Math Help - Solving Rational Equations

1. Solving Rational Equations

Please help me understand how to solve rational equations. I've always had trouble with solving these. I do problems thinking that I would get better but every time there is a new problem that just gets me stuck...
The problem statement, all variables and given/known data
1. Solve - $\frac{7}{x^{2}-5} -\frac{ 2}{x-5} = \frac{4}{x}$
The problem statement, all variables and given/known data
2. Solve - $\frac{2x-3}{2} + \frac{5x}{x+1} = x$
The problem statement, all variables and given/known data
3. A man sold a car for 1000$which is 25% more than what he had paid for the car. How much did the man pay for the car? I also have a question where can I find a good book(s) that will contain problems, rational equations, percents, radicals, graphs, etc... and will in detail show you how to do each problem... Starting from the simplest to the hardest...I need to study for an exam and a lot of the material I have forgotten how to do. 2. Re: Solving Rational Equations Let's look at these one at a time. $\displaystyle \frac{7}{x^{2}-5} -\frac{ 2}{x-5} = \frac{4}{x}$ What happens if you multiply both sides through by $\displaystyle x(x^{2}-5)(x-5)$ ? 3. Re: Solving Rational Equations i first multiplied the two paranthesis together.. is that the right step? $x(x^{3}-5x^{2}-5x+25)$ 4. Re: Solving Rational Equations Not really, you're just making things more difficult. You have $\displaystyle \frac{7}{x^{2}-5} -\frac{ 2}{x-5} = \frac{4}{x}$ And mulitplying through by $\displaystyle x(x^{2}-5)(x-5)$ Gives $\displaystyle \frac{7x(x^{2}-5)(x-5)}{x^{2}-5} -\frac{ 2x(x^{2}-5)(x-5)}{x-5} = \frac{4x(x^{2}-5)(x-5)}{x}$ Now you can cancel some terms $\displaystyle 7x(x-5) - 2x(x^{2}-5) = 4(x^{2}-5)(x-5)$ Which takes the pronumerals out of each denominator. Now you can do some expanding in each term. 5. Re: Solving Rational Equations so you found the least common denominator (LCD) by looking at each denominator of the fraction and then you multiplied each numerator by that LCD? then the denominator of the fractions cancels with something in the top. So, when i multiplied through i got this: $7x^{2}-35x-2x^{3}+10x = 4x^{3}-20x^{2}-20x+100$ 6. Re: Solving Rational Equations Originally Posted by Aravsion so you found the least common denominator (LCD) by looking at each denominator of the fraction and then you multiplied each numerator by that LCD? Correct Now you have a cubic to solve. Do you know how? 7. Re: Solving Rational Equations I don't really know what to do next? can you please show me? 8. Re: Solving Rational Equations Before I continue, just letting you know I have not checked this is correct. Let's continue like it is. $\displaystyle 7x^{2}-35x-2x^{3}+10x = 4x^{3}-20x^{2}-20x+100$ Group some like terms $\displaystyle 7x^{2}-25x-2x^{3} = 4x^{3}-20x^{2}-20x+100$ Then make equal to zero. $\displaystyle 0 = 6x^{3}-27x^{2}+5x+100$ Now you need to employ the factor theorem, but that could be messy. I would use technology to solve it. 9. Re: Solving Rational Equations you mean graph it on a graphing calculator? i started doing the 2nd problem... this is what i got: LCD = 2x+2 $\frac{(2x-3)}{2} + \frac{5x}{(x+1)} = \frac{x}{1}$ $\frac{(2x-3)(2x+2)}{2} + \frac{5x(2x+2)}{x+1} = \frac{x(2x+2)}{1}$ $\frac{4x^{2}+4x-6x-6}{2} + \frac{10x^{2}+10x}{x+1} = \frac{2x^{2}+2x}{1}$ is this correct? 10. Re: Solving Rational Equations In the second term the x+1 should be cancelled from the denominator $\displaystyle \frac{5x(2x+2)}{x+1} = \frac{5x\times 2(x+1)}{x+1} = 10x$ 11. Re: Solving Rational Equations how come you can take out the 2 from the paranthesis? this is how i did it: $\frac{4x^{2}+4x-6x-6}{2} + \frac{10x^{2}+10x}{x+1} = \frac{2x^{2}+2x}{1}$ $\frac{4x^{2}-2x-6}{2} + \frac{10x(x+1)}{(x+1)} = \frac{2x(x+1)}{1}$ $\frac{(2x+2)(2x-3)}{2} + \frac{10x}{1} = \frac{2x(x+1)}{1}$ can you check if this is correct? 12. Re: Solving Rational Equations Originally Posted by Aravsion how come you can take out the 2 from the paranthesis? This is called factorisation, you should have a strong understanding of this before attempting this type of problems. Remember that if a(b+c) = ab+ac then the reverse is true i.e ab+ac = a(b+c) 13. Re: Solving Rational Equations Based on the mis-matched parentheses, I'm going with "no. Based on the different solutions compared to where it started, I's say you wandered off a bit. Please be more careful. By the way, start a new thread when you start a new problem. Note: You should also worry about Domain issues. In the original problem statement x = 1 is NOT and will not be a solution, no matter what else happens. Do you see why? 14. Re: Solving Rational Equations where did i mess up though? In the original problem statement x - 1 is NOT and will not be a solution, no matter what else happens. Do you see why? yeah, because if x = -1 and in the denominator the factor is (x+1) that will make the dividing by zero (undefined)... 15. Re: Solving Rational Equations Originally Posted by Aravsion 3. A man sold a car for 1000$ which is 25% more than what he had paid for the car. How much did the man pay for the car?
Lets call x the amount he paid for it.

Therefore 1.25x = 1000

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