Question

question
simplify the expression below using order of operations.
\\(\frac{(-4)^{2}+(-6)}{-4 - (-2)}-\frac{2-(-6)}{-7 + 8}\\)

Explanation:

Step1: Simplify numerators and denominators separately

First, simplify $(-4)^2+(-6)$ in the first - fraction's numerator: $(-4)^2 = 16$, so $(-4)^2+(-6)=16 - 6=10$.
The first - fraction's denominator: $-4-(-2)=-4 + 2=-2$.
The second - fraction's numerator: $2-(-6)=2 + 6 = 8$.
The second - fraction's denominator: $-7 + 8=1$.

Step2: Calculate the two fractions

The first fraction $\frac{(-4)^2+(-6)}{-4-(-2)}=\frac{10}{-2}=-5$.
The second fraction $\frac{2-(-6)}{-7 + 8}=\frac{8}{1}=8$.

Step3: Subtract the two results

$-5-8=-13$.

Answer:

$-13$