How do you combine #\frac { 5u } { u ^ { 2} + 8u + 7} - \frac { 15} { u ^ { 2} + 9u + 14} - \frac { 2} { u ^ { 2} + 3u + 2}# into one term?

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Answer 1 · 2018-06-09T18:16:05

#(5u^2-17u-19)/((u+1)(u+2)(u+7))#

You need to find a common denominator, so factor each denominator:

#(5u)/((u+1)(u+7))-15/((u+2)(u+7))-2/((u+1)(u+2))#

So the LCD is: #(u+1)(u+2)(u+7)#

#(u+2)/(u+2)*(5u)/((u+1)(u+7))-(u+1)/(u+1)*15/((u+2)(u+7))-(u+7)/(u+7)*2/((u+1)(u+2))#

#=(5u(u+2)-15(u+1)-2(u+7))/((u+1)(u+2)(u+7))#

#=(5u^2+10-15u-15-2u-14)/((u+1)(u+2)(u+7))#

#=(5u^2-17u-19)/((u+1)(u+2)(u+7))#

Answer 2 · 2018-06-09T18:17:49

#(5u^2-7u-29)/((u+1)(u+2)(u+7))#

Note that
#u^2+8u+7=(u+1)(u+7)#
#u^2+9u+14=(u+2)(u+7)#
#u^2+3u+2=(u+1)(u+2)#

so we get

#(5u(u+2)-15(u+1)-2(u+7))/((u+1)(u+2)(u+7))#
simplifying we get

#(5u^2-7u-29)/((u+1)(u+2)(u+7))#