Question

Solve the following pair of linear equations by the elimination method and find the value of $p+q$ rounded off to the nearest integer.

$4p=3q+5;p-q=\frac{7}{6}$

Answer 1

$4p=3q+5\dots \left(i\right)$
$p-q=\frac{7}{6}\dots \left(ii\right)$

Rewriting $\left(i\right)$ we get,
$4p-3q=5\dots \left(iii\right)$

Multiplying $\left(ii\right)$ by $4$ we get,
$4p-4q=\frac{14}{3}\dots \left(iv\right)$

Subtracting $\left(iv\right)$ from $\left(iii\right)$ we get,
$\begin{array}{c}\underset{–}{\begin{array}{ccccc}4p& -& 3q& =& 5\\ 4p& -& 4q& =& \frac{14}{3}\\ -& +& & & -\end{array}}\\ \begin{array}{ccccc}& & q& =& \frac{1}{3}\end{array}\end{array}$

Re substituting $q=\frac{1}{3}$ in $\left(i\right)$ we get,
$\begin{array}{cc}4p& =3\left(\frac{1}{3}\right)+5\\ 4p& =1+5\\ 4p& =6\\ p& =\frac{3}{2}\end{array}$

So,

$\begin{array}{cc}p+q& =\frac{3}{2}+\frac{1}{3}\\ & =\frac{9+2}{6}\\ & =\frac{11}{6}\\ & \approx 2\end{array}$