## Algebra: A Combined Approach (4th Edition)

$\dfrac{|6-2|+3}{8+2\cdot5}=\dfrac{7}{18}$
$\dfrac{|6-2|+3}{8+2\cdot5}$ First, perform the operation inside the absolute value bars in the numerator and the product in the denominator: $\dfrac{|6-2|+3}{8+2\cdot5}=\dfrac{|4|+3}{8+10}=...$ Now, evaluate the sums both in the numerator and the denominator: $...=\dfrac{4+3}{8+10}=\dfrac{7}{18}$