Question

How do I solve this equation for x?

Answer 1

\[\frac{ Q }{ A }(1-x)-k(x-1)=0\]

Answer 2

All of the letters are just constants?

Answer 3

Yeah they are. I tried multiplying out the brackets but I am sure what do after that

Answer 4

Umm I would probably start by multiplying both sides by A, fractions are no fun.\[\Large\rm Q(1-x)-Ak(x-1)=0\]Factor a -1 out of the second term,\[\Large\rm Q(1-x)+Ak(1-x)=0\]Factor a (1-x) out of each term,\[\Large\rm (1-x)\left[Q-Ak\right]=0\]Is this method too fancy you think?
If you prefer expanding out the brackets we can do it that way instead.

Answer 5

Woops, last step should be,\[\Large\rm (1-x)\left[Q+Ak\right]=0\]

Answer 6

I'm not the best at factorising unfortunately!

Answer 7

Ya it's a little tricky :o
Did the multiplication step make sense?
After that we're looking at it like this:\[\Large\rm \left[Q\color{royalblue}{(1-x)}+Ak\color{royalblue}{(1-x)}\right]=0\]And pulling a \(\Large\rm \color{royalblue}{(1-x)}\) out of each term gives us,\[\Large\rm \color{royalblue}{(1-x)}\left[Q+Ak\right]=0\]And from there, divide each side by [Q+Ak],\[\Large\rm 1-x=0\]

Answer 8

But the 2nd term was x-1 not 1-x

Answer 9

I pulled a negative out of the brackets and it changed to addition in front.
\(\Large\rm (x-1)=-(1-x)\)

Too fancy? :d
Fine fine fine, let's see what happens when we expand instead.

Answer 10

\[\Large\rm \frac{Q}{A}(1-x)-k(x-1)=0\]
\[\Large\rm \frac{Q}{A}-\frac{Q}{A}x-kx+k=0\]Subtract the constant terms to the other side,\[\Large\rm -\frac{Q}{A}x-kx=-\frac{Q}{A}-k\]Well, taking this route, you still are left with some factoring to do :( lol
Factoring out an x gives you,\[\Large\rm x\left(-\frac{Q}{A}-k\right)=-\frac{Q}{A}-k\]

Answer 11

I understand now!

Answer 12

If I want to rearrange the equation in terms of x, I can just divide by the 2nd term, and subtract 1 from both sides

Answer 13

and then change the signs to get x=blah blah blah

Answer 14

Yah there ya go :)
Should give you a nice simple answer x=1, yes?

(Remember, when you divide the [Q+Ak] to the other side, you're dividing 0 by that quantity, so you just get zero there.

Answer 15

Yeah my original equation I posted above was only one side of a much bigger equation I am trying to solve. There was so many variables, It got confusing!

Answer 16

Ah yes I see it :o
Lot of goofy stuff to deal with!

Answer 17

NB: you can only divide by [Q+Ak] if you can assume that [Q+Ak]≠0