Question

Find the value of x. Given $\frac{3x+5y}{95-7z}=1;3y=3z-21;z-3=6$.

• A. $\frac{22}{7}$
• B. $\frac{22}{5}$
• C. $\frac{22}{3}$
• D. $\frac{22}{9}$

Answer 1

The correct option is C $\frac{22}{3}$

Given
$z-3=6z=6+3=9$
and also
$3y=3z-21y=z-7\left[\text{On dividing by 3}\right]=9-7=2$

Now, by substituting the value of $z$ and $y$ in the given equation
$\frac{3x+5y}{95-7z}$
$\Rightarrow \frac{3x+5\times 2}{95-7\times 9}=1\Rightarrow 3x+10=95-63\Rightarrow 3x=95-63-10=22\Rightarrow x=\frac{22}{3}$
So, the value of $x$ is $\frac{22}{3}$.