How do you solve #(a+4)/2 = (a+4)/a#?

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Answer 1 · 2016-06-07T19:21:21

2, -4.

By first cross-multiplying 7 then trasposing, the eqn. becomes,
a(a+4)-2(a+4)=0, i.e., (a+4)(a-2)=0.

Hence the soln. : a=-4, a=2.

Answer 2 · 2016-06-07T19:21:52

#a=2#

As the top values (numerators) are of the same value we could disregard them and simply write #a=2#

However, I suspect the that question poser is looking for you to use algebraic manipulation. So hear goes!

Invert both sides giving:

#2/(a+4)=a/(a+4)#

Multiply both sides by #(a+4)# giving:

#2xx(a+4)/(a+4)=axx(a+4)/(a+4)#

But #(a+4)/(a+4)=1#

#2 xx1=axx1#

#a=2#