Question

(a) solve for v.
4(v + 1)+v = 5(v - 1)+9
no solution
v =
all real numbers are solutions
(b) solve for u.
3(u - 2)-4u = 2(u - 9)
no solution
u =
all real numbers are solutions

Explanation:

Step1: Expand both sides of the first - equation

Expand $4(v + 1)+v$ and $5(v - 1)+9$.
$4(v + 1)+v=4v+4 + v=5v + 4$, and $5(v - 1)+9=5v-5 + 9=5v + 4$.
Since $5v + 4=5v + 4$ for all real - valued $v$, all real numbers are solutions.

Step2: Expand both sides of the second - equation

Expand $3(u - 2)-4u$ and $2(u - 9)$.
$3(u - 2)-4u=3u-6-4u=-u - 6$, and $2(u - 9)=2u-18$.

Step3: Solve the second - equation for $u$

Set $-u - 6=2u-18$.
Add $u$ to both sides: $-6 = 3u-18$.
Add 18 to both sides: $3u=12$.
Divide both sides by 3: $u = 4$.

Answer:

(a) All real numbers are solutions
(b) $u = 4$