1. ## algebra Qs

hi people, im working on rearranging formulae at the mo, and im tuck on two questions. you couldnt give me a hand could you???

1. make y the subject:
2/y = x-3y/2y

2. make x the subject:
4a = b-3c/c

could you take me through the steps to do these. im sure its all very logical!!

ryan xx

2. Originally Posted by diablo
1. make y the subject:
2/y = x-3y/2y
You mean "solve for y"? Also, please use parenthesis. This could mean
2/y = x - (3y)/(2y)
2/y = (x - 3y)/(2y)
and a few other things.

I will assume your equation is:
$\frac{2}{y} = \frac{x - 2y}{2y}$ Multiply both sides by 2y:

$(2y) \frac{2}{y} = (2y)\frac{x - 2y}{2y}$

$2 \cdot 2 = x - 2y$ Add 2y to both sides:

$2y + 4 = x - 2y + 2y$

$2y + 4 = x$ Add -4 to both sides:

$2y + 4 - 4 = x - 4$

$2y = x - 4$ Divide both sides by 2:

$\frac{2y}{2} = \frac{x - 4}{2}$

$y = \frac{x-4}{2}$

(I should mention that we can't let x = 4 in this expression, since that would make y = 0 and the original expression forbids this.)

Originally Posted by diablo
2. make x the subject:
4a = b-3c/c
This equation has no x in it!!

-Dan

3. thanks for the help topsquark. one thing i dont understand is the first step you did, you multiplied both sides by 2y and on the left side, (2y)2/y became 2 . 2 what does this mean??

sorry about the 2nd equation. it was 'make c the subject'. if u got a mo, could u help me out with that.

top quality stuff,
ryan

4. Originally Posted by diablo
thanks for the help topsquark. one thing i dont understand is the first step you did, you multiplied both sides by 2y and on the left side, (2y)2/y became 2 . 2 what does this mean??
That step in detail then:

$(2y) \frac{2}{y} = \frac{(2y) \cdot 2}{y} = \frac{4y}{y} = 4$

-Dan

5. Originally Posted by diablo
2. make c the subject:
4a = b-3c/c
Again, I must make an assumption as to what you mean. Is this
4a = (b - 3c)/c ?

Actually it is very much like the other one.

$4a = \frac{b - 3c}{c}$ Multiply both sides by c:

$c \cdot 4a = c \cdot \frac{b - 3c}{c}$

$4ac = b - 3c$ Add 3c to both sides:

$4ac + 3c = b - 3c + 3c$

$4ac + 3c = b$ New trick. We're going to factor the common "c" from the RHS:

$(4a+3)c = b$ Now divide both sides by 4a + 3:

$\frac{(4a+3)c}{4a+3} = \frac{b}{4a+3}$

$c = \frac{b}{4a+3}$

(For this one we ought to mention that we can't have b = 0, or a = -3/4 either.)

-Dan

6. good stuff, thanks a lot for all the help!!!