How do you graph -2x + 3y = 12 on a coordinate graph?

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How do you graph -2x + 3y = 12 on a coordinate graph?

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Answer 1 · 2015-08-15T06:03:33

Convert the equation to slope-intercept form. Use the equation to find two points. Plot the points and draw a straight line through the points.

Explanation:

$- 2 x + 3 y = 12$ $-2x+3y=12$ follows the standard form for a linear equation, $A x + B y = C$ $Ax+By=C$ .

In order to graph this equation, you need to convert it to the slope-intercept form, and solve fore $y$ $y$ . The slope-intercept form for a linear equation is $y = m x + b$ $y=mx+b$ , where $m$ $m$ is the slope, and $b$ $b$ is the slope-intercept.

Convert the Standard Equation to Slope-intercept Form

$- 2 x + 3 y = 12$ $-2x+3y=12$

Add $2 x$ $2x$ to both sides of the equation.

$3 y = 2 x + 12$ $3y=2x+12$

Divide both sides by $3$ $3$ .

$y = \frac{2}{3} x + \frac{12}{3}$ $y=2/3x+12/3$ =

$y = \frac{2}{3} x + 4$ $y=2/3x+4$

Now use the equation to find two points on the line. Plot them, then draw a straight line through the points.

Point A: $(0, 4)$ $(0,4)$

If $x = 0, y = 4$ $x=0, y=4$

$y = \frac{2}{3} x + 4$ $y=2/3x+4$ =

Substitute $0$ $0$ for $x$ $x$ .

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

$y = 0 + 4 = 4$ $y=0+4=4$ =

$y = 4$ $y=4$

Point B: $(- 6, 0)$ $(-6,0)$

If $y = 0, y = - 6$ $y=0, y=-6$

$y = \frac{2}{3} x + 4$ $y=2/3x+4$

Substitute $0$ $0$ for $y$ $y$ .

$0 = \frac{2}{3} x + 4$ $0=2/3x+4$

Subtract $4$ $4$ from both sides.

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

Divide both sides by $\frac{2}{3}$ $2/3$ .

$- \frac{4}{1} \div \frac{2}{3} = x$ $-4/1-:2/3=x$ =

$- \frac{4}{1} \times \frac{3}{2} = x$ $-4/1xx3/2=x$ =

$- \frac{12}{2} = x$ $-12/2=x$ =

$- 6 = x$ $-6=x$

graph{y=2/3x+4 [-10, 10, -5, 5]}

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How do you graph -2x + 3y = 12 on a coordinate graph?

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