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How do you solve #\frac { 5} { y - 2} = \frac { y } { 3}#?

Answer 1

See a solution process below:

First, multiply each side of the equation by #color(red)(3)color(blue)((y - 2))# to eliminate the fractions while keeping the equation balanced:

#color(red)(3)color(blue)((y - 2)) xx 5/(y - 2) = color(red)(3)color(blue)((y - 2)) xx y/3#

#color(red)(3)cancel(color(blue)((y - 2))) xx 5/color(blue)(cancel(color(black)(y - 2))) = cancel(color(red)(3))color(blue)((y - 2)) xx y/color(red)(cancel(color(black)(3)))#

#color(red)(3) xx 5 = color(blue)((y - 2)) xx y#

#15 = y^2 - 2y#

Next, put the equation in standard form:

#15 - color(red)(15) = y^2 - 2y - color(red)(15)#

#0 = y^2 - 2y - 15#

#y^2 - 2y - 15 = 0#

Then factor the left side of the equation as:

#(y - 5)(y + 3) = 0#

Now, solve each term on the left for #0#:

Solution 1:

#y - 5 = 0#

#y - 5 + color(red)(5) = 0 + color(red)(5)#

#y - 0 = 5#

#y = 5#

Solution 2:

#y + 3 = 0#

#y + 3 - color(red)(3) = 0 - color(red)(3)#

#y + 0 = -3#

#y = -3#

The Solutions Are:

#y = {-3, 5}#

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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.