How do you solve $(7a)/(3a+3)-5/(4a-4)=(3a)/(2a+2)$ ?

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Answer 1 · 2018-03-07T12:50:40

Solution : $a=3 and a= -0.5$

$(7a)/(3a+3)-5/(4a-4)=(3a)/(2a+2)$ or

$(7a)/(3(a+1))-5/(4(a-1))=(3a)/(2(a+1))$ or

$(7a)/(3(a+1))-(3a)/(2(a+1))=5/(4(a-1))$ or

$(14a-9a)/(6(a+1))=5/(4(a-1))$ or

$(5a)/(6(a+1))=5/(4(a-1))$ or

$a/(6(a+1))=1/(4(a-1))$ or

$4a^2-4a=6a+6 or 4a^2-10a-6=0$ or

$2a^2-5a-3=0 or 2a^2- 6a+a-3=0$ or

$2a(a-3)+1(a-3)=0 or (a-3)(2a+1)=0$

$:.a=3 and a= -1/2=-0.5$

Solution : $a=3 and a= -0.5$ [Ans]

How do you solve (7a)/(3a+3)-5/(4a-4)=(3a)/(2a+2)(7a)/(3a+3)-5/(4a-4)=(3a)/(2a+2)?