How do you solve 1/ (x-2) + (2x)/ ((x-2)(x-8)) = x/ (2(x-8)) 1x−2+2x(x−2)(x−8)=x2(x−8)?

1 Answer
Douglas K.
Apr 22, 2017

Multiply both sides by the least common multiple of the numerators.
Solve the resulting quadratic.
Check.

Explanation:

Given: 1/ (x-2) + (2x)/ ((x-2)(x-8)) = x/ (2(x-8));x!=2,x!=81x−2+2x(x−2)(x−8)=x2(x−8);x≠2,x≠8

Multiply both sides by 2(x-2)(x-8)2(x−2)(x−8)

2(x-8) + 2(2x) = x(x-2);x!=2,x!=82(x−8)+2(2x)=x(x−2);x≠2,x≠8

2x-16 + 4x = x^2-2x;x!=2,x!=82x−16+4x=x2−2x;x≠2,x≠8

Combine like terms:

0 = x^2-8x+16;x!=2,x!=80=x2−8x+16;x≠2,x≠8

Neither 2 nor 8 are roots, therefore, we drop the restrictions:

0 = x^2-8x+160=x2−8x+16

The above factors into a perfect square:

(x - 4)^2 = 0(x−4)2=0

x = 4x=4

check:

1/ (4-2) + (2(4))/ ((4-2)(4-8)) = 4/ (2(4-8))14−2+2(4)(4−2)(4−8)=42(4−8)
1/2 + 8/ ((2)(-4)) = 2/(-4)12+8(2)(−4)=2−4
-1/2 = -1/2−12=−12

This checks.

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