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

Answer 1

Julia Adams

#y=2+-sqrt14#

#(y+2)/y = 1/(y-5)#

#(y+2)(y-5) = y#

#y^2-3y-10=y#

#y^2-4y-10=0#

Using quadratic formula,

#y=(-b+-sqrt(b^2-4ac))/(2a)#

#y=(4+-sqrt((-4)^2-4(1)(-10)))/(2times1)#

#y=(4+-sqrt(16+40))/2#

#y=(4+-sqrt56)/2#

#y=(4+-2sqrt14)/2#

#y=2+-sqrt14#

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Answer 2

Avery Alexander

To solve the rational equation (y+2)/y = 1/(y-5), you can start by cross-multiplying to eliminate the fractions. This gives you (y+2)(y-5) = y. Expanding the left side of the equation, you get y^2 - 3y - 10 = y. Rearranging the equation, you have y^2 - 4y - 10 = 0. To solve this quadratic equation, you can use factoring, completing the square, or the quadratic formula. In this case, the quadratic equation cannot be factored easily, so you can use the quadratic formula: y = (-b ± √(b^2 - 4ac))/(2a), where a = 1, b = -4, and c = -10. Plugging in these values, you get y = (4 ± √(16 + 40))/2. Simplifying further, you have y = (4 ± √56)/2. The square root of 56 can be simplified as √(56) = √(4 * 14) = 2√14. Therefore, the solutions to the equation are y = (4 + 2√14)/2 and y = (4 - 2√14)/2, which can be further simplified as 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.