OpenStudy (anonymous):

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.

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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\[x=-\frac{ 5 }{ 2 }ANSWER\]

4 years ago