Question

for each part below, solve the equation.
(a) solve for y.
6(y - 2)+1 = 2(3y - 5)
no solution
y =
all real numbers are solutions
(b) solve for u.
4(u - 3)-6u=-2(u + 6)
no solution
u =
all real numbers are solutions

Explanation:

Step1: Expand both sides of the equation in part (a)

Expand $6(y - 2)+1$ to get $6y-12 + 1=6y-11$, and expand $2(3y - 5)$ to get $6y-10$. So the equation becomes $6y-11=6y - 10$.

Step2: Subtract $6y$ from both sides

$6y-6y-11=6y-6y - 10$, which simplifies to $-11=-10$. Since this is a false statement, there is no solution for $y$.

Step3: Expand both sides of the equation in part (b)

Expand $4(u - 3)-6u$ to get $4u-12-6u=-2u-12$, and expand $-2(u + 6)$ to get $-2u-12$. So the equation becomes $-2u-12=-2u-12$.

Step4: Add $2u$ to both sides

$-2u + 2u-12=-2u+2u-12$, which simplifies to $-12=-12$. This is a true statement for all real - values of $u$, so all real numbers are solutions for $u$.

Answer:

(a) No solution
(b) All real numbers are solutions