SOLUTION: Find all real or imaginary solutions for the following equation: (1/x) + 1/(x-1) = (3/2) I have really been struggling with using imaginaries and fractions. Help is greatly a

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Question 992286: Find all real or imaginary solutions for the following equation:
(1/x) + 1/(x-1) = (3/2)
I have really been struggling with using imaginaries and fractions. Help is greatly appreciated.
Thank you!
Found 4 solutions by satyareddy22, solver91311, macston, MathTherapy:
Answer by satyareddy22(84)   (Show Source): You can put this solution on YOUR website!
(1/x) + 1/(x-1) = (3/2)
{(x-1)+x}/x(x-1)=3/2
(2x-1/x^2-x)=3/2
2(2x-1)=3(x^2-x)
4x-2=3x^2-3x
3x^2-3x-4x+2=0
3x^2-7x+2=0
3x^2-6x-x+2=0
3x(x-2)-1(x-2)=0
(x-2)(3x-1)=0
=>x-2=0,3x-1=0
=>x=2,x=1/3
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!


Remember how to add fractions with different denominators. You find the common denominator, convert, and add the numerators. If the two denominators have no factors in common, then the lowest common denominator is simply the product of the two denominators.



Now you have a simple proportion. Cross-multiply, collect like terms, and then solve the resulting quadratic. Said quadratic is factorable and therefore has two real roots.

John

My calculator said it, I believe it, that settles it

Answer by macston(5194)   (Show Source): You can put this solution on YOUR website!
.
Multiply through by x.
.
Multiply through by (x-1)
Multiply each side by (2/3).
Subtract (4/3)x from each side.
Add 2/3 to each side
Multiply by 3.
Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!

Find all real or imaginary solutions for the following equation:
(1/x) + 1/(x-1) = (3/2)
I have really been struggling with using imaginaries and fractions. Help is greatly appreciated.
Thank you!


2(x - 1) + 2x = 3x(x - 1) ------ Multiplying through by LCD: 2x(x - 1)









(3x - 1)(x - 2) = 0 ------ Factoring trinomial to determine solutions

3x - 1 = 0        OR       x - 2 = 0

     OR       

Both satisfy the original equation so both are solutions. 

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