# Math Help - Solving for x in terms of other letters (pls help)

1. ## Solving for x in terms of other letters (pls help)

My problem is:

Solve for x: ax + b = bx - a

I found a similar problem in the book, and wondered if I'm on the right track or not.

I rewrote:

ax - bx = -a - b

x(a - b) = -a - b

Code:

x = -a – b
-------
a – b
x = -1 ?

I guess because each letter is only equal to 1 my only possible answer is then 1? Is that how we get this answer? I'm only guessing...

2. Originally Posted by lilrhino
My problem is:

Solve for x: ax + b = bx - a

I found a similar problem in the book, and wondered if I'm on the right track or not.

I rewrote:

ax - bx = -a - b

x(a - b) = -a - b

Code:

x = -a – b
-------
a – b
x = -1 ?

I guess because each letter is only equal to 1 my only possible answer is then 1? Is that how we get this answer? I'm only guessing...
no. the answer is $x = \frac {-a - b}{a - b}$ as you have. this is not -1. to have -1 we would need it to be: $x = \frac {- (a - b)}{a - b} = \frac {b - a}{a - b}$

3. Originally Posted by Jhevon
no. the answer is $x = \frac {-a - b}{a - b}$ as you have. this is not -1. to have -1 we would need it to be: $x = \frac {- (a - b)}{a - b} = \frac {b - a}{a - b}$
I thought it was supposed to equal a number not the resulting fraction. It's a little confusing anyway. I'm not sure why I was trying to make x equal a number .

Can you please explain how this one equals 1 and the other one results to a fraction?

ax + b = bx + a

Thanks again! Just trying to understand this, so hopefully I'll retain it.

4. Originally Posted by lilrhino
I thought it was supposed to equal a number not the resulting fraction. It's a little confusing anyway. I'm not sure why I was trying to make x equal a number .

Can you please explain how this one equals 1 and the other one results to a fraction?

ax + b = bx + a

Thanks again! Just trying to understand this, so hopefully I'll retain it.
ok, so there were no instructions that said $x$ must be a number. it said we should simply solve for $x$ in terms of the other letters.

Now recall, a fraction is equal to 1 if the numerator and denomenator are equal--and of course the denominator can't be zero. We get -1 if the numerator is the negative of the denominator, which was the case in the example i choose that we would need to get -1.

$x = \frac {-(a - b)}{a - b} =-1 \times \frac {a - b}{a - b} = -1 \times 1 = -1$

Now on to the next question:

$ax + b = bx + a$ ..........get all the $x$'s on one side

$\Rightarrow ax - bx = a - b$ ........now factor out the common $x$

$\Rightarrow x(a - b) = a - b$ .......now divide both sides by $a - b$

$\Rightarrow x = \frac {a - b}{a - b}$

we have a number divided by itself, provided $a \neq b$ we will get 1

So, $x = 1$

5. Originally Posted by Jhevon
ok, so there were no instructions that said $x$ must be a number. it said we should simply solve for $x$ in terms of the other letters.

Now recall, a fraction is equal to 1 if the numerator and denomenator are equal--and of course the denominator can't be zero. We get -1 if the numerator is the negative of the denominator, which was the case in the example i choose that we would need to get -1.

$x = \frac {-(a - b)}{a - b} =-1 \times \frac {a - b}{a - b} = -1 \times 1 = -1$

Now on to the next question:

$ax + b = bx + a$ ..........get all the $x$'s on one side

$\Rightarrow ax - bx = a - b$ ........now factor out the common $x$

$\Rightarrow x(a - b) = a - b$ .......now divide both sides by $a - b$

$\Rightarrow x = \frac {a - b}{a - b}$

we have a number divided by itself, provided $a \neq b$ we will get 1

So, $x = 1$
Thanks for the explanation Jhevon!