Question

$\frac{4+z-(3+2z)}{6}=\frac{-z-3(5-2)}{7}$.
What is the value of $z$ in the equation?

• A. $-61$
• B. $-\frac{61}{27}$
• C. $\frac{61}{27}$
• D. $61$

Answer 1

The correct option is D $61$

Given, $\frac{4+z-(3+2z)}{6}=\frac{-z-3(5-2)}{7}$

$\Rightarrow \frac{4+z-3-2z}{6}=\frac{-z-3\left(3\right)}{7}$ ...[By Distribution Law]

$\Rightarrow \frac{1-z}{6}=\frac{-z-9}{7}$ ...[On simplifying]

$\Rightarrow 7(1-z)=6(-z-9)$ …[Cross-multiplying the denominators]

$\Rightarrow 7-7z=-6z-54$ ...[By Distribution Law].

Transposing $z$ terms to one side, we get,

$\Rightarrow 7z-6z=7+54$

$\Rightarrow z=61$.

So, option $\text{D}$ is correct.