anonymous
  • anonymous
simplify this for a medal: 2x/3x+3 - 2/x+5
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\frac{2x}{3x+3} = \frac{-2}{x+5}\] is that the question?
anonymous
  • anonymous
no there is a subtraction sign in the middle instead of the equal sign, and the two is positive
anonymous
  • anonymous
OK, the answer should be \[x ^{2} + 2x + 3\] @Luis_Rivera can you please double check my work?

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anonymous
  • anonymous
how did you get that?
anonymous
  • anonymous
or @amistre64 could you check when you're free?
amistre64
  • amistre64
\[\frac{2x}{3x+3} - \frac{2}{x+5}\] \[\frac{2x(x+5)}{3x+3} - \frac{2(3x+3)}{x+5}\] \[\frac{2x(x+5)-2(3x+3)}{(3x+3)(x+5)}\] \[\frac{2x^2+10x-6x-6}{(3x+3)(x+5)}\] \[\frac{2x^2+4x-6}{(3x+3)(x+5)}\] \[\frac{(x+6/2)(x-2/2)}{(3x+3)(x+5)}\] \[\frac{2(x+3)(x-1)}{3(x+1)(x+5)}\]whatd i miss anything?
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=+2x%2F%283x%2B3%29+-+2%2F%28x%2B5%29 if thats what you gave us, thats what we get :/
anonymous
  • anonymous
I'm not sure if i simplified it right, but i went from: \[\frac{2x}{3x+3} - \frac{2}{x+5}\] then you multiply each part by each denominator to cancel out each denominator: \[3x + 3 \times \frac{2x}{3x+3} - \frac{2}{x+5} \times 3x+3\] \[\rightarrow 2x - \frac{2}{x+5}\times 3x + 3\] \[x + 5 \times 2x - x + 5 \times \frac{2}{x+5} \times 3x + 3 = 2x(x+5) - 2(3x + 3)\] \[\rightarrow 2x ^{2}+10x - 6x + 6 \rightarrow 2x ^{2}+4x+6\] Did i do it right? and now we simplify by dividing by 2???
anonymous
  • anonymous
@amistre how did you get \[\frac{(x+6/2)(x-2/2)}{3x+3)(x+5)}\] ? Can you please explain, sorry.
amistre64
  • amistre64
thats just a method i use to factor a trinomial. Ideally i should have factored out the 2 first but since i didnt i was lead to a result that was off by a factor of 2. So i adjusted for that on the next step. but given that ax^2 +bx + c has no common factors; and that it will factor into something real; i simple set it up as (x+ /a) (x+ /a) then i multiply the first to the last, ac, and factor that to sum up to the middle term b; say ac = mn, such that m+n = b. I then insert m and n into my setup (x+ m/a) (x+ n/a) form there i reduce the fraction to simpliest form and if there still remains a denominator i stick it in front: (x+ p) (x+ q/r) --> (x+ p) (rx+ q)
anonymous
  • anonymous
hey thanks a lot for explaining, i really appreciate it
amistre64
  • amistre64
yep; i think i see where your setup went wrong at; its ini your attempt to "clear" the fractions. take for example some numbers we can see:\[\frac{3}{4}-\frac13\] we know this should equal:\[\frac34\frac33-\frac13\frac44=\frac5{12}\] your method is actually changeing the values since you are not multiplying by a useful for of "1". \[\frac34\frac41-\frac13\frac31=3-1 = 2\] \[2\ne\frac{5}{12}\]
anonymous
  • anonymous
So multiplying by the denominator to clear them like i tried to do, does that only work for when it's an equation? when the minus sign is an equal sign otherwise i have to treat these like fractions, right?
amistre64
  • amistre64
correct.
anonymous
  • anonymous
Thank you! :)

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