- Thread starter vindiou
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The appearance seems a bit pointless. Can you solve \(\displaystyle 0.45 = \dfrac{a - xH}{a+x(J-H)}\), for x?Hi everyone,

0.45 = (a - x(bcd + bce + bf)) / (a + x(bcg - bcd -bce -bf))

x = ?

I'm stuck and I need help! (I'm a 56 yo guy and school classes are very far...)

Thanks!

Thanks, I solved it, x = 0.55a / (0.45bcg + 0.55b(cd + ce + f)), merry xmasThe appearance seems a bit pointless. Can you solve \(\displaystyle 0.45 = \dfrac{a - xH}{a+x(J-H)}\), for x?

Ahh, I do not think it pointless at all. Finding the relatively simple in the seemingly complex is one of the skills that math teaches you.The appearance seems a bit pointless. Can you solve \(\displaystyle 0.45 = \dfrac{a - xH}{a+x(J-H)}\), for x?

Which obviously entails that it cannot be pointless.Well, Sir Vindiou he no seem too interested...edited his post to a point!!

Since you're solving for x, then the rest is assumed to be givens; so simplify:0.45 = (a - x(bcd + bce + bf)) / (a + x(bcg - bcd -bce -bf))

x = ?

u = bcd + bce + bf

v = (bcg - bcd -bce -bf

Equation becomes:

(a - xu) / (a + xv) = 9/20

Let us know if yer still alive Sir V !!