Thread: 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 -
The problem statement, all variables and given/known data
2. Solve -
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.

Let's look at these one at a time.



What happens if you multiply both sides through by ?

i first multiplied the two paranthesis together.. is that the right step?

Not really, you're just making things more difficult.

You have



And mulitplying through by



Gives



Now you can cancel some terms



Which takes the pronumerals out of each denominator.

Now you can do some expanding in each term.

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:

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?

I don't really know what to do next? can you please show me?

Before I continue, just letting you know I have not checked this is correct. Let's continue like it is.



Group some like terms



Then make equal to zero.



Now you need to employ the factor theorem, but that could be messy.

I would use technology to solve it.

you mean graph it on a graphing calculator?
i started doing the 2nd problem... this is what i got:
LCD = 2x+2



is this correct?

In the second term the x+1 should be cancelled from the denominator

how come you can take out the 2 from the paranthesis?
this is how i did it:



can you check if this is correct?

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)

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?

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)...

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