SOLUTION: Please help me out with this problem: (1/x-4) + (4/4x-4) = 5/4 I don't know what the LCD would be, so I am unable to find the solution(s). If you could please help me,

Question 240686: Please help me out with this problem:
(1/x-4) + (4/4x-4) = 5/4

I don't know what the LCD would be, so I am unable to find the solution(s).
If you could please help me, it would be greatly appreciated!
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
your equation is:

1/(x-4) + 4/(4x-4) = 5/4

factor out the 4 in 4/(4x-4) to get:

1/(x-4) + 1/(x-1) = 5/4

your common denominator is going to be (x-4)*(x-1)

multiply 1/(x-4) * (x-1)/(x-1) to get (x-1)/((x-4)*(x-1))

multiply 1/(x-1) * (x-4)/(x-4) to get (x-4)/(x-4)*(x-1)))

your equation becomes:

(x-1)/((x-4)*(x-1)) + (x-4)/((x-4)*x-1)) = 5/4

combine the two fractions with the common denominator to get:

((x-1) + (x-4)) / ((x-4)*(x-1)) = 5/4

simplify by combining like terms to get:

(2x-5)/ ((x-4)*(x-1)) = 5/4

multiply both sides of this equation by 4 to get:

(4 * (2x-5))/ ((x-4)*(x-1)) = 5

simplify by removing parentheses in the numerator on the left hand side of the equation to get:

(8x-20) / ((x-4)*(x-1)) = 5

multiply both sides of this equation by ((x-4)*(x-1)) to get:

8x-20 = 5 * ((x-4)*(x-1))

simplify by performing indicated operations and removing parentheses to get:

8x-20 = 5x^2 - 25x + 20

subtract 8x from both sides of the equation and add 20 to both sides of the equation to get:

0 = 5x^2 - 33x + 40 which is the same as:

5x^2 - 33x + 40 = 0

use the quadratic equation to solve this equation to get:

x = 5
or
x = 1.6

substitute for x in the original equation to confirm the result.

I did that and the results are confirmed so these values are your answer.

The quadratic formula is:

the standard form of the quadratic equation is:

ax^2 + bx + c = 0

your equation is:

5x^2 - 33x + 40 = 0

a = 5
b = -33
c = 40

plug those values into the quadratic formula and you get:

which becomes:

which becomes:

The result is x = 5 or x = 1.6

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