Solve the Inequality for a 4(2a+3)<3(a-1)

$4 (2 a + 3) < 3 (a - 1)$

Simplify $4 (2 a + 3)$ .

Start simplifying.

$0 + 0 + 4 (2 a + 3) < 3 (a - 1)$

Simplify by adding zeros.

$4 (2 a + 3) < 3 (a - 1)$

Apply the distributive property.

$4 (2 a) + 4 \cdot 3 < 3 (a - 1)$

Multiply.

Multiply $2$ by $4$ .

$8 a + 4 \cdot 3 < 3 (a - 1)$

Multiply $4$ by $3$ .

$8 a + 12 < 3 (a - 1)$

Simplify $3 (a - 1)$ .

Apply the distributive property.

$8 a + 12 < 3 a + 3 \cdot - 1$

Multiply $3$ by $- 1$ .

$8 a + 12 < 3 a - 3$

Move all terms containing $a$ to the left side of the inequality.

Subtract $3 a$ from both sides of the inequality.

$8 a + 12 - 3 a < - 3$

Subtract $3 a$ from $8 a$ .

$5 a + 12 < - 3$

Move all terms not containing $a$ to the right side of the inequality.

Subtract $12$ from both sides of the inequality.

$5 a < - 3 - 12$

Subtract $12$ from $- 3$ .

$5 a < - 15$

Divide each term in $5 a < - 15$ by $5$ and simplify.

Divide each term in $5 a < - 15$ by $5$ .

$\frac{5 a}{5} < \frac{- 15}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 a}{5} < \frac{- 15}{5}$

Divide $a$ by $1$ .

$a < \frac{- 15}{5}$

Simplify the right side.

Divide $- 15$ by $5$ .

$a < - 3$

The result can be shown in multiple forms.

Inequality Form:

$a < - 3$

Interval Notation:

$(- \infty, - 3)$