Graph 4x+3y>12

$4 x + 3 y > 12$

Step 1

Write in $y = m x + b$ form.

Solve for $y$ .

Subtract $4 x$ from both sides of the inequality.

$3 y > 12 - 4 x$

Divide each term in $3 y > 12 - 4 x$ by $3$ and simplify.

Divide each term in $3 y > 12 - 4 x$ by $3$ .

$\frac{3 y}{3} > \frac{12}{3} + \frac{- 4 x}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} > \frac{12}{3} + \frac{- 4 x}{3}$

Divide $y$ by $1$ .

$y > \frac{12}{3} + \frac{- 4 x}{3}$

Simplify the right side.

Simplify each term.

Divide $12$ by $3$ .

$y > 4 + \frac{- 4 x}{3}$

Move the negative in front of the fraction.

$y > 4 - \frac{4 x}{3}$

Rearrange terms.

$y > - \frac{4 x}{3} + 4$

Reorder terms.

$y > - (\frac{4}{3} x) + 4$

Remove parentheses.

$y > - \frac{4}{3} x + 4$

Step 2

Use the slope-intercept form to find the slope and y-intercept.

Find the values of $m$ and $b$ using the form $y = m x + b$ .

$m = - \frac{4}{3}$

$b = 4$

The slope of the line is the value of $m$ , and the y-intercept is the value of $b$ .

Slope: $- \frac{4}{3}$

y-intercept: $(0, 4)$

Slope: $- \frac{4}{3}$

y-intercept: $(0, 4)$

Step 3

Graph a dashed line, then shade the area above the boundary line since $y$ is greater than $- \frac{4}{3} x + 4$ .

$y > - \frac{4}{3} x + 4$

Step 4