How do you find the x and y intercept of $3 x + 2 y = 6$ $3x+2y=6$ ?

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How do you find the x and y intercept of $3 x + 2 y = 6$ $3x+2y=6$ ?

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Answer 1 · 2017-04-05T12:59:59

$y_{intercept} = 3 \to (x, y) = (0, 3)$ $y_("intercept")=3 ->(x,y)=(0,3)$

$x_{intercept} = 2 \to (x, y) = (2, 0)$ $x_("intercept")=2->(x,y)=(2,0)$

Explanation:

Subtract $3 x$ $3x$ from both sides giving:

$2 y = - 3 x + 6$ $2y=-3x+6$

Divide each side by 2

$y = - \frac{3}{2} x + 3$ $y=-3/2 x+3$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The y-intercept is at $x = 0$ $x=0$ giving:

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

$y_{intercept} = 3 \to (x, y) = (0, 3)$ $y_("intercept")=3 ->(x,y)=(0,3)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The x-intercept is at $y = 0$ $y=0$ giving:

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

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

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

$x_{intercept} = 2 \to (x, y) = (2, 0)$ $x_("intercept")=2->(x,y)=(2,0)$

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