solving fractions w/ variables 2: {4x/(x^2-64)}+2={2/(x+8)}

A

anm2007

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{4x/(x^2-64)} + 2 = {2/(x+8)}

i already factored the x^2-64....what do i do now?

btw when i factored that i got (x+8)(x-8)
 
\(\displaystyle \frac{4x}{(x+8)(x-8)} + 2 = \frac{2}{x+8}\)

The way I would do it is to multiply both sides by a common factor so that I will get rid of all the denominators. So, what can you multiply to both sides to get rid of the denominators? Hint: Look at the first fraction on the left side of the equation
 
No no. If you multiplied 4 to both sides, you would still have all those fractions. What is the common denominator of the fractions? Once you find that, you can multiply both sides by it so that you'll get an easier equation to deal with.

Here's a link that might help you. The example on the bottom of the page (2nd method as they call it) is similar to your problem in that you have to find a common denominator. See how that applies and come back when you get stuck again:
http://www.purplemath.com/modules/solvrtnl.htm
 
the common denominator is (x+8) so that means that i take that x+8 and multiply the 4x with it. then i multiply the top and the bottom of the fraction on the other side of the equal sign?
 
Actually - you need to multiply with "lowest Common Multiple (LCM) of the denominators, to get rid of the fraction.

In your case the lowest common multiple is --> \(\displaystyle (x+8)(x-8)\)

Please try to understand why you are doing this. Go back to addition like

1/2 + 3/7 = ?

What did you do there - You found the common denominator (which is the lowest common multiple of the denominators) - in this case it was 14.

So you made all the denominators 14.

Here you are doing similar thing - but multiplying with the common denominator - to eradicate fraction.
 
Subhotosh Khan said:
Actually - you need to multiply with "lowest Common Multiple (LCM) of the denominators, to get rid of the fraction.

In your case the lowest common multiple is --> \(\displaystyle (x+8)(x-8)\)

Please try to understand why you are doing this. Go back to addition like

1/2 + 3/7 = ?

What did you do there - You found the common denominator (which is the lowest common multiple of the denominators) - in this case it was 14.

So you made all the denominators 14.

Here you are doing similar thing - but multiplying with the common denominator - to eradicate fraction.
Click to expand...

ok so mr. khan i do basically what i did with the last problem we just did. except after i get what "x" is i add +2 right?
 
anm2007 said:
ok so mr. khan i do basically what i did with the last problem we just did. except after i get what "x" is i add +2 right?
Click to expand...
No - absolutely not.

Whatever gave you that idea?

\(\displaystyle \frac{4x}{(x+8)(x-8)} + 2 = \frac{2}{x+8}\)

multiply both sides by \(\displaystyle (x+8)(x-8)\) - to get

\(\displaystyle 4x+ 2(x+8)(x-8)= 2(x-8)\)

\(\displaystyle 2x^2 + 4x - 128 = 2x - 16\)

\(\displaystyle 2x^2 + 2x - 112 = 0\)

Above is a quadratic equation - solve it by your favorite method. Have you been taught to solve quadratic equation?
 
Subhotosh Khan said:
\(\displaystyle \frac{4x}{(x+8)(x-8)} + 2 = \frac{2}{x+8}\)

multiply both sides by \(\displaystyle (x+8)(x-8)\) - to get

\(\displaystyle 4x+ 2(x+8)(x-8)= 2(x-8)\)

\(\displaystyle 2x^2 + 4x - 128 = 2x - 16\)

\(\displaystyle 2x^2 + 2x - 112 = 0\)

Above is a quadratic equation - solve it by your favorite method. Have you been taught to solve quadratic equation?
Click to expand...
ok so now i factored that out and got {2(x+7)(x-8)}=0

then i take 2(x+7)=0 which equals -7
then i take (x-8)=0 which equals 8

so the two answers are x=-7 and x=8

right? this is the way our teacher told us to do it
 
anm2007 said:
ok so now i factored that out and got {2(x+7)(x-8)}=0 incorrect This should be 2(x-7)(x+8)

then i take 2(x+7)=0 which equals -7
then i take (x-8)=0 which equals 8

so the two answers are x=7 and x=-8 CORRECT - with the correction shown
Click to expand...
 
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