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)
-1 = x^4 + 6x^3
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:)
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