Graph 3x+4y>12 y>2x-1

$3 x + 4 y > 12$ $y > 2 x - 1$

Graph $3 x + 4 y > 12$ .

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

Solve for $y$ .

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

$4 y > 12 - 3 x$

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

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

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

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Divide $12$ by $4$ .

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

Move the negative in front of the fraction.

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

Rearrange terms.

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

Reorder terms.

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

Remove parentheses.

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

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

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

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

$b = 3$

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

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

y-intercept: $(0, 3)$

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

y-intercept: $(0, 3)$

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

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

Graph $y > 2 x - 1$ .

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

The slope-intercept form is $y = m x + b$ , where $m$ is the slope and $b$ is the y-intercept.

$y = m x + b$

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

$m = 2$

$b = - 1$

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

Slope: $2$

y-intercept: $(0, - 1)$

Slope: $2$

y-intercept: $(0, - 1)$

Graph a dashed line, then shade the area above the boundary line since $y$ is greater than $2 x - 1$ .

$y > 2 x - 1$

Plot each graph on the same coordinate system.

$3 x + 4 y > 12$

$y > 2 x - 1$

Graph 3x+4y>12 y>2x-1

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