Question

One of the values given in brackets next to each equation satisfies the equation . Find that value .

$\frac{2a+7}{3}=5$ $\left(15,8,4,2\right)$

Answer 1

Given , the linear equation is $\frac{2a+7}{3}=5-------\left(1\right)$

Putting $a=15$ in equation $\left(1\right)$ we get ,

LHS$=\frac{2\times 15+7}{3}$

$=\frac{30+7}{3}$

$=\frac{37}{3}$

Since, LHS $\ne$RHS $\ne$$5$

So, $a=15$ is not solution of the given equation .

Putting $a=8$ in equation $\left(1\right)$ we get ,

LHS$=\frac{2\times 8+7}{3}$

$=\frac{16+7}{3}$

$=\frac{23}{3}$

Since, LHS $\ne$RHS $\ne$$5$

So, $a=8$ is not solution of the given equation .

Putting $a=4$ in equation $\left(1\right)$ we get ,

LHS$=\frac{2\times 4+7}{3}$

$=\frac{8+7}{3}$

$=\frac{15}{3}$

$=5$

Since, LHS $=$RHS $=$$5$

Hence , $a=4$ is satisfies the given equation . So , $a=4$ is the required value .