# Thread: how to solve for X

1. ## how to solve for X

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Can someone please tell how do you solve this i don't even know how to even start. I would really appreciate it.

2. " i don't even know how to even start"

Ah, my favorite words. If this is true, you should not be in this class and you should not have been given this assignment.

The idea is to use the same techniques you used to solve the 3000 other problems you have been given before you got to this point. Get x alone - simple as that.

Y = (b+d)/(1-x)

1) Notice that x = 1 would be bad. Why?
2) x in the denominator is no good. How do you get it out? Multiply by 1-x

Y(1-x) = (b+d)

3) Pesky Y is hanging around, attached by multiplication. Divide by it.

(1-x) = (b+d)/Y

You should notice that Y = 0 would be bad, as well. Why?

4) Simplfy

1-x = (b+d)/Y

5) Pesky 1 is hanging around, attached by addition. Subtract it.

-x = [(b+d)/Y]-1

5) Pesky "-" is hanging around, attached by multiplication. Divide by it.

x = -{[(b+d)/Y]-1}

6) Simplify

x = -{[(b+d)/Y]-1} = -[(b+d)/Y]+1 = 1-[(b+d)/Y]

That's about it. You must see that nothing went on here that you have not previously seen. Things are not where we want them, so get them there using appropriate arithmetic.

You do the next one and let's see what you get.

3. so for the second one i get bx+b3-bd=ax-a2+ac.

I am not sure what do to next can i substract the ax-a2+ac from the other side or would that be wrong.

4. Originally Posted by abc
so for the second one i get bx+b3-bd=ax-a2+ac.

I am not sure what do to next can i substract the ax-a2+ac from the other side or would that be wrong.
Wrong!
The equation is
$\displaystyle \frac{x+3}{a} - d = \frac{x-2}{b} + c$
not $\displaystyle \frac{x+3-d}{a} = \frac{x-2+c}{b}$

5. Actually i think the last working of mine is wrong.
Isn't is bx-ax=Ac-2a-3b+bd but now can i divide both sides by b and then by a to get rid of the a and b on the x side.

6. Originally Posted by Isomorphism
Wrong!
The equation is
$\displaystyle \frac{x+3}{a} - d = \frac{x-2}{b} + c$
not $\displaystyle \frac{x+3-d}{a} = \frac{x-2+c}{b}$
Okay thanks so would i start off like this then:

[bx+b3/ba]-bd = ax-a2/ab+ac

7. Originally Posted by abc
Actually i think the last working of mine is wrong.
Isn't is bx-ax=Ac-2a-3b+bd but now can i divide both sides by b and then by a to get rid of the a and b on the x side.
First get rid of the denominator terms >CAREFULLY<
You are missing the additive terms -d and c .

Let me do the first step (You can do it many different ways though)
$\displaystyle \frac{x+3}{a} - d = \frac{x-2}{b} + c$
Hammer this equation by multiplying a throughout, followed by b.
$\displaystyle bx + 3b - abd = ax - 2a + abc$

8. Originally Posted by abc
Okay thanks so would i start off like this then:

[bx+b3/ba]-bd = (ax-a2)/ab+ac
Careful, the ones marked in red should not be there.

Question to ponder:
ok say $\displaystyle \frac{x}{y} = \frac{a}{b}$

Now if I say multiply throughout by b, what will you get?

9. Originally Posted by Isomorphism
Careful, the ones marked in red should not be there.

Question to ponder:
ok say $\displaystyle \frac{x}{y} = \frac{a}{b}$

Now if I say multiply throughout by b, what will you get?
bx/by=ba/bb

10. Hammer this equation by multiplying a throughout, followed by b.

Okay this is helpful didn't know this. So now would the next stage be
bx-ax=abd+abc-3b-2a

11. Originally Posted by abc
bx/by=ba/bb
I thought you would say that.
What you did is not wrong, but not very useful!
Multiplying above and below as well generally does not help when there is equal to sign. It is more fruitful to multiply on both sides of equal to.
$\displaystyle \frac{x}{y} = \frac{a}{b}$
Mostly when people say multiply throughout by 'b', they mean
$\displaystyle \frac{bx}{y} = \frac{ba}{b} = a$

12. Originally Posted by abc
Okay this is helpful didn't know this. So now would the next stage be
bx-ax=abd+abc-3b-2a
Yes! You are on the right track

Notice that on the left hand side, x is common. So taking x common out we get....??

13. Originally Posted by Isomorphism
Yes! You are on the right track

Notice that on the left hand side, x is common. So taking x common out we get....??
don't you factorise it. So X(a-b). If this is correct would the next step be divide a-b from both sides so X = abd +abc-3b-2a / a-b

14. Originally Posted by abc
don't you factorise it. So X(a-b). If this is correct would the next step be divide a-b from both sides so X = abd +abc-3b-2a / a-b
Yes! You are done!
Congrats

let me document it clearly
$\displaystyle x = \frac{abd +abc-3b-2a}{a-b}$

15. Originally Posted by abc
so for the second one i get bx+b3-bd=ax-a2+ac.

I am not sure what do to next can i substract the ax-a2+ac from the other side or would that be wrong.