Graph y<=-1/3x+1

$y \leq - \frac{1}{3} x + 1$

Step 1

Find the slope and the y-intercept for the boundary line.

Rewrite in slope-intercept form.

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

$y = m x + b$

Simplify the right side.

Combine $x$ and $\frac{1}{3}$ .

$y \leq - \frac{x}{3} + 1$

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

Reorder terms.

$y \leq - (\frac{1}{3} x) + 1$

Remove parentheses.

$y \leq - \frac{1}{3} x + 1$

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

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

$m = - \frac{1}{3}$

$b = 1$

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

Slope: $- \frac{1}{3}$

y-intercept: $(0, 1)$

Slope: $- \frac{1}{3}$

y-intercept: $(0, 1)$

Slope: $- \frac{1}{3}$

y-intercept: $(0, 1)$

Step 2

Graph a solid line, then shade the area below the boundary line since $y$ is less than $- \frac{1}{3} x + 1$ .

$y \leq - \frac{1}{3} x + 1$

Step 3