anonymous
  • anonymous
simplify
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\frac{ 4x }{ 9 }+\frac{ x-10 }{ 9 }\]
anonymous
  • anonymous
so easy tati
anonymous
  • anonymous
LCM=9

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More answers

hartnn
  • hartnn
common denominator so \(\huge \frac{4x+x-10}{9}\)
hartnn
  • hartnn
whats 4x+x ?
anonymous
  • anonymous
thank i got it
anonymous
  • anonymous
\[\frac{ 5x-10 }{ 9 }\] ?
anonymous
  • anonymous
@hartnn
hartnn
  • hartnn
yup, thats correct :)
anonymous
  • anonymous
thank you can you help me with like 5 more ?
hartnn
  • hartnn
only 5 ?
anonymous
  • anonymous
yes
hartnn
  • hartnn
whats next ?
anonymous
  • anonymous
\[\frac{ t^{2}+6 }{ t-1 }-\frac{ 7 }{ t-1 }\]
hartnn
  • hartnn
again common denominator, \(\frac{t^2+6-7}{t-1}\)
anonymous
  • anonymous
sorry the 6 has a t next to it
hartnn
  • hartnn
whats 6-7 ?
hartnn
  • hartnn
so can u factor t^2+6t-7 ?
anonymous
  • anonymous
umm yes ?
hartnn
  • hartnn
then tell me the two factors....
anonymous
  • anonymous
ik you can factors that .. wouldnt one of the numbers be negative
hartnn
  • hartnn
tell me 2 numbers whose sum is 6 and product is -7 yes,one of them is negative.
anonymous
  • anonymous
(t-6)(t+1) ??
hartnn
  • hartnn
nice, its (t+7)(t-1) then cancel out t-1 from numerator and denominator what do u get finally ?
anonymous
  • anonymous
(t+7)
hartnn
  • hartnn
thats it. :)
anonymous
  • anonymous
okay the next one \[\frac{ 3 }{ x }+\frac{ 13 }{ x-11 }\]
anonymous
  • anonymous
wouldnt the top be 16
hartnn
  • hartnn
nopes,its not the same denominator now.
hartnn
  • hartnn
u need to cross multiply..... can u ?
anonymous
  • anonymous
3x-33+13x
hartnn
  • hartnn
yup,thats the correct numerator.... which will be 16x-33 and what will be denominator?
anonymous
  • anonymous
uhhh i dont know i thought the denominator would cancel out cross we crossed multiplied
hartnn
  • hartnn
nopes it would be \(\huge \frac{16x-33}{x(x-11)}\)
anonymous
  • anonymous
ohh okay ... \[\frac{ 5 }{ 4x ^{2}y }-\frac{ y }{ 14xy }\]
anonymous
  • anonymous
??
hartnn
  • hartnn
sorry, i kinda lost connection :( tell me the LCM of 4x^y and 14xy ?
anonymous
  • anonymous
oh okay it would be 2xy
hartnn
  • hartnn
right, so can u solve that further ?
anonymous
  • anonymous
what do you mean?
hartnn
  • hartnn
ok, i'll write next step, see it that makes sense \(\huge \frac{5(7)}{28x^2y}-\frac{y(2x)}{28x^2y}\)
anonymous
  • anonymous
ohh okay .. \[\frac{ 35 }{28x ^{2}y }-\frac{ 2xy }{ 28x ^{2} y}\]
anonymous
  • anonymous
then combine the denominator
hartnn
  • hartnn
but did u get how i got that ? yes,then combine denom.
anonymous
  • anonymous
yea i saw you got that,
hartnn
  • hartnn
so finally u get (35- 2xy)/(28x^2y) ok?
anonymous
  • anonymous
\[\frac{ x ^{2}+2x }{12x+54 }-\frac{ 3x }{ 8x+36 }\]
anonymous
  • anonymous
would i combine the like terms on the bottom & top
hartnn
  • hartnn
nopes, first can u write 12x+54 as 6(2x+9) ??
anonymous
  • anonymous
you can cause that would equal the original equation
hartnn
  • hartnn
so how would u write 8x+36 ?
anonymous
  • anonymous
8 doesn't go into 36 tho
anonymous
  • anonymous
36 doesnt go into 8 i mean
hartnn
  • hartnn
what about 4 ? 36 and 8 both go into 4.
anonymous
  • anonymous
4(2x+9)
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 ?
anonymous
  • 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)
anonymous
  • anonymous
i see it i see it it makes sense so then you would take away the denominator?
hartnn
  • hartnn
take away ? since the denominator is common u combine the numerators. (2x^2+4x-9x)\(12(2x-9))
anonymous
  • anonymous
yea thats what i meant
anonymous
  • anonymous
would the answer be 2x^2-5x/(12(2x-9)
hartnn
  • hartnn
yup, thats correct. (2x^2-5x)/(12(2x-9))
anonymous
  • anonymous
awesome .. there is one more
hartnn
  • hartnn
ok, ask.
anonymous
  • anonymous
\[\frac{ 2 }{ x ^{2}8x+15 } +\frac{ 1 }{ x ^{2} +11x+30}\]
anonymous
  • anonymous
x^2 + 8 **
hartnn
  • hartnn
can u factor both the quadratic polynomials in denom. ?
anonymous
  • anonymous
i dont think so, im not sure
hartnn
  • hartnn
x^2+8x+15 = (x+5)(x+3) and x^2+11x+30=(x+5)(x+6) ok?
anonymous
  • anonymous
ohhhhhh okay
anonymous
  • anonymous
\[\frac{ 3 }{ (x+3)(x+6) } or \frac{ 2 }{ (x+3) }+\frac{ 1 }{ (x+6)}\]
hartnn
  • hartnn
\(\large\frac{ 2 }{(x+5)(x+3)} +\frac{ 1 }{(x+5)(x+6)}=\frac{ 2(x+6) }{(x+6)(x+5)(x+3)} +\frac{ (x+3) }{(x+3)(x+5)(x+6)}\) the idea is to make the denominator same, so that u can combine numerator....
anonymous
  • anonymous
oh okay
hartnn
  • hartnn
so after combining numerators , u get 2x+12+x+3 = ?
anonymous
  • anonymous
3x+15
hartnn
  • hartnn
can u take out something common from that?
anonymous
  • anonymous
5
hartnn
  • hartnn
i would say 3 3(x+5)
anonymous
  • anonymous
so thats the answer?
hartnn
  • hartnn
the numerator's (x+5) cancels with denominator's (x+5) to give u \(\huge \frac{3}{(x+3)(x+6)}\)
anonymous
  • anonymous
thank you soo much, you're amazing!
hartnn
  • hartnn
welcome :) u too are awesome ;)
anonymous
  • anonymous
thanks ^_^

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