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How do you solve #(y+2)/y=1/(y-5)#?

Answer 1

#y = 2 +sqrt(14)#

First multiply out to get rid of denominators: #(y+2)(y-5) = 1*y# #y^2 -3y - 10 -y = 0# #y^2 - 4y - 10 = 0# Using completing the squares #(y - 2)^2 - 10 - 4 = 0# #(y-2)^2 =14# #y = 2 +sqrt(14)#

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Answer 2

Eva Abramson

To solve the equation (y+2)/y=1/(y-5), we can cross-multiply to eliminate the fractions. This gives us (y+2)(y-5) = y. Expanding the left side of the equation, we get y^2 - 3y - 10 = y. Rearranging the equation, we have y^2 - 4y - 10 = 0. To solve this quadratic equation, we can use the quadratic formula: y = (-b ± √(b^2 - 4ac))/(2a), where a = 1, b = -4, and c = -10. Plugging in these values, we get y = (4 ± √(16 + 40))/2. Simplifying further, we have y = (4 ± √56)/2. This can be simplified to y = 2 ± √14. Therefore, the solutions to the equation are y = 2 + √14 and y = 2 - √14.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.