How do you solve the equation and identify any extraneous solutions for #1/(x-1) + 4/(4x-4) = 2#?

Question

Hazel Anglin · Accepted Answer

Extraneous would be if one of the numerator would be #=0#
In both fractions that means #x!=1# We can divide top and bottom of the second fraction by #4#: #1/(x-1)+1/(x-1)=2/(x-1)=2-># #1/(x-1)=1->x-1=1->x=2# And this is allowed.

Eva Adamson · Answer

undefined To solve the equation 1/(x-1) + 4/(4x-4) = 2 and identify any extraneous solutions, we can follow these steps:

1. Start by finding a common denominator for the fractions on the left side of the equation. In this case, the common denominator is (x-1)(4x-4).

2. Multiply each term by the common denominator to eliminate the fractions. This gives us (4x-4) + 4(x-1) = 2(x-1)(4x-4).

3. Simplify the equation by distributing and combining like terms. This results in 4x - 4 + 4x - 4 = 8x^2 - 8x - 8.

4. Combine like terms again to get 8x - 8 = 8x^2 - 8x - 8.

5. Move all terms to one side of the equation to obtain 8x^2 - 16x = 0.

6. Factor out the greatest common factor, which is 8x, to get 8x(x - 2) = 0.

7. Set each factor equal to zero and solve for x. This gives us two possible solutions: x = 0 and x = 2.

8. Check each solution by substituting them back into the original equation. For x = 0, the original equation becomes 1/(-1) + 4/(-4) = 2, which simplifies to -1 - 1 = 2. This is not true, so x = 0 is an extraneous solution.

9. For x = 2, the original equation becomes 1/(2-1) + 4/(4(2)-4) = 2, which simplifies to 1/1 + 4/4 = 2. This is true, so x = 2 is the only valid solution.

Therefore, the solution to the equation 1/(x-1) + 4/(4x-4) = 2 is x = 2, and there are no extraneous solutions.