Question

for each part below, solve the equation.
(a) solve for v.
$6(v + 1) - 10 = 4(v - 1) + 2v$
no solution
$v = \square$
all real numbers are solutions
(b) solve for u.
$5(2 + u) - u = 10 + 4(u + 1)$
no solution
$u = \square$
all real numbers are solutions

Explanation:

Part (a)

Step1: Expand both sides

Left side: \(6(v + 1)-10=6v+6 - 10=6v - 4\)
Right side: \(4(v - 1)+2v=4v-4 + 2v=6v - 4\)

Step2: Compare both sides

We have \(6v - 4=6v - 4\). Subtracting \(6v\) from both sides gives \(- 4=-4\), which is always true. So all real numbers are solutions.

Part (b)

Step1: Expand both sides

Left side: \(5(2 + u)-u = 10+5u - u=10 + 4u\)
Right side: \(10+4(u + 1)=10+4u+4=4u + 14\)

Step2: Solve for \(u\)

Set \(10 + 4u=4u + 14\). Subtract \(4u\) from both sides: \(10=14\), which is false. So there is no solution.

Answer:

All real numbers are solutions