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Question
Solve the following equation: `1/("x" - 1) + 2/("x" - 1) = 6/"x" , (x ≠ 0)`
Solution
`1/("x" - 1) + 2/("x" - 1) = 6/"x" , (x ≠ 0)`
`(("x" - 1) + 2 ("x" - 2))/(("x" - 2)("x" - 1)) = 6/"x"`
x(x + 1) + 2x(x - 2) = 6(x - 2)(x - 1)
x2 - x + 2x2 - 4x = 6(x2 - 2x - x + 2)
3x2 - 5x = 6x2 - 18x + 12
3x2 - 13x + 12 = 0
`"x"^2 - 13/3 "x" + 4 = 0`
`"x"^2 - 3"x" - 4/3 "x" + 4 = 0`
`"x"("x" - 3) - 4/3 ("x" - 3) = 0`
`("x" - 3)("x" - 4/3) = 0`
x = 3 , x = `4/3`
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