Question 277318: My son has 2 problems that seem impossible.
(1) 1/(x-1)-(2(3x+1)/5x-2)+1/(1-x)=0
(2)(7x+5)/6-(5(x-1))/4=1/3+8/(1-x)
Thank you for your help Found 2 solutions by stanbon, mananth:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Thank you for your prompt reply. We are still not sure how in problem 1 you can modify the equation 1/(x-1)...+ 1/(1-x)to your modification of 1/(x-1)...- 1/(x-1).
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(1)
1/(x-1)-(2(3x+1)/5x-2)+1/(1-x)=0
Modify the 3rd term: +1/(1-x)
Multiply numerator and denominator by -1 to get -1/(x-1)
Also apply the distributive law on the 2nd term.
[1/(x-1)] - [(6x+2)/(5x-2)] - [(1/(x-1)] = 0
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Notice that the 1st and 3rd terms add up to zero.
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(6x+2)/(5x-2) = 0
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The fraction will be zero when the numerator is zero:
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6x+2 = 0
6x = -2
x = -2/6
x = -1/3
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The same question arises in problem 2 when you go from ... 1/3 + 8/(1-x) to ...1/3 - 8/(x-1). Perhaps you can explain that.
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you have 8/(1-x)
Multiply numerator and denominator by -1 to get:
-8/(x-1)
(2)
[(7x+5)/6]-[(5(x-1))/4] = [1/3]+[8/(1-x)]
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[(7x+5)/6]-[(5(x-1))/4] = [1/3]-[8/(x-1)]
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Multiply thru by 12(x-1) to get rid of the denominators:
[2(x-1)(7x+5)] - [15(x-1)^2] = [4(x-1)] - [12*8]
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Modify:
[2(7x^2-2x-5] - [15(x^2-2x+1)] = [4x-4]-[96]
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14x^2-4x-10-15x^2+30x-15 = 4x-100
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-x^2+26x-15 = 4x-100
-x^2+22x+85 = 0
x^2-22x-85 = 0
Factor:
(x-17)(x-5) = 0
x = 17 or x = 5
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Cheers,
Stan H.
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