Graph 5x+10y-15>=0

$5 x + 10 y - 15 \geq 0$

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

Solve for $y$ .

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

Subtract $5 x$ from both sides of the inequality.

$10 y - 15 \geq - 5 x$

Add $15$ to both sides of the inequality.

$10 y \geq - 5 x + 15$

Divide each term in $10 y \geq - 5 x + 15$ by $10$ and simplify.

Divide each term in $10 y \geq - 5 x + 15$ by $10$ .

$\frac{10 y}{10} \geq \frac{- 5 x}{10} + \frac{15}{10}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 y}{10} \geq \frac{- 5 x}{10} + \frac{15}{10}$

Divide $y$ by $1$ .

$y \geq \frac{- 5 x}{10} + \frac{15}{10}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 5$ and $10$ .

Factor $5$ out of $- 5 x$ .

$y \geq \frac{5 (- x)}{10} + \frac{15}{10}$

Cancel the common factors.

Factor $5$ out of $10$ .

$y \geq \frac{5 (- x)}{5 (2)} + \frac{15}{10}$

Cancel the common factor.

$y \geq \frac{5 (- x)}{5 \cdot 2} + \frac{15}{10}$

Rewrite the expression.

$y \geq \frac{- x}{2} + \frac{15}{10}$

Move the negative in front of the fraction.

$y \geq - \frac{x}{2} + \frac{15}{10}$

Cancel the common factor of $15$ and $10$ .

Factor $5$ out of $15$ .

$y \geq - \frac{x}{2} + \frac{5 (3)}{10}$

Cancel the common factors.

Factor $5$ out of $10$ .

$y \geq - \frac{x}{2} + \frac{5 \cdot 3}{5 \cdot 2}$

Cancel the common factor.

$y \geq - \frac{x}{2} + \frac{5 \cdot 3}{5 \cdot 2}$

Rewrite the expression.

$y \geq - \frac{x}{2} + \frac{3}{2}$

Reorder terms.

$y \geq - (\frac{1}{2} x) + \frac{3}{2}$

Remove parentheses.

$y \geq - \frac{1}{2} x + \frac{3}{2}$

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

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

$m = - \frac{1}{2}$

$b = \frac{3}{2}$

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

Slope: $- \frac{1}{2}$

y-intercept: $(0, \frac{3}{2})$

Slope: $- \frac{1}{2}$

y-intercept: $(0, \frac{3}{2})$

Graph a solid line, then shade the area above the boundary line since $y$ is greater than $- \frac{1}{2} x + \frac{3}{2}$ .

$y \geq - \frac{1}{2} x + \frac{3}{2}$

Graph 5x+10y-15>=0

Related Questions

Support LCMGCF.com by disabling your adblocker.