How do you evaluate #\frac { 20} { x ^ { 2} - 8x } - \frac { x - 10} { x ^ { 2} + 8x } = \frac { 36} { x ^ { 2} - 64}#?

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Answer 1 · 2017-12-13T14:35:04

#x=10#

#20/(x^2-8x)-(x-10)/(x^2+8x)=36/(x^2-64)#

#[20*(x+8)-(x-10)*(x-8)]/(x^3-64x)=(36x)/(x^3-64x)#

#20*(x+8)-(x-10)*(x-8)=36x#

#20x+160-(x^2-18x+80)=36x#

#36x+x^2-18x+80-20x-160=0#

#x^2-2x-80=0#

#(x+8)*(x-10)=0#

Hence #x_1=-8# and #x_2=10#

However, #x=-8# don't provide a solution due to it undefines original equation. Thus, solution of it, #x=10#