If 4t/3x = 2/3y + 2/5y and xy is not equal to 0, is t ...

[steilbergauf]

Hi guys,

im struggling with the following problem:

If 4t/3x = 2/3y + 2/5y and xy is not equal to 0, is t equal to 1?

(1) y = 5

(2) x = 5y/4


I dont understand how I can get the common denominator 15xy in order te duce the fraction?! Can you help please?

Thanks in advance!

[RonPurewal]

Well, you know that the common denominator is 15xy, so at least that's a big step.

Do you know how to make common denominators with integers? e.g., if you had x/3 + y/4 = 5/6, would you know how to make everything have a denominator of 12?
If so, it's the same procedure here. There's an "x" in there making faces at you, but the procedure is analogous.

[steilbergauf]

Hi Ron,

thanks for your quick reply.

It was kinda obvious this morning, so I guess it was just late yesterday.

However, I think my confusion derived partly from problems in which you cant find a common denominator so easily.

Lets say:

4t/3x = 2/11y + 2/13y

In order to find a common denominator, I would have to multiply as following:

4t/3x(11y/11y)(13y/13y) = 2/11y(3x/3x)(13y/13y) + 2/13(3x/3x)(11y/11y)

Is my thinking correct?

Thanks for your help in advance!

[RonPurewal]

Correct except for the extra y's. You don't need two y's.

So you'd multiply the left-hand part by just 11, 13, and y (*not* 11y AND 13y).

The middle one you'd multiply by x, 3, and 13, but not y. Likewise with the last one -- x, 3, and 11, but not y.

[shradhagrover18]

Hi Ron,

I was wondering if we could take the LCM of 5y and 3y which i thought will be 15 y. with this approach we end up eliminating y from the equation and to know the value of t, we only need the value of x. However using this method i am getting the incorrect answer. Can you help me understand why exactly we cant do this ?

[RonPurewal]

how would you "end up eliminating y from the equation"?

if you add 2/(3y) + 2/(5y), that becomes
10/(15y) + 6/(15y)
= 16/(15y)

the "y" is very much still there. what would make you think you could eliminate it?