Slope-Intercept and Standard Form...HELP!!!!

I understand how to convert Standard Form to Slope-Intercept form:

Ax+By=C becomes y = -A/Bx + C/B

But I can't figure out how to converst an equation in Slope-Intercept form to Standard form. PLEASE HELP!!!

y = -A/Bx + C/B is correct but you might want to write it as
y = -(A/B)x + C/B
just so there is no chance of confusion.

The reverse of what you did to get that is
y = (-Ax + C)/B
Multiply by B
By = -Ax + C
Ax + By = C

An example of one that is bothering you might be more helpful.

This is one that I can't figure out...I know the answer but not how to get there:

Write y = (2/3)x - 4 in Standard Form.

I know the answer is 2x - 3y = 12. Thanks.

Does "standard form" have fractions? No? Then multiply through to get rid of them.

Does "standard form" have variables on either side of the equation? No? Then rearrange to orient the variable-containing terms correctly.

Does "standard form" have a "minus" sign on the x-containing term? No? Then multiply through by -1, if necessary, to get the right sign.

Does "standard form" have the constant term (the plain number) on the same side as the variable-containing terms? No? Then rearrange, as necessary.

If you get stuck, please reply showing how far you have gotten in working these steps. Thank you.

Eliz.

lincoln20 said:

This is one that I can't figure out...I know the answer but not how to get there:
Write y = (2/3)x - 4 in Standard Form.
I know the answer is 2x - 3y = 12. Thanks.

Click to expand...

In case (2/3)x is what is confusing you:
that's same as 2x / 3

2/3 * x/1 = 2*x / 3*1 = 2x / 3

so equation is: y = 2x / 3 - 4

Thanks, I get it now!