How do you find x and y intercepts of $- 6 x + 3 y = - 9$ $-6x + 3y = -9$ ?

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Answer 1 · 2016-12-17T21:41:56

x-intercept is $x = \frac{3}{2}$ $x = 3/2$ or #(3/2, 0).

y-intercept is $y = - 3$ $y = -3$ or (0, -3)

To find the x-intercept, set $y$ $y$ equal to $0$ $0$ and solve for $x$ $x$ :

$- 6 x + 3 \cdot 0 = - 9$ $-6x + 3*0 = -9$

$- 6 x + 0 = - 9$ $-6x + 0 = -9$

$- 6 x = - 9$ $-6x = -9$

$\frac{- 6 x}{- 6} = \frac{- 9}{- 6}$ $(-6x)/-6 = (-9)/-6$

$(cancel(-6)x)/cancel(-6) = 3/3 * (-3)/-2$

$x = 1 * 3/2$

$x = 3/2$ or #(3/2, 0).

To find the y-intercept, set $x$ equal to $0$ and solve for $y$ :

$-6*0 + 3y = -9$

$0 + 3y = -9$

$3y = -9$

$(3y)/3 = (-9)/3$

$(cancel(3)y)/cancel(3) = -3$

$y = -3$ or (0, -3)

How do you find x and y intercepts of -6x + 3y = -9−6x+3y=−9 -6x + 3y = -9?