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Question
Simplify: `("y"/6 + "2y"/3) div ("y" + ("2y" - 1)/3)`
Sum
Solution
`("y"/6 + "2y"/3) div ("y" + ("2y" - 1)/3)`
`= (("y" + "4y")/6) div (("3y + 2y - 1")/3)`
`= "5y"/6 div ("5y" - 1)/3`
`= "5y"/6 xx 3/("5y" - 1)`
`= ("5y")/(2("5y" - 1)) = "5y"/(10"y" - 2)`
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