How do you subtract #\frac { 5x - 45} { x ^ { 2} - 81} - \frac { 9- x } { 81- x ^ { 2} }#?

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Answer 1 · 2016-12-27T04:45:11

#4/(x+9)#

#(5x - 45)/(x^2 - 81) -(9-x)/(81-x^2)#

#(5x - 45)/(x^2 - 81) - (9 -x)/-(x^2-81)#

change the sign for #(81-x^2) to -(x^2-81)#

#(5x -45)/(x^2-81) + (9-x)/(x^2-81)#

since they have a same denominators, we can simplify as

#(5x-45+9-x)/(x^2-81)#

#(4x-36)/(x^2-81)#

then we can factorize as

#(4(x-9))/((x+9)(x-9))#

= #4/(x+9)#