solve for x: 5/x-3 - 30/x^2-9 = 1?
回答 (11)
2008-12-26 8:02 am
✔ 最佳答案
5/x-3-30/x^2-9=1[5/x-3-30/(x+3)(x-3)=1](x+3)(x-3) lcd (x+3)(x-3)
5(x+3)-30=x^2-9
5x+15-30=x^2-9
x^2-9+15-5x=0
x^2-5x+6=0
(x-2)(x-3)=0
x-2=0 x-3=0
x=2 x=3 answer {2,3} @(^_^)@
2008-12-26 8:02 am
x = 2 plug it in 5/(2-3) - 30/(2^2-9) = 1
5/(-1) = -5
30/(4-9) = 30/(-5) = -6
-5 - -6 = 1
therefore x must equal 2
5/(-1) = -5
30/(4-9) = 30/(-5) = -6
-5 - -6 = 1
therefore x must equal 2
2008-12-26 7:54 am
first, x^2 - 9 can be written as (x-3)(x+3). [identity- (a+b)(a-b) = a^2 - b^2]
so, taking LCM, we have
[5(x+3) - 30] / [x^2 - 9] = 1
5x + 15 -30 = x^2 - 9
x^2 - 5x + 6 = 0
x^2 - 2x - 3x + 6 = 0
x(x-2) -3(x-2) = 0
(x-3)(x-2) = 0
so, x = 3 and x = 2 are the solutions.
even then, i doubt everyone's answers, including mine.
the equation is not determined for x = +3 or x = -3.
It also doesnt work for x = 4 or x = 9, as another answerer suggested.
so far, the only solution that works is x = 2. Since, this is a quadratic equation, it must have two solutions of which one is x = 2 for sure.
so, taking LCM, we have
[5(x+3) - 30] / [x^2 - 9] = 1
5x + 15 -30 = x^2 - 9
x^2 - 5x + 6 = 0
x^2 - 2x - 3x + 6 = 0
x(x-2) -3(x-2) = 0
(x-3)(x-2) = 0
so, x = 3 and x = 2 are the solutions.
even then, i doubt everyone's answers, including mine.
the equation is not determined for x = +3 or x = -3.
It also doesnt work for x = 4 or x = 9, as another answerer suggested.
so far, the only solution that works is x = 2. Since, this is a quadratic equation, it must have two solutions of which one is x = 2 for sure.
2008-12-26 7:50 am
First you have to see that (x² - 9) can be reduced to (x + 3) (x - 3). After that, it all comes very easily :
[5 / (x-3)] - [30 / (x - 3)(x + 3)] = 1
Get common denominators:
[5(x+3) / (x - 3) (x + 3)] - [30 / (x - 3)(x + 3)] = 1
To make it easy I'll multiply by the denominator to bring it to the other side.
5(x + 3) - 30 = (x - 3)(x + 3)
5x + 15 - 30 = (x - 3)(x + 3)
5x - 15 = (x - 3)(x + 3)
5(x - 3) = (x - 3)(x + 3)
5 = (x + 3)
x = 2
[5 / (x-3)] - [30 / (x - 3)(x + 3)] = 1
Get common denominators:
[5(x+3) / (x - 3) (x + 3)] - [30 / (x - 3)(x + 3)] = 1
To make it easy I'll multiply by the denominator to bring it to the other side.
