Question

1. solve $\left|\frac{1}{2}x + 3\

ight| = x + 3$ for x.
\bigcirc\\; $x = 0$
\bigcirc\\; no solutions
\bigcirc\\; $x = -4$ and $x = 0$
\bigcirc\\; $x = -4$

Explanation:

Step1: Split absolute value equation

For $|A|=B$, solve $A=B$ and $A=-B$

1. $\frac{1}{2}x + 3 = x + 3$
2. $\frac{1}{2}x + 3 = -(x + 3)$

Step2: Solve first equation

Subtract $\frac{1}{2}x$ and 3 from both sides:
$\frac{1}{2}x + 3 - \frac{1}{2}x - 3 = x + 3 - \frac{1}{2}x - 3$
$\implies 0 = \frac{1}{2}x$
$\implies x = 0$

Step3: Solve second equation

Expand and rearrange terms:
$\frac{1}{2}x + 3 = -x - 3$
Add $x$ and subtract 3 from both sides:
$\frac{1}{2}x + x + 3 - 3 = -x + x - 3 - 3$
$\implies \frac{3}{2}x = -6$
Multiply both sides by $\frac{2}{3}$:
$x = -6 \times \frac{2}{3} = -4$

Step4: Verify solutions

• For $x=0$: $|\frac{1}{2}(0)+3|=|3|=3$, and $0+3=3$. Valid.
• For $x=-4$: $|\frac{1}{2}(-4)+3|=|-2+3|=1$, and $-4+3=-1$. $1

eq -1$, so invalid.

Answer:

$x = 0$