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

$x_1=28/3$

$x_2=7$

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