1. Algebraic fraction

Hi,
We have to express $\displaystyle X$ in terms of$\displaystyle Y$.

The correct answer is $\displaystyle x = \frac{5y - 2}{y + 3}$

Multiplying both sides by $\displaystyle (5 - X)$ we get

$\displaystyle 5y - xy = 3x + 2$; $\displaystyle 5y - 2 = 3x + xy$; $\displaystyle \frac{5y - 2}{3} = x + xy$

I am going to stop there as I am on the wrong track. I would appreciate your help in finding the solution.

Cheers,
Sean.

2. Re: Algebraic fraction

Originally Posted by Seaniboy
Hi,
We have to express $\displaystyle X$ in terms of$\displaystyle Y$.

The correct answer is $\displaystyle x = \frac{5y - 2}{y + 3}$

Multiplying both sides by $\displaystyle (5 - X)$ we get

$\displaystyle 5y - xy = 3x + 2$; $\displaystyle 5y - 2 = 3x + xy$; $\displaystyle \frac{5y - 2}{3} = x + xy$

I am going to stop there as I am on the wrong track. I would appreciate your help in finding the solution.

Cheers,
Sean.
What's the original equation?

3. Re: Algebraic fraction

From your other posts I assume you started with $\displaystyle y = \frac {3x+2}{5-x}$, and want to rearrange to make x the subject. By the way - to avoid confusion iit would have been better to add your attempt to your previous thread rather than start a new one - that way we wouldn't have to guess at what the original problem was.

Originally Posted by Seaniboy
Multiplying both sides by $\displaystyle (5 - X)$ we get

$\displaystyle 5y - xy = 3x + 2$; $\displaystyle 5y - 2 = 3x + xy$;
So far so good

Originally Posted by Seaniboy
$\displaystyle \frac{5y - 2}{3} = x + xy$
No - if you divide both sides by 3 the right hand side becomes $\displaystyle \frac {3x +xy} 3 = x + \frac {xy}3$. This is not the right approach. Instead from the previous expression you can gather the x terms:

$\displaystyle 5y-2 = 3x + xy = x(3 + y)$

Now divide both sides by $\displaystyle (3+y)$ and you're done.

4. Re: Algebraic fraction

Thanks. Yes, I have regularly given myself less than a pass grade in the past.

5. Re: Algebraic fraction

Hi,
Apologies for the confusion and well deduced.
Thanks once again for your help.
Cheers,
Sean.

6. Re: Algebraic fraction

can any one help me in this type of sum.. I cn't got it.