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solving algebra 2 problems

Mathematics
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solve x^-9 divided by x^-7x+10 = 2x^-8x+6 divided by x^-9x+10 +3 

Dec 9th, 2014

I'm inferring that your equation is (x^-9)/(x^-7)(x + 10) = (2x^-8)(x + 6)/(x^-9)(x +10)  +3

If this is the case then first you need to use the property of exponents, which makes (x^-9) = 1/x^9 and (1/x^-7) = x^7 and (2x^-8) becomes 1/(2x^8)

Your equation now looks like this: (x^7)/(x^9)(x+10) = (x+6)(x^9)/(2x^8)(x+10)  +3

Now you can cross-cancel x's.

1/(x^2)(x+10) = (x+6)(x)/2(x+10)  +3

now multiply both sides by 2(x+10)(x^2) to cancel your denominators

2 = (x+6)(x)(x^2)  +3

-1 = (x+6)(x)(x^2)

now distribute

-1 = x^4 + 6x^3

+1                      +1

x^4 + 6x^3 +1 = 0

you can then split 6x^3 into 3x^3 + 3x^3

x^4 + 3x^3 + 3x^3 +1 = 0           Notice: since there is an x with a power of 4, there will be 4 solutions.

Using the grouping method:

x^3(x+3)(3x^3 + 1) = 0

x^3 = 0 or x+3 = 0 or 3x^3 +1 = 0

x=0 or x=-3 or 3x^3 = -1

                        x^3 = (-1/3)

                        x = 3rd power SQRT(-1/3) 

When you have a negative SQRT, you seperate it into SQRT-1 and SQRT3

then SQRT-1 becomes i

                       x= (+or-) i(3rd powerSQRT3)/3

So the solutions are x=0,-3,i(3rdpowerSQRT3)/3, -i(3rdpowerSQRT3)/3


Hope this helps! If it does please mark it as best answer! Thanks:)

Dec 9th, 2014

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