Question 195474
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Missed it by <b><i>that</i></b> much.


Simply set the two functions equal to each other and solve the resulting quadratic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y}{y^2 + 7y + 10}+\frac{y}{y^2 - 4} = \frac{y}{y^2 +3y -10}]


Factor all three denominators:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y}{(y + 2)(y + 5)}+\frac{y}{(y + 2)(y - 2)} = \frac{y}{(y - 2)(y + 5)}]


Three different factors in all the denominators, so the LCD is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (y + 2)(y - 2)(y + 5)]


Apply the LCD:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y(y - 2) + y(y + 5)}{(y + 2)(y - 2)(y + 5)} = \frac{y(y + 2)}{(y + 2)(y - 2)(y + 5)}]


Multiply both sides by the LCD to eliminate the denominators:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(y - 2) + y(y + 5) = y(y + 2)]


Distribute to remove parentheses:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2 - 2y + y^2 + 5y = y^2 + 2y]


Collect like terms on the left:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2 + y = 0]


Factor:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(y + 1) = 0]


Use the Zero Product Rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = 0] or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = -1]


Verification of all elements of the solution set is left as an exercise for the student.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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