# simplify the following a. 1/y^2-4 + 3/y^2+y-6 divided by 1/y^2-4 + 2/y^2+3y+2 b. 4a/2a^2-1 - 4/a-1 divided by 1/a-1 + 6/2a+1help!!! algebra

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### 1 Answer

## 1/y^2-4 + 3/y^2+y-6 divided by 1/y^2-4 + 2/y^2+3y+2

1/(y^2-4) + 3/(y^2+y-6) divided by 1/(y^2-4) + 2/(y^2+3y+2)

First we will factor the equations.

1/((y-2)(y+2) + 3/((y+3)(y-2)) divided by 1/((y-2)(y+2)) + 2/((y+2)(y+1))

Now find the LCD of both sides

(y-2)(y+2)(y+3) and (y-2)(y+2)(y+1) are the LCD's, now get the same denominator for both fractions on both equations.

1/((y-2)(y+2))*(y+3)/(y+3) + 3/((y+3)(y-2))*(y+2)/(y+2) and

1/((y-2)(y+2))*(y+1)/(y+1)+2/((y+2)(y+1))*(y-2)/(y-2)

to get

(y+3)/((y-2)(y+2)(y+3)) + 3(y+2)/((y-2)(y+2)(y+3)) and

(y+1)/((y-2)(y+2)(y+1)) + 2(y-2)/((y-2)(y+2)(y+1))

Now that we have the same denominators we add the numerators

((y+3)+3(y+2))/((y-2)(y+2)(y+3)) and ((y+1)+2(y-2))/((y-2)(y+2)(y+1))

And simplify to get

(4y+9)/((y-2)(y+2)(y+3)) and (3y-3)/((y-2)(y+2)(y+1))

Now we divide by changing the division to a multiplication by the reciprocal

(4y+9)/((y-2)(y+2)(y+3)) * ((y-2)(y+2)(y+1))/(3y-3)

Cancel y+2 and y-2 to get

((4y+9)(y+1))/(3(y-1)(y+3))

And our final answer is

((4y+9)(y+1))/(3(y-1)(y+3))