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