# Basic algebra question, confused

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• Sep 3rd 2010, 03:06 PM
NecroWinter
Basic algebra question, confused
I didnt know how to name this topic, but hopefully someone can help

heres the equation:
(one-fourth y minus two thirds y equals 5)
1/4y - 2/3y = 5

I have the answer, which is -12. But can someone explain to me HOW to get the answer without plugging -12 in? When I start moving numbers to the other side I end up getting a number thats waay off by the end of it.

Thank you!
• Sep 3rd 2010, 03:10 PM
Plato
Multiply every term by 12.
Then solve the equation you get.
• Sep 3rd 2010, 03:12 PM
yeKciM
Quote:

Originally Posted by NecroWinter
I didnt know how to name this topic, but hopefully someone can help

heres the equation:
(one-fourth y minus two thirds y equals 5)
1/4y - 2/3y = 5

I have the answer, which is -12. But can someone explain to me HOW to get the answer without plugging -12 in? When I start moving numbers to the other side I end up getting a number thats waay off by the end of it.

Thank you!

$\displaystyle \displaystyle \frac {y}{4} -\frac {2y}{3} = 5$

multiply all by 12

$\displaystyle 3y-8y=60 \Rightarrow 5y=-60 \Rightarrow y = -12$

or is it

$\displaystyle \displaystyle \frac {1}{4y} - \frac {2}{3y} = 5$

Edit : Sorry Plato, didn't see that you have posted answer ...
• Sep 3rd 2010, 03:44 PM
Archie Meade
Quote:

Originally Posted by NecroWinter
I didnt know how to name this topic, but hopefully someone can help

heres the equation:
(one-fourth y minus two thirds y equals 5)
1/4y - 2/3y = 5

I have the answer, which is -12. But can someone explain to me HOW to get the answer without plugging -12 in? When I start moving numbers to the other side I end up getting a number thats waay off by the end of it.

Thank you!

Two-thirds of some number subtracted from a quarter of the same number equals 5.

$\displaystyle \displaystyle\ y\left(\frac{1}{4}\right)-y\left(\frac{2}{3}\right)=5\Rightarrow\ y\left(\frac{1}{4}-\frac{2}{3}\right)=5$

A quarter is just a value where the denominator is 4 times the numerator.

$\displaystyle \displaystyle\frac{1}{4}=\frac{2}{8}=\frac{3}{12}= \frac{4}{16}=.....$

Also

$\displaystyle \displaystyle\frac{2}{3}=\frac{2}{2}\cdot\frac{2}{ 3}=\frac{4}{6}=\frac{8}{12}=....$

We can add and subtract fractions with the same denominator...

$\displaystyle \displaystyle\ y\left(\frac{3}{12}-\frac{8}{12}\right)=y\left(-\frac{5}{12}\right)$

$\displaystyle \displaystyle\ -\frac{5}{12}\left(-\frac{12}{5}\right)=1$

Therefore...

$\displaystyle \displaystyle\ y=\left(-\frac{12}{5}\right)5$

Since LHS=RHS, then 12(LHS)=12(RHS)

so a shortcut is to multiply both sides by 12 to avoid having to combine the fractions.