Find the x and y Intercepts y+2=-3/2(x-4)

$y + 2 = - \frac{3}{2} (x - 4)$

Find the x-intercepts.

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$(0) + 2 = - \frac{3}{2} \cdot (x - 4)$

Solve the equation.

Rewrite the equation as $- \frac{3}{2} \cdot (x - 4) = (0) + 2$ .

$- \frac{3}{2} \cdot (x - 4) = (0) + 2$

Multiply both sides of the equation by $- \frac{2}{3}$ .

$- \frac{2}{3} (- \frac{3}{2} \cdot (x - 4)) = - \frac{2}{3} ((0) + 2)$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- \frac{2}{3} (- \frac{3}{2} \cdot (x - 4))$ .

Simplify terms.

Apply the distributive property.

$- \frac{2}{3} (- \frac{3}{2} x - \frac{3}{2} \cdot - 4) = - \frac{2}{3} ((0) + 2)$

Combine $x$ and $\frac{3}{2}$ .

$- \frac{2}{3} (- \frac{x \cdot 3}{2} - \frac{3}{2} \cdot - 4) = - \frac{2}{3} ((0) + 2)$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$- \frac{2}{3} (- \frac{x \cdot 3}{2} + \frac{- 3}{2} \cdot - 4) = - \frac{2}{3} ((0) + 2)$

Factor $2$ out of $- 4$ .

$- \frac{2}{3} (- \frac{x \cdot 3}{2} + \frac{- 3}{2} \cdot (2 (- 2))) = - \frac{2}{3} ((0) + 2)$

Cancel the common factor.

$- \frac{2}{3} (- \frac{x \cdot 3}{2} + \frac{- 3}{2} \cdot (2 \cdot - 2)) = - \frac{2}{3} ((0) + 2)$

Rewrite the expression.

$- \frac{2}{3} (- \frac{x \cdot 3}{2} - 3 \cdot - 2) = - \frac{2}{3} ((0) + 2)$

Multiply $- 3$ by $- 2$ .

$- \frac{2}{3} (- \frac{x \cdot 3}{2} + 6) = - \frac{2}{3} ((0) + 2)$

Move $3$ to the left of $x$ .

$- \frac{2}{3} (- \frac{3 x}{2} + 6) = - \frac{2}{3} ((0) + 2)$

Simplify terms.

Apply the distributive property.

$- \frac{2}{3} (- \frac{3 x}{2}) - \frac{2}{3} \cdot 6 = - \frac{2}{3} ((0) + 2)$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{2}{3}$ into the numerator.

$\frac{- 2}{3} (- \frac{3 x}{2}) - \frac{2}{3} \cdot 6 = - \frac{2}{3} ((0) + 2)$

Move the leading negative in $- \frac{3 x}{2}$ into the numerator.

$\frac{- 2}{3} \cdot \frac{- 3 x}{2} - \frac{2}{3} \cdot 6 = - \frac{2}{3} ((0) + 2)$

Factor $2$ out of $- 2$ .

$\frac{2 (- 1)}{3} \cdot \frac{- 3 x}{2} - \frac{2}{3} \cdot 6 = - \frac{2}{3} ((0) + 2)$

Cancel the common factor.

$\frac{2 \cdot - 1}{3} \cdot \frac{- 3 x}{2} - \frac{2}{3} \cdot 6 = - \frac{2}{3} ((0) + 2)$

Rewrite the expression.

$\frac{- 1}{3} (- 3 x) - \frac{2}{3} \cdot 6 = - \frac{2}{3} ((0) + 2)$

Cancel the common factor of $3$ .

Factor $3$ out of $- 3 x$ .

$\frac{- 1}{3} (3 (- x)) - \frac{2}{3} \cdot 6 = - \frac{2}{3} ((0) + 2)$

Cancel the common factor.

$\frac{- 1}{3} (3 (- x)) - \frac{2}{3} \cdot 6 = - \frac{2}{3} ((0) + 2)$

Rewrite the expression.

$- - x - \frac{2}{3} \cdot 6 = - \frac{2}{3} ((0) + 2)$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 x - \frac{2}{3} \cdot 6 = - \frac{2}{3} ((0) + 2)$

Multiply $x$ by $1$ .

$x - \frac{2}{3} \cdot 6 = - \frac{2}{3} ((0) + 2)$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{2}{3}$ into the numerator.

$x + \frac{- 2}{3} \cdot 6 = - \frac{2}{3} ((0) + 2)$

Factor $3$ out of $6$ .

$x + \frac{- 2}{3} \cdot (3 (2)) = - \frac{2}{3} ((0) + 2)$

Cancel the common factor.

$x + \frac{- 2}{3} \cdot (3 \cdot 2) = - \frac{2}{3} ((0) + 2)$

Rewrite the expression.

$x - 2 \cdot 2 = - \frac{2}{3} ((0) + 2)$

Multiply $- 2$ by $2$ .

$x - 4 = - \frac{2}{3} ((0) + 2)$

Simplify the right side.

Simplify $- \frac{2}{3} ((0) + 2)$ .

Add $0$ and $2$ .

$x - 4 = - \frac{2}{3} \cdot 2$

Multiply $- \frac{2}{3} \cdot 2$ .

Multiply $2$ by $- 1$ .

$x - 4 = - 2 (\frac{2}{3})$

Combine $- 2$ and $\frac{2}{3}$ .

$x - 4 = \frac{- 2 \cdot 2}{3}$

Multiply $- 2$ by $2$ .

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

Move the negative in front of the fraction.

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

Move all terms not containing $x$ to the right side of the equation.

Add $4$ to both sides of the equation.

$x = - \frac{4}{3} + 4$

To write $4$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = - \frac{4}{3} + 4 \cdot \frac{3}{3}$

Combine $4$ and $\frac{3}{3}$ .

$x = - \frac{4}{3} + \frac{4 \cdot 3}{3}$

Combine the numerators over the common denominator.

$x = \frac{- 4 + 4 \cdot 3}{3}$

Simplify the numerator.

Multiply $4$ by $3$ .

$x = \frac{- 4 + 12}{3}$

Add $- 4$ and $12$ .

$x = \frac{8}{3}$

x-intercept(s) in point form.

x-intercept(s): $(\frac{8}{3}, 0)$

Find the y-intercepts.

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y + 2 = - \frac{3}{2} \cdot ((0) - 4)$

Solve the equation.

Simplify $- \frac{3}{2} \cdot ((0) - 4)$ .

Subtract $4$ from $0$ .

$y + 2 = - \frac{3}{2} \cdot - 4$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$y + 2 = \frac{- 3}{2} \cdot - 4$

Factor $2$ out of $- 4$ .

$y + 2 = \frac{- 3}{2} \cdot (2 (- 2))$

Cancel the common factor.

$y + 2 = \frac{- 3}{2} \cdot (2 \cdot - 2)$

Rewrite the expression.

$y + 2 = - 3 \cdot - 2$

Multiply $- 3$ by $- 2$ .

$y + 2 = 6$

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

Subtract $2$ from both sides of the equation.

$y = 6 - 2$

Subtract $2$ from $6$ .

$y = 4$

y-intercept(s) in point form.

y-intercept(s): $(0, 4)$

List the intersections.

x-intercept(s): $(\frac{8}{3}, 0)$

y-intercept(s): $(0, 4)$

Find the x and y Intercepts y+2=-3/2(x-4)

Related Questions

Support LCMGCF.com by disabling your adblocker.