Algebra help, Rational Equations

In summary: I was getting a little lost. In summary, the student is trying to solve a problem involving rational expressions, but is having trouble. He has been trying different methods, but has not been getting anywhere. He has been helped by the other participants, and is close to solving the problem.
  • #1
Mhorton91
42
1

Homework Statement



Solving formulas involving Rational expressions
for each excersise, solve for the indicated variable

Homework Equations



(x-y)/xy = z

Solve for x

The Attempt at a Solution



I started by multiplying both sides by xy, to which I got

(x-y) = xyz

Then I tried multiplying both sides by yz

(x-y)/yz = x

And from there I just get stuck, I have tried everything I can think of, and I keep losing variables... I have done a bunch of other problems of the same type... and for some reason this one (the last one) has had me stuck for the past 45 minutes.

Also, according to the back of the book, the answer is

x = y/1-yz or x = -y/yz-1

and even having the answers I can't figure out how to get there...
Thank everyone for any help, I feel like I'm going to smack myself when I finally do figure it out.Marshall.
 
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  • #2
Hint: put terms with "x" one the one side of equation and terms without "x" on the other
 
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  • #3
Thank you, I tried that. However, after I get:

(x-y)/yz = x

I can't seem to move the (x-y) without losing a variable somewhere, or, I could try backing up to step 2 and going to:

(x-y)/x = yz


Are either of these methods on the right track?



EDIT

Ok, I tried something else, which I feel is getting me closer,

I started here as usual:

x-y = (x)(y)(z)

Then I moved the x from the left, over to the right, and the (y)(z) from the right to the left, giving me:

-y/(y)(z) = x(-x)


But then again, this could be moving in the wrong direction...
 
Last edited:
  • #4
szynkasz said:
Hint: put terms with "x" one the one side of equation and terms without "x" on the other

Mhorton91 said:
Thank you, I tried that. However, after I get:

(x-y)/yz = x

I can't seem to move the (x-y) without losing a variable somewhere, or, I could try backing up to step 2 and going to:

(x-y)/x = yz


Are either of these methods on the right track?



EDIT

Ok, I tried something else, which I feel is getting me closer,

I started here as usual:

x-y = (x)(y)(z)

Then I moved the x from the left, over to the right, and the (y)(z) from the right to the left, giving me:

-y/(y)(z) = x(-x)


But then again, this could be moving in the wrong direction...

You are on the right track but going off-track. You must concentrate on TERMS OF x, and get them all on one side, and everything else on the other side, and then you can finish in one step. Look back again on szynkasz's suggestion. That is really all you need.
 
  • #5
Mhorton91, how did you get:

##-\frac{y}{yz}=x-x##

from:

##-y=xyz-x##

You can't divide one term on the right side by ##yz## and leave the other unchanged.
 
  • #6
szynkasz said:
Mhorton91, how did you get:

##-\frac{y}{yz}=x-x##

from:

##-y=xyz-x##

You can't divide one term on the right side by ##yz## and leave the other unchanged.

when I had -y = (x)(y)(z)-x I divided both sides by (y)(z)

Which gave me the -y/(y)(z) = (x) -x... is that not correct?
 
  • #7
Got it!

x-y/xy = z

x-y = xyz

-y = (xyz - x)

-y = x(yz-1)

-y/(yz-1) = xThank you for the hints!
 

Related to Algebra help, Rational Equations

What is a rational equation?

A rational equation is an equation that contains one or more rational expressions. A rational expression is a fraction in which the numerator and denominator are polynomials.

How do I solve rational equations?

To solve a rational equation, follow these steps:

  1. Clear the equation of fractions by multiplying both sides by the least common denominator.
  2. Simplify the resulting equation by combining like terms.
  3. Use algebraic methods to solve for the variable.
  4. Check your solution by plugging it back into the original equation.

What is the common mistake when solving rational equations?

The most common mistake when solving rational equations is forgetting to check for extraneous solutions. These are solutions that satisfy the final equation, but do not satisfy the original equation.

When should I use the LCD method to solve rational equations?

The LCD method should be used when the equation contains two or more fractions with different denominators. It simplifies the equation and makes it easier to solve.

What are some real-life applications of rational equations?

Rational equations are commonly used in fields such as physics, engineering, and economics. They can be used to calculate rates, proportions, and proportions. For example, they can be used to calculate how much medicine to give a patient based on their weight and the prescribed dosage.

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