How do you multiply #8/(x-4) -3=1/(x-10)#?

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Answer 1 · 2018-07-10T17:34:49

#x=28/3# or #x=7#

we have to solve

#8/(x-4)-3=1/(x-10)#
simplifying the left-Hand side:

#8/(x-4)-3=(8-3(x-4))/(x-4)=(8-3x+12)/(x-4)=(20-3x)/(x-4)#
so we have

#(20-3x)/(x-4)=1/(x-10)#
by cross multiplication we get

#(20-3x)(x-10)=x-4#

multiplying out we get

#20x-3x^2-200+30x=x-4#

simplifying we have

#-3x^2+49x-196=0#
dividing by #-3#

#x^2-49/3+196/3=0#

#x_(1,2)=49/6pmsqrt((49/6)^2-196/3)#

note that

#(49/6)^2-196/3=49/36#

so we get

#x_1=28/3#

#x_2=7#