5(x + 3) - 30 = (x - 3)(x + 3)
5x + 15 - 30 = (x - 3)(x + 3)
5x - 15 = (x - 3)(x + 3)
5(x - 3) = (x - 3)(x + 3)
5 = (x + 3)
x = 2
2008-12-26 8:07 am
(5/[x - 3]) - (30/[x² - 9]) = 1
([5{x + 3}] - 30)/(x² - 9) = 1
5x + 15 - 30 = x² - 9
x² - 5x = - 6
x² - 5/2x = - 6 + (- 5/2)²
x² - 5/2x = - 24/4 + 25/4
(x - 5/2)² = 1/4
x - 5/2 = 1/2
x = 1/2 + 5/2, x = 6/2, x = 3
x = - 1/2 + 5/2, x = 4/2, x = 2
Answer: x = 3, 2
([5{x + 3}] - 30)/(x² - 9) = 1
5x + 15 - 30 = x² - 9
x² - 5x = - 6
x² - 5/2x = - 6 + (- 5/2)²
x² - 5/2x = - 24/4 + 25/4
(x - 5/2)² = 1/4
x - 5/2 = 1/2
x = 1/2 + 5/2, x = 6/2, x = 3
x = - 1/2 + 5/2, x = 4/2, x = 2
Answer: x = 3, 2
2008-12-26 2:55 pm
5/(x - 3) - 30/(x^2 - 9) = 1
[x - 3][x + 3][5/(x - 3) - 30/(x + 3)(x - 3)] = [x - 3][x + 3][1]
5(x + 3) - 30 = (x - 3)(x + 3)
5*x + 5*3 - 30 = x*x - 3*x + x*3 - 3*3
5x + 15 - 30 = x^2 - 3x + 3x - 9
5x - 15 = x^2 - 9
x^2 - 5x + 15 - 9 = 0
x^2 - 5x + 6 = 0
x^2 - 2x - 3x + 6 = 0
(x^2 - 2x) - (3x - 6) = 0
x(x - 2) - 3(x - 2) = 0
(x - 2)(x - 3) = 0
x - 2 = 0
x = 2
x - 3 = 0
x = 3
∴ x = 2 , 3
[x - 3][x + 3][5/(x - 3) - 30/(x + 3)(x - 3)] = [x - 3][x + 3][1]
5(x + 3) - 30 = (x - 3)(x + 3)
5*x + 5*3 - 30 = x*x - 3*x + x*3 - 3*3
5x + 15 - 30 = x^2 - 3x + 3x - 9
5x - 15 = x^2 - 9
x^2 - 5x + 15 - 9 = 0
x^2 - 5x + 6 = 0
x^2 - 2x - 3x + 6 = 0
(x^2 - 2x) - (3x - 6) = 0
x(x - 2) - 3(x - 2) = 0
(x - 2)(x - 3) = 0
x - 2 = 0
x = 2
x - 3 = 0
x = 3
∴ x = 2 , 3
2008-12-26 8:27 am
5 / (x - 3 ) - 30 / ( x^2 - 9 ) = 1
multiplying throughout by x^2 - 9,
we get,
5 (x + 3 ) - 30 = x^2 - 9
5x + 15 - 30 = x^2 - 9
x^2 - 5x - 9 - 15 + 30 = 0
x^2 - 5x + 6 = 0
x^2 - 3x - 2x + 6 = 0
x ( x - 3 ) - 2 ( x - 3 ) = 0
( x - 3 ) (x - 2 ) = 0
so either
x - 3 = 0 or x - 2 = 0
x = 3 or x = 2
multiplying throughout by x^2 - 9,
we get,
5 (x + 3 ) - 30 = x^2 - 9
5x + 15 - 30 = x^2 - 9
x^2 - 5x - 9 - 15 + 30 = 0
x^2 - 5x + 6 = 0
x^2 - 3x - 2x + 6 = 0
x ( x - 3 ) - 2 ( x - 3 ) = 0
( x - 3 ) (x - 2 ) = 0
so either
x - 3 = 0 or x - 2 = 0
x = 3 or x = 2
2008-12-26 8:49 am
As presented, question is INCORRECT.
Should be shown as :-
5 / (x - 3) - 30 / (x² - 9) = 1
5 / (x - 3) - 30 / (x - 3)(x + 3) = 1
5 (x + 3) - 30 = (x - 3)(x + 3)
5x + 15 - 30 = x² - 9
5x - 15 = x² - 9
x² - 5x + 6 = 0
(x - 3)(x - 2) = 0
x = 2 , x = 3
Should be shown as :-
5 / (x - 3) - 30 / (x² - 9) = 1
5 / (x - 3) - 30 / (x - 3)(x + 3) = 1
5 (x + 3) - 30 = (x - 3)(x + 3)
5x + 15 - 30 = x² - 9
5x - 15 = x² - 9
x² - 5x + 6 = 0
(x - 3)(x - 2) = 0
x = 2 , x = 3
2008-12-26 8:34 am
as we know x^2 - 9 can simply to term (x-3)(x+3).
Thus we both sides we multiply factor (x-3)(x+3), and the expression can be simply easily as below:
5(x+3) - 30 = (x-3)(x+3)
5x+15-30 -x^2 + 9 = 0
x^2 - 5x + 6 = 0
(x -3)(x-2)=0
x=2 &3
Thus we both sides we multiply factor (x-3)(x+3), and the expression can be simply easily as below:
5(x+3) - 30 = (x-3)(x+3)
5x+15-30 -x^2 + 9 = 0
x^2 - 5x + 6 = 0
(x -3)(x-2)=0
x=2 &3
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