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jackaleclipse · Jul 31, 2009, 02:39 PM

I do not understand how to solve the following problems:

(4/(x-1)) + (6/(3x+1)) = (15/(3x+1))

(x-5)^(3/2) = 8

galactus · Jul 31, 2009, 02:52 PM

(4/(x-1)) + (6/(3x+1)) = (15/(3x+1))

$\frac{4}{x-1}+\frac{6}{3x+1}=\frac{15}{3x+1}$

The LCD is (x-1)(3x+1). As with any fraction, we make the denominators the same in order to add.

$\frac{3x+1}{3x+1}\cdot \frac{4}{x-1}+\frac{x-1}{x-1}\cdot \frac{6}{3x+1}=\frac{15}{3x+1}\cdot \frac{x-1}{x-1}$

$\frac{4(3x+1)+6(x-1)}{(3x+1)(x-1)}=\frac{15}{(3x+1)(x-1)}$

Now that the denominators are the same, concentrate on the numerators.

$4(3x+1)+6(x-1)=15(x-1)$

Can you finish now?

Here is another way.

Rewrite as $\frac{4}{x-1}=\frac{15}{3x+1}-\frac{6}{3x+1}$

See? The denominators on the right are the same.

$\frac{4}{x-1}=\frac{15-6}{3x+1}$

Cross multiply:

$4(3x+1)=9(x-1)$

Now. Solve away. Did that seem a little easier than the other way?

(x-5)^(3/2) = 8

$(x-5)=8^{\frac{2}{3}}$

Take that 3/2 power on the left, flip it, and make that the power on the right.

What is $8^{\frac{2}{3}}$ you ask? That is the same as $\sqrt[3]{8^{2}}=\sqrt[3]{64}=4$

Finish?