Solve for y -4y-3=1/3(12y-9)-8y

$- 4 y - 3 = \frac{1}{3} (12 y - 9) - 8 y$

Step 1

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

$\frac{1}{3} \cdot (12 y - 9) - 8 y = - 4 y - 3$

Step 2

Simplify $\frac{1}{3} \cdot (12 y - 9) - 8 y$ .

Simplify each term.

Apply the distributive property.

$\frac{1}{3} (12 y) + \frac{1}{3} \cdot - 9 - 8 y = - 4 y - 3$

Cancel the common factor of $3$ .

Factor $3$ out of $12 y$ .

$\frac{1}{3} (3 (4 y)) + \frac{1}{3} \cdot - 9 - 8 y = - 4 y - 3$

Cancel the common factor.

$\frac{1}{3} (3 (4 y)) + \frac{1}{3} \cdot - 9 - 8 y = - 4 y - 3$

Rewrite the expression.

$4 y + \frac{1}{3} \cdot - 9 - 8 y = - 4 y - 3$

Cancel the common factor of $3$ .

Factor $3$ out of $- 9$ .

$4 y + \frac{1}{3} \cdot (3 (- 3)) - 8 y = - 4 y - 3$

Cancel the common factor.

$4 y + \frac{1}{3} \cdot (3 \cdot - 3) - 8 y = - 4 y - 3$

Rewrite the expression.

$4 y - 3 - 8 y = - 4 y - 3$

Subtract $8 y$ from $4 y$ .

$- 4 y - 3 = - 4 y - 3$

Step 3

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

Add $4 y$ to both sides of the equation.

$- 4 y - 3 + 4 y = - 3$

Combine the opposite terms in $- 4 y - 3 + 4 y$ .

Add $- 4 y$ and $4 y$ .

$0 - 3 = - 3$

Subtract $3$ from $0$ .

$- 3 = - 3$

Step 4

Since $- 3 = - 3$ , the equation will always be true for any value of $y$ .

All real numbers

Step 5

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$