anonymous
  • anonymous
how to simplify (1x-3)/(5x^2 + 5x -60) ???
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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lgbasallote
  • lgbasallote
try factoring out the numerator and the denominator..what factors would you get?
anonymous
  • anonymous
see if that denominator can be factored first. take out a 5...
anonymous
  • anonymous
yeah 5 so (x-3)/ 5(x^2 + x - 12) ?

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lgbasallote
  • lgbasallote
seems x^2 + x -3 can be factored as well...try factoring it out..
anonymous
  • anonymous
im not sure how but would you take the x out? D:
lgbasallote
  • lgbasallote
no..you factor out binomials..
anonymous
  • anonymous
hmm
perl
  • perl
its a trinomial in the denominator
perl
  • perl
(1x-3)/(5(x-3)(x+4))
anonymous
  • anonymous
ohhhhh
perl
  • perl
(x-3) -------------- 5 ( x-3)(x+4)
perl
  • perl
now see if you can cancel. (please give me medal. )
perl
  • perl
yay!!
anonymous
  • anonymous
okay so 5(x +4)?
anonymous
  • anonymous
wait 1 / that?
anonymous
  • anonymous
yay it is thanks guys!
perl
  • perl
anything else?
anonymous
  • anonymous
it says express this as single fractions in its simplest form : 1/x + 2/(x+4)
perl
  • perl
we need to find a common denominator first,
anonymous
  • anonymous
i simplified it and i got 3x + 4 but the answer is 3x +4 / (x(x+4)) :(
anonymous
  • anonymous
okay
anonymous
  • anonymous
??
perl
  • perl
so we know that
perl
  • perl
the common denominator is x(x+4) , agreed?
anonymous
  • anonymous
yep :)
perl
  • perl
ok so now we have to make each term have that denominator
anonymous
  • anonymous
yep
perl
  • perl
we know that we can multiply the first fraction by (x+4)/(x+4) which is the same thing as multiplying by 1 (so its a legal operation)
perl
  • perl
and we can multiply the second term by x/x ,again the same thing as multiplying by 1. so its legal
anonymous
  • anonymous
hmmm
perl
  • perl
1 (x+4) 2 ( x) -- * ---- + ---- * ---- x (x+4) (x+4) ( x)
perl
  • perl
its legal in the sense that multiplying by 1 does not change the number , we are only changing it's 'form' , to make it easier to manipulate or add
anonymous
  • anonymous
oh okay yep
perl
  • perl
|dw:1333352449764:dw|
perl
  • perl
collect like terms on the top, and you may or may not choose to expand the bottom by distributing. somtimes it is better to leave denominator factored
anonymous
  • anonymous
yep so 3x +4/ (x X x+4)
anonymous
  • anonymous
woo thanks :)
perl
  • perl
:):)

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