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

Question

2210 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-09T18:24:35

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

#(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)#

Answer 2 · 2017-02-09T18:30:12

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

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.