# Thread: Making x the subject (HARD)

1. ## Making x the subject (HARD)

Don't get how you do this!

I am sure you don't start working it out by multiplying the bracets but i dunno!

here it is:

5(x-3) = y(4-3x)

Thanks a bubble

2. Originally Posted by Danielisew
Don't get how you do this!

I am sure you don't start working it out by multiplying the bracets but i dunno!

here it is:

5(x-3) = y(4-3x)

Thanks a bubble
You are solving for x?
$\displaystyle 5(x - 3) = y(4 - 3x)$

$\displaystyle 5x - 15 = 4y - 3xy$<-- Distribute the parentheses

$\displaystyle 5x + 3xy = 4y + 15$<-- Get all x terms on one side

$\displaystyle (5 + 3y)x = 4y + 15$<-- Factor the LHS

$\displaystyle x = \frac{4y + 15}{3y + 5}$

-Dan

3. Originally Posted by Danielisew
Don't get how you do this!

I am sure you don't start working it out by multiplying the bracets but i dunno!

here it is:

5(x-3) = y(4-3x)

Thanks a bubble
what did you try?

$\displaystyle 5(x - 3) = y(4 - 3x)$ ...........expand

$\displaystyle \Rightarrow 5x - 15 = 4y - 3xy$.....now get all the terms with $\displaystyle x$'s on one side

$\displaystyle \Rightarrow 5x + 3xy = 4y + 15$

Can you take it from here?

EDIT: Oh, well, Dan finished the problem already, forget it

4. Thanks both of you..
But how can you just get x by itself when you factorised it? Technically, isn't it still multiplying the bracets, so how did you just take it by itself and make it the subject of what was there ..

Thanks

5. Originally Posted by Danielisew
Thanks both of you..
But how can you just get x by itself when you factorised it? Technically, isn't it still multiplying the bracets, so how did you just take it by itself and make it the subject of what was there ..

Thanks
maybe i'm answering the wrong question, but once you factor out the x, you just divide both sides by what is multiplying the x. this gets the x by itself

This is my 2222 post! It's not a multiple of 100, but i think it looks cool nonetheless

6. lol nice

Thanks a lot