Try eliminating those denominators:
(x-1) * [3/(x - 1) + 2/x] = 4 * (x-1)
--> 3 + 2(x-1)/x = 4*(x-1)
x * [3 + 2(x-1)/x] = [4*(x-1)] * x
--> 3x + 2(x-1) = 4x(x-1)
Multiply them out:
3x + 2x-2 = 4x² -4x
--> 5x - 2 = 4x² -4x
Make one side of the = sign zero:
(5x-2) + 2 = (4x²-4x) + 2
--> 5x = 4x²-4x+2
--> (5x) - 5x = (4x²-4x+2) - 5x
--> 0 = 4x²-9x+2 (same as 4x²-9x+2 = 0)
When ax²+bx+c = 0, remember the quadratic formula x = [-b +/- sqrt(b² - 4ac)] / 2a?
In this case:
x = [-(-9) +/- sqrt((-9)² - 4(4)(2))] / 2(4)
= [9 +/- sqrt(81 - 32)] / 8
= [9 +/- sqrt(49)] / 8
= (9 +/- 7) / 8
= (9 + 7) / 8 , (9 - 7) / 8
= 16/8 , 2/8
= 2 , 1/4
2007-03-19 11:58:15
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answer #1
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answered by Anonymous
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3/(x-1) + 2/x = 4
x = x
3x/(x-1) + 2 = 4x multiplied
(x - 1) = (x - 1)
3x + 2(x - 1) = 4x (x - 1) multiplied
3x + 2x - 2 = 4x^2 - 4x simplify
0 = 4x^2 - 4x - 3x - 2x + 2
0 = 4x^2 - 9x + 2
Plug it into the quadratic formula, you get
x = (9 + 7) /8 and x = (9 - 7) / 8
x = 2 and x = 1/4
Plug these back into the original equation 3/(x-1) + 2/x = 4
3/(2-1) + 2/2 = 4 YUP! So X=2 is a valid answer.
3/(1/4 - 1) + 2/(1/4) = 4
3/(- 3/4) + 8 = 4
12/(-3) + 8 = 4
-4 + 8 = 4 YUP! So X=1/4 is also a valid answer.
Um, you do know the quadratic formula, don't you?
2007-03-19 11:34:33
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answer #2
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answered by Anonymous
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3 / (x-1) + 2/x=4
so u want to distribute first andputt all the x's together and the #'s together.
so, it would be 3 /x - 3/1 + 2/x
3/x + 2/x = 5/x, so it would be
5/x -3/1 =4
3/1= 3, so
5/x-3 =4
+ 3 to both sides
5/x=7
*5 to bothsides, and u have
x=35!!!!!!!!!
2007-03-19 12:23:54
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answer #3
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answered by ♥me, myself and i♥ 3
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5/7
2007-03-19 11:07:29
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answer #4
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answered by CanProf 7
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3x-3+2x=4
5x-3=4
5x=7
x=1.4
hey smart people if i did this wrong what did i do wrong because we all got different answers
2007-03-19 11:11:38
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answer #5
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answered by That/Cool/Person 2
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