How do you find the product of (y^2-1)/(y^2-49)*(y-7)/(y+1)?

2 Answers
Feb 9, 2017

(y-1)/(y+7)

Explanation:

(y^2-1)/(y^2-49) * (y-7)/(y+1)

difference of two squares rule:

(a+b)*(a-b) = a^2 - b^2

in the same way:

(y+1)(y-1) = y^2 - 1^2 = y^2 - 1

(y+7)(y-7) = y^2 - 7^2 = y^2 - 49

therefore (y^2-1)/(y^2-49) * (y-7)/(y+1)

= ((y+1)(y-1))/((y+7)(y-7)) * (y-7)/(y+1)

cancel out fractions:

((y+1)(y-1))/(y+1) = (y-1)
((y+7)(y-7))/(y-7) = (y+7)

therefore (y^2-1)/(y^2-49) * (y-7)/(y+1)

= (y-1)/(y+7) * 1/1

=(y-1)/(y+7)

Feb 9, 2017

=(y-1)/(y+7)

Explanation:

We need to remember two things here. First, for a,b,c,d in RR
and b,d != 0, we have:

a/b * c/d = (ac)/(bd)

Then, we must know how to spot and factor a difference of squares:

a^2 - b^2 = (a-b)(a+b) for any a,b in RR.

Using this relationship, the first fraction is converted to:

(y^2-1)/(y^2 - 49) = ((y-1)(y+1))/((y-7)(y+7))

Then, multiply that with the other fraction:

((y-1)(y+1))/((y-7)(y+7)) * (y-7)/(y+1) = ((y-1)cancel((y+1))cancel((y-7)))/(cancel((y-7))(y+7)cancel((y+1))) = (y-1)/(y+7)

Of course, we have the restrictions y != +-7 and y != -1, otherwise the denominator would have been 0.

Remember these two things to help you spot and factor a difference of squares:

1) See if you can transform the terms into squares, for example:

x^2 = (x)^2 (obviously)
81x^2 = (9x)^2
256 = (16)^2
etc.

2) Remember that 1 = 1^2, so if one of the terms is a 1, you can still factor. For example:

x^2 - 1 = x^2 - 1^2 = (x-1)(x+1)
9y^4 - 1 = (3y^2)^2 - 1^2 = (3y^2 - 1)(3y^2 + 1)
etc.