Algebraic Equations

[afvatcha]

Keep getting this problem wrong.

y(3x-5/2)=y

I can think of a number of different ways to solve this problem, but only one way is right. I don't know what makes that the right way.

It appears you should first divide both sides by y, so 3x-5/2=1. Then solve the rest. But why can't you first multiply by 2. So, 2y(3x-5)=2y...3x-5=1. This obviously changes the answer. Why is the first way correct?

[nitin_prakash_khanna]

I suggest you post the whole problem (and the source).

If its DS question then, there could be multiple interpretation to the eq.
like...
y(3x-5/2)=y
could mean
y(3x-5/2)-y=0
y(3x-7/2)=0
so either y=0 or x=7/6 will satisfy the above.

so please post the complete problem.

One small comment on what you mentioned below. Even if you multiply it by 2 first.
and then divide by y. it will be 6x-5=2 not 3x-5=1 as you have written.

[afvatcha]

Hi,

That is the entire problem. It's not a ds prob.

Quick question about your response. Why would it be 6x-5=y. If I multiply both sides of the equation by two wouldn't the denominator cancel. ie.(3x-5)/2=y which is 3x-5=2y...(removed the one y from the problem for clarity sake).

But most of all thanks for your help.

[pritesh.suvarna]

If the problem is
y(3x-5/2)=y

3x-5/2 = 1 (hope the qst. means 5/2 and NOT (3x-5)/2 )
(6x-5)/2 = 1 (equalise denominator)
x = 7/6

Thanks
Pritesh

[jeetdan]

y(x-5/2) = y

y(x-5/2) - y =0

y(x - 5/2-1) = 0

thus
either : y =0

or : x= 7/2

If we were given that y is positive integer we would have directly cancelled out y on both sides. But since we aren't told that y is positive we have to substract y on both sides and so on.

[Ben Ku]

y(x-5/2) = y

y(x-5/2) - y =0

y(x - 5/2-1) = 0

thus
either : y =0

or : x= 7/2

If we were given that y is positive integer we would have directly cancelled out y on both sides. But since we aren't told that y is positive we have to substract y on both sides and so on.

Jeetdan's response is correct. NEVER divide both sides by a variable unless you know it's not zero. The correct approach is to bring all terms to one side of the equation, then factor.