1. ## Simplifying

(x-a)/b-(x+c)/d=0

2. ## Re: Simplifying

Do you mean $\displaystyle \displaystyle \frac{(x - a)}{b} - \frac{(x - c)}{d} = 0$?

If so...

\displaystyle \displaystyle \begin{align*} \frac{(x - a)}{b} - \frac{(x - c)}{d} &= 0 \\ \frac{(x - a)}{b} &= \frac{(x - c)}{d} \\ d(x - a) &= b(x - c) \end{align*}

Now expand the brackets and see what you can do from there...

3. ## Re: Simplifying

Yes, except it's x+c. According to my math book, the answer is x=(ad+bc)/(d-b)

4. ## Re: Simplifying

Your maths book would be right

As Prove It says (with the amended $\displaystyle (x+c)$:

$\displaystyle [(x-a)/b]-[(x+c)/d]=0$

add $\displaystyle (x+c)/d$ to both sides

$\displaystyle (x-a)/b=(x+c)/d$

multiply top and bottom of lhs by $\displaystyle d$
multiply top and bottom of rhs by $\displaystyle b$
gives

$\displaystyle d(x-a)/bd=b(x+c)/bd$

multiplying both sides by $\displaystyle bd$ gives

$\displaystyle d(x-a)=b(x+c)$

You should be able to complete it from there

hth

btw Prove It, How do you get the horizontal line for divide?

thx

Pro

5. ## Re: Simplifying

I'm sorry, but I don't get the last bit. dx-bx=ad+bc, then divide both sides with d-b? However, I still have two x:es

6. ## Re: Simplifying

dx-bx=ad+bc, then divide both sides with d-b?
You're getting there

$\displaystyle dx-bx=ad+bc$

$\displaystyle dx-bx$ is the same as $\displaystyle x(d-b)$

dividing both sides by (d-b) gives

$\displaystyle x=(ad+bc)/(d-b)$

8. ## Re: Simplifying

Thanks alot=)

9. ## Re: Simplifying

You're welcome

10. ## Re: Simplifying

Originally Posted by cristianodoni72
(x-a)/b-(x+c)/d=0
Keep it simple; first, cross multiply:

d(x - a) = b(x + c) ; expand:

xd - ad = xb + bc ; get x's together:

xd - xb = ad + bc ; take out the x:

x(d - b ) = ad + bc ; finish off:

x = (ad + bc) / (d - b)

11. ## Re: Simplifying

Originally Posted by procyon
btw Prove It, How do you get the horizontal line for divide?

thx

Pro
$$\dfrac{ax+b}{cx+d}$$ is the same as $\displaystyle \dfrac{ax+b}{cx+d}$

12. ## Re: Simplifying

Hi e^(i*pi),

$\displaystyle \dfrac{Thank}{you}$
That's great, many thanks

Pro