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OpenStudy (anonymous):
\[x ^{2}+6x-16/3x \times6x-18/x ^{2}-x-2\]
OpenStudy (anonymous):
could u draw it or write it in eqn form ?
OpenStudy (anonymous):
I don't know how to..
OpenStudy (anonymous):
\[\frac{x^2+6x-16}{3x}*\frac{6x-18}{x^3-x-2}\]
this is your question
right?
OpenStudy (anonymous):
@sakigirl
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OpenStudy (anonymous):
3x^2-9x, not just 3x. My bad
OpenStudy (anonymous):
ok
\[\frac{x^2+6x-16}{3x^2-9x}*\frac{6x-18}{x^2-x-2}\]
this is correct now?
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
ok
factor
x^2+6x-16
x^2+8x-2x-16
x(x+8)-2(x+8)
(x+8)(x-2)
now factors the
x^2-x-2
x^2-2x+x-2
x(x-2)+1(x-2)
(x-2)(x+1)
are you fallowing this?
OpenStudy (anonymous):
yes
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OpenStudy (anonymous):
so now put the factor at the respective place
\[\frac{(x+8)(x-2)}{3x^2-9x}*\frac{6x-18}{(x-2)(x+1)}\]
so the x-2 in numerator and denominator gets cancelled
OpenStudy (anonymous):
now
write 6x-18 as 6(x-3)
and
3x^2-9x as 3x(x-3)
so it becomes
\[\frac{x+8}{3x(x-3)}*\frac{6(x-3)}{x+1}\]
this time x-3 from both numerator and denominator will gets cancelled
OpenStudy (anonymous):
now it becomes
\[\frac{x+8}{3x}*\frac{6}{x+1}\]
OpenStudy (anonymous):
can you do this from here ??
OpenStudy (anonymous):
@sakigirl
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