Solve the Inequality for a 4a+3b<12

$4 a + 3 b < 12$

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

$4 a < 12 - 3 b$

Divide each term in $4 a < 12 - 3 b$ by $4$ and simplify.

Divide each term in $4 a < 12 - 3 b$ by $4$ .

$\frac{4 a}{4} < \frac{12}{4} + \frac{- 3 b}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 a}{4} < \frac{12}{4} + \frac{- 3 b}{4}$

Divide $a$ by $1$ .

$a < \frac{12}{4} + \frac{- 3 b}{4}$

Simplify the right side.

Simplify each term.

Divide $12$ by $4$ .

$a < 3 + \frac{- 3 b}{4}$

Move the negative in front of the fraction.

$a < 3 - \frac{3 b}{4}$