Solving an equation with a fraction in it.

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cgarcia71

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Mar 18, 2006
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Hi I have this problem.

y+1/2 = -1/3 (x+1/2)

I am thinking that I need to first do this

y + 1/2 = -1/3x - 1/6

then maybe

y + 1/2 - 1/6 = - 1/3 x

This is where I get lost.

Can someone help? Thanks you all have been great today!
 
suckatmath71 said:
Hi I have this problem.

y+1/2 = -1/3 (x+1/2)
Click to expand...
Very nicely written.

y + 1/2 = -1/3x - 1/6
Click to expand...
Not clear. Perhaps...
y + 1/2 = (-1/3)x - 1/6 -- This keeps the 'x' out of the denominator.

y + 1/2 - 1/6 = (-1/3)x
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How did that happen? If you are using addition, you should have added 1/6.
 
Sorry for not being clear...

I started out with this...

y+1/2 = -1/3 (x+1/2)

then I did this

y + 1/2 = -1/3x - 1/6 (I multipled -1/3 * x and then -1/3 * 1/2 to get -1/6)

is the part in parentheses incorrect?


This where I get lost.

Thanks for your help.
 
I am sorry.. I need to solve for Y. I missed that part in the instructions. I am still lost.
 
cgarcia71 said:
I am sorry.. I need to solve for Y. I missed that part in the instructions. I am still lost.
Click to expand...
So you want to solve y + 1/2 = -1/3 (x + 1/2) for y.

Really, you can stop after subtracting 1/2 from each side:
y = -1/3 (x + 1/2) - 1/2
If you're simply asked to solve for y, that would be sufficient.

HOWEVER, you are PROBABLY expected to solve for y AND simplify.

Start with the multiplication:
-1/3 (x + 1/2)
= -x/3 - 1/6 ; RULE: 1/a * b = b/a

So we now have:
y = -x/3 - 1/6 - 1/2
since -1/6 - 1/2 = -2/3, then:
y = -x/3 - 2/3
y = (-x - 2) / 3

You can also show that this way:
y = -(x + 2) / 3
(some teachers may give you an extra point for this :wink: )
 
cgarcia71 said:
Thanks so much! You are hilarious.... =)
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Yes, I know...BUT I'm presently taking special treatment for that disease :roll:
 
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