but did u get how i got that ?
yes,then combine denom.
OpenStudy (anonymous):
yea i saw you got that,
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hartnn (hartnn):
so finally u get (35- 2xy)/(28x^2y) ok?
OpenStudy (anonymous):
\[\frac{ x ^{2}+2x }{12x+54 }-\frac{ 3x }{ 8x+36 }\]
OpenStudy (anonymous):
would i combine the like terms on the bottom & top
hartnn (hartnn):
nopes,
first can u write 12x+54 as 6(2x+9) ??
OpenStudy (anonymous):
you can cause that would equal the original equation
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hartnn (hartnn):
so how would u write 8x+36 ?
OpenStudy (anonymous):
8 doesn't go into 36 tho
OpenStudy (anonymous):
36 doesnt go into 8 i mean
hartnn (hartnn):
what about 4 ?
36 and 8 both go into 4.
OpenStudy (anonymous):
4(2x+9)
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hartnn (hartnn):
right so u have
\(\huge\frac{ x ^{2}+2x }{12x+54 }-\frac{ 3x }{ 8x+36 }=\frac{ x ^{2}+2x }{6(2x+9) }-\frac{ 3x }{ 4(2x+9) }\)
can u figure out next step ?
OpenStudy (anonymous):
you would factor the top
hartnn (hartnn):
before that...
\(\huge\frac{ x ^{2}+2x }{6(2x+9) }-\frac{ 3x }{ 4(2x+9) }=\frac{2( x ^{2}+2x) }{12(2x+9) }-\frac{ 4(3x) }{ 12(2x+9) }\)
u see how i tried to make the denominator same ?
hartnn (hartnn):
sorry it would be 3 in the numerator of 2nd fraction
3(3x)
OpenStudy (anonymous):
i see it i see it it makes sense so then you would take away the denominator?
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hartnn (hartnn):
take away ?
since the denominator is common u combine the numerators.
(2x^2+4x-9x)\(12(2x-9))
OpenStudy (anonymous):
yea thats what i meant
OpenStudy (anonymous):
would the answer be 2x^2-5x/(12(2x-9)
hartnn (hartnn):
yup, thats correct.
(2x^2-5x)/(12(2x-9))
OpenStudy (anonymous):
awesome .. there is one more
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hartnn (hartnn):
ok, ask.
OpenStudy (anonymous):
\[\frac{ 2 }{ x ^{2}8x+15 } +\frac{ 1 }{ x ^{2} +11x+30}\]
OpenStudy (anonymous):
x^2 + 8 **
hartnn (hartnn):
can u factor both the quadratic polynomials in denom. ?
OpenStudy (anonymous):
i dont think so, im not sure
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hartnn (hartnn):
x^2+8x+15 = (x+5)(x+3)
and
x^2+11x+30=(x+5)(x+6)
ok?
\(\large\frac{ 2 }{(x+5)(x+3)} +\frac{ 1 }{(x+5)(x+6)}=\frac{ 2(x+6) }{(x+6)(x+5)(x+3)} +\frac{ (x+3) }{(x+3)(x+5)(x+6)}\)
the idea is to make the denominator same, so that u can combine numerator....
OpenStudy (anonymous):
oh okay
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hartnn (hartnn):
so after combining numerators , u get 2x+12+x+3 = ?
OpenStudy (anonymous):
3x+15
hartnn (hartnn):
can u take out something common from that?
OpenStudy (anonymous):
5
hartnn (hartnn):
i would say 3
3(x+5)
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OpenStudy (anonymous):
so thats the answer?
hartnn (hartnn):
the numerator's (x+5) cancels with denominator's (x+5) to give u
\(\huge \frac{3}{(x+3)(x+6)}\)