Solve for y 10+3y-2=4y-y+8

$10 + 3 y - 2 = 4 y - y + 8$

Since $y$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$4 y - y + 8 = 10 + 3 y - 2$

Subtract $y$ from $4 y$ .

$3 y + 8 = 10 + 3 y - 2$

Subtract $2$ from $10$ .

$3 y + 8 = 3 y + 8$

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

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

$3 y + 8 - 3 y = 8$

Combine the opposite terms in $3 y + 8 - 3 y$ .

Subtract $3 y$ from $3 y$ .

$0 + 8 = 8$

Add $0$ and $8$ .

$8 = 8$

Since $8 = 8$ , the equation will always be true for any value of $y$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$