-8q=-52\\ \overline{16n=-32}\\ n=\frac{-32}{16}\\ \boxed{n=-2}\end{array}$

Now putting value of $n$ in equation 1 to find value of $q$ :

  $\begin{array}{l}6\left(-2\right)+8q=20\\ -12+8q=20\\ 8q=20+12\\ 8q=32\\ q=\frac{32}{8}\\ \boxed{q=4}\end{array}$