Question

Solve for x:
\[\frac{ 4 }{ x+4 } - \frac{ 3 }{ x+3 } = \frac{ 2 }{ x+2 } - \frac{ 1 }{ x+1 }\]

I solved this question. Now please tell me which one is correct? Either method#1 or method#2?

Below there are two solutions.

Answer 1

Method#1
\[\frac{ 4 }{ x+4 } - \frac{ 3 }{ x+3 } = \frac{ 2 }{ x+2 } - \frac{ 1 }{ x+1 }\]
\[\frac{ 4x+12-3x-12 }{ (x+4)(x+3) } = \frac{ 2x+2-x-2 }{ (x+2)(x+1) }\]
\[\frac{ x }{ (x+4)(x+3) } = \frac{ x }{ (x+2)(x+1) }\]
\[(x+4)(x+3) = (x+2)(x+1)\]
\[x^{2}+3x+4x+12 = x ^{2}+x+2x+2\]
\[7x+12 = 3x+2\]
\[7x-3x+12-2 = 0\]
\[4x+6 = 0\]
\[2x+3 = 0\]
\[x = -\frac{ 3 }{ 2 }\] ANSWER

Answer 2

Method#2
\[\frac{ 4 }{ x+4 } - \frac{ 3 }{ x+3 } = \frac{ 2 }{ x+2 } - \frac{ 1 }{ x+1 }\]
\[\frac{ 4x+12-3x-12 }{ (x+4)(x+3) } = \frac{ 2x+2-x-2 }{ (x+2)(x+1) }\]
\[\frac{ x }{ x ^{2} +3x+4x+12} = \frac{ x }{ x ^{2} +x+2x+2}\]
\[x(x ^{2}+3x+2) = x(x ^{2}+7x+12)\]
\[3x ^{2}+2x = 7x ^{2}+12x\]
\[4x ^{2}+10x=0\]
\[2x ^{2}+5x=0\]
\[x(2x+5)=0\]
\[x = 0ANSWER\]and\[x=-\frac{ 5 }{ 2 }ANSWER\]