How do you write $y = - \frac{3}{2} x + 3$ $y = -3/2x + 3$ in standard form?

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Answer 1 · 2017-06-14T08:48:41

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

Standard Form looks like :
$a x + b y = c$ $ax+by=c$
where $a, b$ $a,b$ and $c$ $c$ are constants.

So we need to rearrange the formula to get both $x$ $x$ and $y$ $y$ to one side, and a number on the other side.

So from:

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

We should get rid of the fraction first. To do this, multiply everything by 2:

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

$\Rightarrow 2 y = \frac{2 \cdot - 3}{2} x + 6$ $=>2y=(2*-3)/2x+6$

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

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

Then we bring -3x to the other side by adding 3x to both sides:

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

$=>3x+2y=cancel(-3x)+6+cancel(3x)$

$=>3x+2y=6$

And you've got it in standard from now

How do you write y = -3/2x + 3y=−32x+3y = -3/2x + 3 in standard form?