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Solve for x:
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Multiply all terms by the common denominator which is 21x(x+4)
Remember that x cannot be 0 or -4
then no solution
I got x = 3
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lest's satrt with the left hand side L.H.S=(3x+12+3x)/(3x^2+12x) then we multiply l.h.s with r.h.s to get 21x^2+84x+21x^2=30x^2+120x therefore 42x^2+84x=30x^2+120x so 12x^2=36x ................ therefore........... 12x^2-36x=0 so.... 12x(x-3)=0 therfore (X=0) Refused.............or........... x=3
\[21x(x+4)\times\frac{3}{3x}+21x(x+4)\times\frac{1}{x+4}=\frac{10}{7x}\times \times21x(x+4)\]
\[7(x+4)(3)+21x(1)=10(3)(x+4)\]
\[21x+84+21x=30x+120\]
\[42x+84=30x=120\]
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\[12x=36\]
\[x=3\]
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