Question #8a1f1

Question

1351 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-14T23:02:11

Your coefficient is right, but you treated #x# and #y# as the same variable.

#(6x^-8)/(7y^-7)#

is the same thing as

#(6/7)(y^7/x^8)#

Then we can proceed with the exponent:

#[((6y^7)/(7x^8))]^(-2)#

#= [(7x^8)/(6y^7)]^2#

#= (49x^16)/(36y^14)#

Now, we're done!