Question

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

$\frac{y}{2}+3=7$ $\left(4,8,10,12\right)$

Answer 1

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

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

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

$=2+3$

$=5$

Since, LHS $\ne$RHS $\ne$$7$

So, $y=4$ is not solution of the given equation .

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

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

$=4+3$

$=7$

Since, LHS $=$RHS $=$$43$

Hence , $y=8$ is satisfies the given equation . So , $y=8$ is the required value .