anonymous
  • anonymous
Solve the equation for x. -1+((2x)/(x+3))=((-2)/(x+2))
Algebra
katieb
  • katieb
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anonymous
  • anonymous
Multiply the equation by the LCD.
anonymous
  • anonymous
Sorry I should have stated to solve the equation for x.
anonymous
  • anonymous
K, let's start again.....

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anonymous
  • anonymous
okay
anonymous
  • anonymous
What is it u want to do?
anonymous
  • anonymous
|dw:1350674953315:dw|
anonymous
  • anonymous
Yes, so what is it u want to do?
anonymous
  • anonymous
Multiply (-1), (2x/x+3), and (-2x/x+2) by the LCD of (x+3) and (x+2).
anonymous
  • anonymous
OK, that is the long way around of solving for x. You will get to the same place at the end. I guess your teacher has told u to do it that way, right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
LCD of (x+3) and (x+2) is just (x+3)(x+2) I will leave you to it, it is not too hard to do.
anonymous
  • anonymous
Okay. Thank you. I know the answer is (-1,0). I just wanted to make sure I was on the right path.
anonymous
  • anonymous
U should also note that x=-3 and x = -2 are not defined for this equation.
anonymous
  • anonymous
correct
anonymous
  • anonymous
Could you please explain the shorter way of doing this type of equation.
anonymous
  • anonymous
If you add the terms on the left side -> (x-3)/(x+3) = -2/(x+2) Cross multiply -> (x-3)(x+2) = -2(x+3) -> x^2 -x -6 = -2x - 6 -> x^2 + x = 0 -> x( x +1) = 0 -> x = 0 or -1
anonymous
  • anonymous
Wow that is much quicker! Thank you so much
anonymous
  • anonymous
yw:)
anonymous
  • anonymous
The other way reminds you that the original equation is not defined for x=-3 and x = -2 But with some practice you will not need reminding....
anonymous
  • anonymous
Could I use this shorter way for this equation as well? (see drawing)
anonymous
  • anonymous
|dw:1350677688912:dw|
anonymous
  • anonymous
Yes, u can.
anonymous
  • anonymous
Thanks. When I add the terms on the LHS I get (z+1) / (z-6). Is this correct?
anonymous
  • anonymous
No, (z+1) + 1*(z-6) / (z-6) = (2z-5) / (z-6)
anonymous
  • anonymous
Gotcha. Thanks

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