anonymous
  • anonymous
Explain how you calculate the restricted value(s) of x for the equation the fraction the binomial x plus 3 over the binomial x plus 5 plus the fraction x over 6 equals the fraction 5 over 6 and what those value(s) represent. equation included.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\frac{ x+3 }{ x+5 }+\frac{ x }{ 6 }=\frac{ 5 }{ 6 }\]
anonymous
  • anonymous
@Hero ?
Hero
  • Hero
The restricted value of x occurs where the denominator cannot equal zero, so to find this restricted value solve x + 5 = 0.

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More answers

anonymous
  • anonymous
x=-5
Hero
  • Hero
yes, correct.
anonymous
  • anonymous
then it can be anything except -5?
Hero
  • Hero
Yes, correct.
anonymous
  • anonymous
ok thanks!
anonymous
  • anonymous
one more question. would you use cross-multiplication or LCD to solve it?
anonymous
  • anonymous
you there?
anonymous
  • anonymous
@hero?
Hero
  • Hero
If you know what you're doing, you could use either method.
anonymous
  • anonymous
which one is easier?
Hero
  • Hero
That depends on how familiar you are with either method. But in general, the easier method tends to be the method that requires the least amount of steps to solve.
anonymous
  • anonymous
which is...?
Hero
  • Hero
I'd say it's different for different people. Me personally, I would use the cross multiplication method after putting the equation in proper form.
anonymous
  • anonymous
how would you do that again?
Hero
  • Hero
There's a particular reason for this. If you subtracted 5/6 from both sides you would get: \[\frac{x + 3}{x + 5} + \frac{x}{6} - \frac{5}{6} = 0\] Which becomes: \[\frac{x + 3}{x + 5} + \frac{x - 5}{6} = 0\] Then after subtracting (x-5)/6 from both sides: \[\frac{x + 3}{x + 5} =- \frac{x - 5}{6}\] You would cross multiply to get \[6(x + 3) = -(x - 5)(x + 5)\] The right hand side represents a difference of squares: \[6x + 18 = -(x^2 - 25)\] \[6x + 18 = -x^2 + 25\] From there, you would re-arrange it into a quadratic equation then solve.
anonymous
  • anonymous
how would it look once it was finished?
anonymous
  • anonymous
@Hero
Hero
  • Hero
Have you attempted to finish solving it?
anonymous
  • anonymous
yesh but you cant put it into quadratic equations because it doesnt have a ^2
Hero
  • Hero
It doesn't? Are you sure about that?
anonymous
  • anonymous
oh sorry. didnt see that! derp
anonymous
  • anonymous
-x^2-6x+7 right?
Hero
  • Hero
Don't forget to set it equal to zero. You want to make sure the leading term is non-negative. It is easier to factor this way. What you do is this: \[-x^2 - 6x + 7 = 0\] Factor out -1: \[-(x^2 + 6x - 7) = 0\] Then divide both sides by -1 \[x^2 + 6x - 7 = 0\] Now you should be able to factor relatively easy
anonymous
  • anonymous
ok thanks! gtg
Hero
  • Hero
Okay, good luck with everything :)

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