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Question
Solve the following equation: `7"x" + 3/"x" = 35 3/5`
Solution
`7"x" + 3/"x" = 35 3/5`
`7"x"^2 + 3 = 178/5 "x"`
`7"x"^2 - 178/5 "x" + 3 = 0`
`"x"^2 - 178/35 "x" + 3/7 = 0`
`"x"^2 - 5"x" - 3/35 "x" + 3/7 = 0`
`"x"("x" - 5) - 3/35 ("x" - 5) = 0`
`("x" - 5)("x" - 3/35) = 0`
x = 5 , x = `3/35`
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