Rational equations

FMMurphy

Junior Member
Joined
Mar 12, 2006
Messages
51
The problem is x/x-5 (first FRACTION) +4 = 1/x+3 (2nd FRACTION)
Is x=4 a solution
 
\(\displaystyle \L\frac{x}{x-5} + \frac{4}{1} = \frac{1}{x+3}\)

To add fractions you need a common denominator...so:

\(\displaystyle \L\frac{x(1)}{x-5} + \frac{4(x-5)}{(x-5)} = \frac{1}{x+3}\)

\(\displaystyle \L\frac{x + 4x-20}{x-5} = \frac{1}{x+3}\)

\(\displaystyle \L\frac{5x-20}{x-5} = \frac{1}{x+3}\)

Cross multiply:

\(\displaystyle \L 1(x-5) = (x+3)(5x-20)\)

FOIL the right side.

\(\displaystyle \L x-5 = 5x^2-5x-60\)

Combine both sides (set the quadratic equal to zero).

\(\displaystyle \L 5x^2-6x-55 = 0\)

Now solve your new polynomial for it's two solutions......

(you can also plug in x = 4 into f(x) = 5x^2-6x-55 to see if it's a zero)


Best of luck!
 
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