Question 3754
multiply both sides by 5t-12, so that the 5t-12 cancels with the 5t-12 on the denominator on the right hand side. And then multiply both sides by t, so that the t on the left hand side will cancel with the denominator. This is basically what people call cross-multiplying:


{{{(2/t) = ((t)/(5t-12))}}}


{{{(2(5t-12)/t) = ((t(5t-12))/(5t-12))}}}


{{{(2(5t-12)/t) = t}}} and then:


{{{(2t(5t-12)/t) = t^2}}}


{{{2(5t-12) = t^2}}}. You should be able to get from your question to this line in one line - i have just shown the whole process.


Now expand and re-order to get a quadratic:


{{{10t-24 = t^2}}}
{{{t^2-10t+24 = 0}}}


Now look to factorise this. If you cannot see the factors in there, then you will have to use the quadratic formula. I can see the factors though:


(x-6)(x-4) = 0


so now, either x-6=0 or x-4=0


so, x=6 or x=4


jon.