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

And you get a simple quadratic equation.

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To solve the equation (y + 2)/y = 1/(y - 5), cross multiply to get (y + 2)(y - 5) = y. Expand and simplify to get y^2 - 3y - 10 = 0. Factor the quadratic equation to get (y - 5)(y + 2) = 0. Set each factor equal to zero and solve for y to find the solutions: y = 5 and y = -2.

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To solve the equation ( \frac{y+2}{y} = \frac{1}{y-5} ), cross multiply to eliminate the denominators, then simplify and solve for ( y ).

[ (y+2)(y-5) = y ]

[ y^2 - 3y - 10 = y ]

[ y^2 - 4y - 10 = 0 ]

This is a quadratic equation. Use the quadratic formula to find the roots of ( y ).

[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]

where ( a = 1 ), ( b = -4 ), and ( c = -10 ).

[ y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-10)}}{2(1)} ]

[ y = \frac{4 \pm \sqrt{16 + 40}}{2} ]

[ y = \frac{4 \pm \sqrt{56}}{2} ]

[ y = \frac{4 \pm 2\sqrt{14}}{2} ]

[ y = 2 \pm \sqrt{14} ]

Therefore, the solutions to the equation are ( y = 2 + \sqrt{14} ) and ( y = 2 - \sqrt{14} ).

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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.

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.

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.

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