## Equation with 3 variables (variables in denominator)

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.
damedi07
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### Equation with 3 variables (variables in denominator)

hello, hopefully i am putting thisin the right forum. i recently took a placement test to go back to school after 20+ years. i hadn't looked at algebra in a long time and had to reteach myself some online to remember. while i did fairly well on the test one question confused me and i can't seem to get it right. hopefully someone here can explain to me how to solve the equation.

the question was:

2/x - c = 2/k
solve for x.

thank you in advance and sorry if this is in the wrong forum. since it was on a placement test i have assumed it is elementary algebra.

maggiemagnet
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Joined: Mon Dec 08, 2008 12:32 am
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### Re: Equation with 3 variables (variables in denominator)

hopefully someone here can explain to me how to solve the equation. the question was:

2/x - c = 2/k
solve for x.
When you have an equation with more than one variable (or "literal") and you're supposed to solve for one of the variables, this is called "solving a literal equation". This site has a lesson on this here. First, though, you do need to be able to solve one-variable equations, because the lessons expects you to know how to do those methods; to brush up, go here. A good first step for your equation (there are different ways to get to the answer) could be to get the x-term by itself by adding the c to the other side: 2/x = c + 2/k. Then multiply both sides by x; this cancels it off on the left side: 2 = x(c + 2/k). Then divide off the non-x stuff to get x by itself: 2/(c + 2/k) = x. They might want you to simplify this, too; it's called a "complex" fraction, and there's a lesson on that here.

damedi07
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Joined: Sun Aug 04, 2013 1:24 pm
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### Re: Equation with 3 variables (variables in denominator)

maggie thnx. i think i actually got this one now. i had found the anwer using an online algebra calculator but it didn't show the steps. after a while i figured it out.

2/x - c = 2/k

multiply by x and get:

2 - cx = 2x/k

multiply by k and get :

2k - cxk = 2x

add cxk to both sides to get all the x's on the same side

2k = 2x + cxk

factor out x

2k = x(2+ck)

divide by 2+ck and answer is:

2k/2+ck = x

so if i did it your way and ended up with 2/(c + 2/k) = x i could then times it all by k and then i get:

2(k)/ c(k) + k(2/k) and that gives me:

2k/ ck+2

is that right?

maggiemagnet
Posts: 358
Joined: Mon Dec 08, 2008 12:32 am
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Yes! Good job!