Find All Complex Solutions 3(y+41)=3y+123

$3 (y + 41) = 3 y + 123$

Step 1

Simplify $3 (y + 41)$ .

Rewrite.

$0 + 0 + 3 (y + 41) = 3 y + 123$

Simplify by adding zeros.

$3 (y + 41) = 3 y + 123$

Apply the distributive property.

$3 y + 3 \cdot 41 = 3 y + 123$

Multiply $3$ by $41$ .

$3 y + 123 = 3 y + 123$

Step 2

Move all terms containing $y$ to the left side of the equation.

Subtract $3 y$ from both sides of the equation.

$3 y + 123 - 3 y = 123$

Combine the opposite terms in $3 y + 123 - 3 y$ .

Subtract $3 y$ from $3 y$ .

$0 + 123 = 123$

Add $0$ and $123$ .

$123 = 123$

Step 3

Since $123 = 123$ , the equation will always be true.

Always true