Question

1. 4(2r - 1) = -2(3r + 16)
2. -(v + 4) + 5 = 4v + 1 - 5v

Explanation:

Step1: Expand both sides of eq4

$4(2r - 1) = -2(3r + 16)$
$8r - 4 = -6r - 32$

Step2: Isolate r terms (eq4)

$8r + 6r = -32 + 4$

Step3: Simplify to solve r (eq4)

$14r = -28$
$r = \frac{-28}{14} = -2$

Step4: Expand/simplify both sides of eq6

$-(v + 4) + 5 = 4v + 1 - 5v$
$-v - 4 + 5 = -v + 1$
$-v + 1 = -v + 1$

Step5: Analyze simplified eq6

$-v + v = 1 - 1$
$0 = 0$

Answer:

For equation 4: $r = -2$
For equation 6: All real numbers are solutions (identity equation)