please help?? cant figure this out

Since the absolute value of any real number is non-negative, it follows that $|x| \geq 0$. Therefore, the equation $|x| = -\frac{1}{2}x + 4$ can only be satisfied if $-\frac{1}{2}x + 4 \geq 0$, which is equivalent to $x \leq 8$.

Now we consider two cases:

Case 1: $x \geq 0$. In this case, $|x| = x$, so the equation becomes $x = -\frac{1}{2}x + 4$. Solving for $x$ gives $x = \frac{8}{3}$.

Case 2: $x < 0$. In this case, $|x| = -x$, so the equation becomes $-x = -\frac{1}{2}x + 4$. Solving for $x$ gives $x = -8$. However, this value does not satisfy the original condition $x < 0$. Therefore, there are no solutions in this case.

Therefore, the only solution is $x = \boxed{\frac{8}{3}}$.  The total number of solutions is 1.

\( |x| = -\dfrac 1 2 x + 4\\ |x| = \frac {-x+8}{2}\\ \text{square both sides}\\ x^2=\frac{x^2-16x+64}{4}\\ 4x^2=x^2-16x+64\\ 3x^2+16x-64=0\\ x=\frac{-16\pm\sqrt{256+768}}{6}\\ x=\frac{-16\pm32}{6}\\ x=\frac{-8\pm16}{3}\\ x=-8\quad or \quad x=\frac{8}{3}\\ \text{BUT these answers need to be checked for validity.} \)

 

Check x=-8

RHS= 4+4=8=LHS     works

 

check x=8/3

RHS= -8/6 +4 = -4/3 +4 = (-4+12)/3 = 8/3 = LHS   also works

 

LaTex

 |x| = -\dfrac 1 2 x + 4\\
 |x| = \frac {-x+8}{2}\\
\text{square both sides}\\
x^2=\frac{x^2-16x+64}{4}\\
4x^2=x^2-16x+64\\
3x^2+16x-64=0\\
x=\frac{-16\pm\sqrt{256+768}}{6}\\
x=\frac{-16\pm32}{6}\\
x=\frac{-8\pm16}{3}\\
x=-8\quad or \quad x=\frac{8}{3}\\
\text{BUT these answers need to be checked for validity.}