Hello, just had a little mind blank.
Was wondering if anyone could please help me in adding the following fraction so its all under one common denominator and showing the steps taken? I just want to know how I could make those denominators even so it can all go together (I have stuff on the numerator however I will just represent it as x for now)...
[x/(a^2 - ab - b^2)] + [x/(a^2 + ab + b^2)]
Thanks in advance!
Was wondering if anyone could please help me in adding the following fraction so its all under one common denominator and showing the steps taken? I just want to know how I could make those denominators even so it can all go together (I have stuff on the numerator however I will just represent it as x for now)...
[x/(a^2 - ab - b^2)] + [x/(a^2 + ab + b^2)]
Thanks in advance!
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There is no lower common denominator then multiplying them together.
[x/(a^2 - ab - b^2)] + [x/(a^2 + ab + b^2)]
[x(a^2 + ab + b^2)/(a^2 - ab - b^2)(a^2 + ab + b^2)] + [x(a^2 - ab - b^2)/(a^2 - ab - b^2)(a^2 + ab + b^2)]
[x(a^2 + ab + b^2) + x(a^2 - ab - b^2)]/[(a^2 - ab - b^2)(a^2 + ab + b^2)]
Thankfully though the numerator mostly cancels with it self and now you get:
(2a^2 x)/[(a^2-a b-b^2)(a^2+a b+b^2)]
[x/(a^2 - ab - b^2)] + [x/(a^2 + ab + b^2)]
[x(a^2 + ab + b^2)/(a^2 - ab - b^2)(a^2 + ab + b^2)] + [x(a^2 - ab - b^2)/(a^2 - ab - b^2)(a^2 + ab + b^2)]
[x(a^2 + ab + b^2) + x(a^2 - ab - b^2)]/[(a^2 - ab - b^2)(a^2 + ab + b^2)]
Thankfully though the numerator mostly cancels with it self and now you get:
(2a^2 x)/[(a^2-a b-b^2)(a^2+a b+b^2)]
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= { x/[a²-b(a+b)] } + {x/[a²+b(a+b)]} note: (a+b) can be cancelled so it remains:
= [ x/(a²-b) ] + [x/(a²+b)]
= [ x/(a²-b) ] + [x/(a²+b)]