Question 7943
you have to write it down correctly on here, anywhere even: maths is a language and if you speak it incorrectly, then others will mis-interpret you: To get the answer you quoted, you need {{{5/(3a-9) - 3/2a + 4/(a^2-3a)}}}


{{{5/(3(a-3)) - 3/2a + 4/(a(a-3))}}}


To alter any fraction, beit a numerical or algebraic fraction, you need to scale it up. We do this by multiplying by "1", to keep the actual fraction unchanged, but we write 1 as something like a/a...


Right then, looking at all the denominators, we need to get them all the same: the LCM is 6a(a-3)...so 


first term is multiplied by {{{(2a)/(2a)}}}
second term is multiplied by {{{(3(a-3))/(3(a-3))}}}
third term is multiplied by {{{(6/6)}}}


so we then get:


{{{(10a - 3(3(a-3)) + 24)/(6a(a-3))}}}
{{{(10a - 3(3a-9) + 24)/(6a(a-3))}}}
{{{(10a - 9a + 27 + 24)/(6a(a-3))}}}
{{{(a + 51)/(6a(a-3))}}}


jon.