Solving for X in a Fractional Equation

  • Thread starter rede96
  • Start date
In summary, the conversation is about solving for x in the formula X ÷ (X + Y) = Z when given values for y and z. The solution involves multiplying both sides of the equation by (x+y) and then factoring out x from the left side. The final solution is X = (Y / (1-Z)) - Y, and the conversation also includes a recommendation to use a website like Wolfram Alpha to check the answer and show the steps.
  • #1
rede96
663
16
Hi,

I should know this but havn't done any transposition of formula for a long, long while :)

If I know y and z, how can I solve for x in the following formula:

X ÷ (X + Y) = Z

Thanks.
 
Last edited:
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  • #2
hi rede96! :smile:

multiply both sides by (x+y) :wink:
 
  • #3
tiny-tim said:
hi rede96! :smile:

multiply both sides by (x+y) :wink:

Thanks tiny-tim.

So I would get X = Z (X + Y) or X = ZX + ZY

But as I am feeling very mathematically challenged today, :smile: can you demonstrate solving for x when y = 28 and z = 0.3 please.
 
  • #4
Step 2: Subtract ZX from both sides of the equation.
 
  • #5
HallsofIvy said:
Step 2: Subtract ZX from both sides of the equation.

Ok, so now I have X - ZX = ZY.

I assume that there are a few more steps but I'll be dammed if I can remember. :blushing:

From memory I know I need to remove the Z from the ZX on left hand side of the equation, but can't figure out how.

Rather than go through this one step at a time, I would very much appreciate it if someone could give me the answer then I can work backwards.

Just by playing with the numbers I figured out that X = (Y / (1-Z)) - Y but have not figured out yet how to get there!
 
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  • #6
Step 3: Factor x from the two terms on the left side.
I'm hopeful you can figure out step 4 (the last step) on your own.

rede96 said:
I assume that there are a few more steps but I'll be dammed if I can remember.
Does that mean you are a river? :biggrin:
 
  • #7
Mark44 said:
Step 3: Factor x from the two terms on the left side.
I'm hopeful you can figure out step 4 (the last step) on your own.

Does that mean you are a river? :biggrin:


Well if you mean do I feel that I am about to burst then yes! :biggrin: So either dammed or damned would do!

Thanks for the help, still not figured it but am really busy at the mo, so will have a look later.

Red.
 
  • #8
When you are done, you can use http://www.wolframalpha.com" to check your answer. You can type in things like "x/(x+y)=z, solve for x". It will also show you the steps in doing so if you choose "show steps".
 
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  • #9
KingNothing said:
When you are done, you can use http://www.wolframalpha.com" to check your answer. You can type in things like "x/(x+y)=z, solve for x". It will also show you the steps in doing so if you choose "show steps".

Cool, thanks. :biggrin: Just wish I could remember all this stuff!
 
Last edited by a moderator:
  • #10
rede96 said:
Ok, so now I have X - ZX = ZY.

Hint: X = 1X (one times X), so this becomes

1X - ZX = ZY

What can you do with the left-hand side now, so you end up with a single X?
 

Related to Solving for X in a Fractional Equation

1. What does "solve for x" mean in this equation?

"Solve for x" means to find the value of x that satisfies the equation. In other words, we are looking for the numerical value of x that makes the equation true.

2. Why is there a fraction in this equation?

The fraction represents a ratio between two quantities, x and x+y. By finding the value of x, we can determine the ratio between x and x+y.

3. What is the purpose of the variable z in this equation?

The variable z represents a constant value that is given in the equation. It helps to define the ratio between x and x+y.

4. Can this equation be solved for more than one value of x?

Yes, this equation can have multiple solutions for x. This is because x can take on different values that satisfy the equation as long as the ratio between x and x+y remains the same.

5. How can I solve this equation for x?

To solve for x, we can use algebraic manipulation to isolate x on one side of the equation. This typically involves performing the inverse operation of any operations that are being performed on x, such as multiplication or division. Once x is isolated, we can plug in the given value for z to find the numerical value of x.

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