##### solve each equation show all work and check for solutions

label Algebra
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2 / x-1 3 / x 1 = 4 / x^2 -1

May 29th, 2015

what does it mean -1 3 / x 1 (??) what does the x 1 stands for?

We can't sole if it is not written accurately...

May 29th, 2015

May 29th, 2015

2/x-1 + 3/x+1 = 4/x^2-1

May 29th, 2015

This looks better . Does the 3/x +1 intended as 3/x +1 or 3/(x+1) ??

May 29th, 2015

2/(x-1) + 3/(x+1) = 4/(x^2-1)

May 29th, 2015

Finally... let me solve it now for you...

May 29th, 2015

since your here, and if you dont mind, can you solve this one to plz:

x/(x+1) + 2/(x-3) = 4/((x+1)(x-3))

May 29th, 2015

1st one: we multiply throughout by x^2-1 assuming x is different from +1 or -1. and note that x^2-1=(x+1)(x-1)

So we have:

2(x+1)+3(x-1)=4

or: 5x-1=4

5x = 5 = > x= 1. This is indeed a solution of 2(x+1)+3(x-1)=4 above, but cannot be a solution of the original equation 2/(x-1) + 3/(x+1) = 4/(x^2-1) because at x=1 both 2/(x-1) and 4/(x^2-1) diverges to infinity. Thus, there is really no solution in the domain of definition of the terms of the equation...

2nd one:

x/(x+1) + 2/(x-3) = 4/((x+1)(x-3))

same method - multiply by (x+1)(x-3) assuming x is not -1 or +3:

x(x-3)+2(x+1)=4

x^2-x-2=0

x= [-1+-sqrt(1+8)]/2 = (-1 +-3)/2

x1=1; x2= -2. Both are acceptable.

May 29th, 2015

thank you so much!!

May 29th, 2015

Sorry - a correction on the last one

May 29th, 2015

its fine, what happened?

May 29th, 2015

x= [+1+-sqrt(1+8)]/2 = (+1 +-3)/2

x1=2; x2= -1. Only x1 is acceptable.

May 29th, 2015

are you sure that number 1 doesnt have a solution? just making sure because we havent seen those in class

May 29th, 2015

I'll check again....1 moment

May 29th, 2015

number 1: (2 / (x-1)) + (3 / (x+1)) = (4 / (x^2 -1))

May 29th, 2015

Yea - that's the way I got it, The x=1 solution simply blow off the 2/(x-1) term and the 4/(x^2-1) so what I said is right. They don't always tell you everything in class, sometime you're supposed to think a little (I guess...)

Also the last seems OK. I think this is the purpose of this exercise to demonstarte unacceptable solutions.

May 29th, 2015

is it alright if you can do one more? because no one else is getting it?

May 29th, 2015

Ok

May 29th, 2015

((7x+7) / (x^2-8x+15)) + ((3x) / (x-5)) = ((-7) / (x-3))

May 29th, 2015

Note that x^2-8x+15=x^2-5x-3x+15=x(x-5)-3(x-5)=(x-3)(x-5)  (the 3 & 5 can also be found by solving x^2-8x+15=0)

Now again the same method as above multiply by (x-3)(x-5)

7x+7+3x(x-3)=-7(x-5)

3x^2+5x-28=0

x= [-5+-sqrt(25+336)]/6=(-5+-13)/6

x1=4/3 x2 = -3 both OK

May 29th, 2015

Correction

x= [-5+-sqrt(25+336)]/6=(-5+-19)/6

x1=7/3 x2 = -4 both OK

May 29th, 2015

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May 29th, 2015
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May 29th, 2015
Oct 21st, 2017
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