SOLUTION: 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
Algebra.Com
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).
------------------------------------
(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
---
Notice that the 1st and 3rd terms add up to zero.
----
(6x+2)/(5x-2) = 0
---
The fraction will be zero when the numerator is zero:
---
6x+2 = 0
6x = -2
x = -2/6
x = -1/3
===================================
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.
---
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)]
---
[(7x+5)/6]-[(5(x-1))/4] = [1/3]-[8/(x-1)]
---
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]
----
Modify:
[2(7x^2-2x-5] - [15(x^2-2x+1)] = [4x-4]-[96]
---
14x^2-4x-10-15x^2+30x-15 = 4x-100
---
-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
=============================
Cheers,
Stan H.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
1/(x-1)-(2(3x+1)/5x-2)+1/(1-x)=0
1/x-1 - 2 (3x+1)/(5x-2) +1/1-x =0
rewrite the equation
1/x-1 + 1/1-x - 2(3x+1) / 5x-2
Solve first two expressions
1-x +x-1 / (x-1)(1-x) - (6x+2) / 5x-2 =0
0 -(6x+2) / 5x-2 =0
-(6x+2)=0
-6x-2=0
-6x = 2
x= 2/-6
x=-1/3
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