Linear equation with fraction... HELP ME!

[blackwell33]

I am stumped on this problem... please help!

2/c+3 - 4/c-4 = 0

I know you are suppose to find the LCD. I am thinking I need to multiple both equations by (c^2 - 12) but I have no Idea how to do that or if that is even right!

Please help! My book only gives one example and theirs works out perfectly and things cancel out.... the book is no help at all.

 

[Aladdin]

2/(c+3)-4/(c-4)=0

Multiply the denomenator and the numerator of the first term by(c-4).
Multiply the denomenator and the numerator of the second term by (c+3)

2(c-4)/(c+3)(c-4) - 4(c+3)/(c-4)(c+3) =0
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[Loren]

2/c+3 - 4/c-4 = 0 means \(\displaystyle \frac{2}{c}+3 - \frac{4}{c}-4 = 0\).

I think you mean 2/(c+3) - 4/(c-4) = 0.

At any rate you are correct to multiply both sides of the equation by the least common denominator. However, the least common denominator is the product of (c+3)(c-4) which is not (c^2 - 12). Rather than multiplying (c+3)(c-4) to get c[sup:1vpncjh5]2[/sup:1vpncjh5]- c - 12 try multiplying leaving them in factored form. That way, it is easier to see what cancels.
\(\displaystyle \frac{(c+3)(c-4)}{1}\cdot \frac{2}{c+3} - \frac{(c+3)(c-4)}{1}\cdot \frac{4}{c-4}= \frac{(c+3)(c-4)}{1}\cdot 0\)
Can you take it from there?

 

[stapel]

the book is no help at all.

For additional lessons on rational equations, try online! :wink: