Intercepts & standard form

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April 27th 2010, 02:07 PM
buck

Intercepts & standard form

Express each of the following in standard form

y= 1/4x - 1/3
-4(-1/4x + y +1/3)=(0)-4
-3(1x - 4y -4/3)=(0)-3
-3x+12y+4=0
-3x+12y=-4

The answer is 3x-12y=4

I don't know what I did wrong, please help.(Itwasntme)
April 27th 2010, 02:17 PM
harish21

Quote:

Originally Posted by buck

Express each of the following in standard form

y= 1/4x - 1/3
-4(-1/4x + y +1/3)=(0)-4
-3(1x - 4y -4/3)=(0)-3
-3x+12y+4=0
-3x+12y=-4

The answer is 3x-12y=4

I don't know what I did wrong, please help.(Itwasntme)

Your ansswer is correct
You have $-3x+12y=-4$

so,

$-(3x-12y) = -4 \implies 3x-12y=4$

another method:

$y = \frac{x}{4} - \frac{1}{3} = \frac{3x-4}{12}$

$\therefore 12y = 3x-4 \implies 3x-12y=4$

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