Question

Solve for x. c-1/x + d= cd

Answer 1

\(\dfrac{c-1}x+d=cd\)?

Answer 2

yes

Answer 3

multiply both sides by x first off

Answer 4

\[\huge \frac{ c-1 }{ x } = cd - d\]
\[\huge \frac{ c-1 }{ cd-d}= x\]

Answer 5

c-1+xd=cxd
c-1=cxd-xd
c-1=x(cd-d)
x=(c-1)/(cd-d)

Answer 6

\[\begin{array}{rl}\frac{c-1}x+d&=&cd\\c-1+xd&=&cxd\\cxd-xd&=&c-1\\x(cd-d)&=&c-1\\x&=&\frac{c-1}{xd-d}\end{array}\]

Answer 7

kc_kennylau you have posted the best answer but you shouldnt leave steps out

Answer 8

it makes it harder for the original poster to follow your solution, never make assumptions

Answer 9

would it be C over D?

Answer 10

sorry i meant \(x=\dfrac{c-1}{cd-d}\)

Answer 11

\[\frac{c-1}{x} + d = cd\]
\[x(\frac{c-1}{x} + d) = x(cd)\]
\[\frac{x}{1}(\frac{c-1}{x} + d) = x(cd)\]
\[\frac{x*d}{1} + \frac{(c-1)x}{1*x} = xcd\]
\[x*d + \frac{(c-1)x}{x} = xcd\]
\[x*d + (c-1)*1 = xcd\]
\[x*d + (c-1) = xcd\]
\[(c-1) = xcd - xd\]
\[(c-1) = (cd - d)x\]
\[\frac{(c-1)}{(cd - d)} = \frac{(cd - d)x}{(cd - d)}\]
\[\frac{(c-1)}{(cd - d)} = 1*x\]
\[\frac{(c-1)}{(cd - d)} =x\]

Answer 12

remember,

\[\frac{x}{1} = x\]

Answer 13

you can divide anything by 1 and it wont change its value

Answer 14

wow @Australopithecus you have done a great job

Answer 15

did you type out all the \(\LaTeX\) yourself or you used the equation button

Answer 16

well I did leave some steps out

Answer 17

I used the equation button

Answer 18

Thanks you guys helped me out a lot.

Answer 19

like I should have written

\[\frac{d}{1}\]
instead of d

Answer 20

but I think he got it at least I hope these are fundamental rules