Question

The solution of the equation $\frac{x}{4}\pm \frac{1}{2}=3$ is/are:

• A. $x=12$
• B. $x=14$
• C. $x=10$
• D. $x=5$

Answer 1

Answer: (B), (C)

$x=10$

Given, the equation is $\frac{x}{4}\pm \frac{1}{2}=3.$

Case 1: Taking positive sign
$\frac{x}{4}+\frac{1}{2}=3$

Step 1: Subtract $\frac{1}{2}$ from both sides of the equation in order to remove the $\frac{1}{2}$ on the left hand side.
$\Rightarrow \frac{x}{4}+\frac{1}{2}-\frac{1}{2}=3-\frac{1}{2}$
$\Rightarrow \frac{x}{4}=3-\frac{1}{2}(\because \frac{1}{2}-\frac{1}{2}=0)$
$\Rightarrow \frac{x}{4}=\frac{3\times 2}{1\times 2}-\frac{1}{2}\left(\text{making the denominators equal for}3\text{and}\frac{1}{2}\right)$
$\Rightarrow \frac{x}{4}=\frac{6}{2}-\frac{1}{2}$
$\Rightarrow \frac{x}{4}=\frac{6-1}{2}$
$\Rightarrow \frac{x}{4}=\frac{5}{2}$

Step 2: Multiply both the sides by 4
$\Rightarrow \frac{x}{4}\times 4=\frac{5}{2}\times 4$
$\Rightarrow x=\frac{5}{2}\times 2\times 2$
$\Rightarrow x=5\times 2$
$\Rightarrow x=10$

Case 2: Taking negative sign
$\frac{x}{4}-\frac{1}{2}=3$

Step 1: Add both sides of the equation by $\frac{1}{2}$ in order to remove the $\frac{1}{2}$ on the left hand side.
$\Rightarrow \frac{x}{4}-\frac{1}{2}+\frac{1}{2}=3+\frac{1}{2}$
$\Rightarrow \frac{x}{4}=\frac{3\times 2}{1\times 2}+\frac{1}{2}$
$\Rightarrow \frac{x}{4}=\frac{6}{2}+\frac{1}{2}$
$\Rightarrow \frac{x}{4}=\frac{6+1}{2}$
$\Rightarrow \frac{x}{4}=\frac{7}{2}$

Step 2: Multiply both the sides by 4
$\Rightarrow \frac{x}{4}\times 4=\frac{7}{2}\times 4$
$\Rightarrow x=\frac{7}{2}\times 2\times 2$
$\Rightarrow x=7\times 2$
$\Rightarrow x=14$

Hence, the solution of the equation $\frac{x}{4}\pm \frac{1}{2}=3$ are $x=10$ and $x=14.$