Thread: Algebra equation I'm trying to solve.

Hi,

I have an equation that I am trying to manipulate for a website project i am working on, but my high school maths is letting me down.

I am trying to get X on its own.

This is how far I have got ;

M*C = (X/V)-O-((X*0.119)+0.2)

(M*C)+O = (X/V)-((X*0.119)+0.2)
(M*C)+O = (X/V)-(X*0.119)-0.2
(M*C)+O+0.2 = (X/V)-(X*0.119)


Any help/pointers will be greatly appreciated!

Thanks!

Robin.

Originally Posted by gingebot

Hi,

I have an equation that I am trying to manipulate for a website project i am working on, but my high school maths is letting me down.

I am trying to get X on its own.

This is how far I have got ;

M*C = (X/V)-O-((X*0.119)+0.2)

(M*C)+O = (X/V)-((X*0.119)+0.2)
(M*C)+O = (X/V)-(X*0.119)-0.2
(M*C)+O+0.2 = (X/V)-(X*0.119)

This last step should be

(M*C)+O-0.2 = (X/V)-(X*0.119).

[EDIT]: Missed a minus sign. OP'er was correct here.

Next step:

(M*C)+O+0.2 = X(1/V-0.119).

Can you continue?

If the equation is:

Write, like you did:




...

((M*C)+O-0.2)/(1/V-0.119) = X


I think!? It's been 12 years since GCSE maths!

Thanks for the help!

Robin.

See my edited post. You were correct and I was incorrect regarding the sign of the 0.2. Yes, you would divide through by the (1/V-0.119) as you did, to find X. Note that V cannot equal 1/0.119, or else that denominator would be zero.

Thanks guys, it's worked a treat!

Thanks again for the help!

Robin.

You're certainly welcome for my contribution, whatever that was.