How do you multiply #yx ^ { 2} \cdot 3y ^ { - 1}#?

Question

484 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-15T10:54:34

#3x^2#

#yx^2 * 3y^-1" "# can be simplified by using a basic law of indices.

"When multiplying, add the indices of like bases"

#x^m xx x^n = x^(m+n)#

#yx^2 * 3y^-1" "# add the indices of y.

=#3x^2y^(1-1)" "larr y^0 = 1#

=#3x^2#

Recall the law of indices for negative indices:
#x^-m = 1/x^m#

#yx^2 * 3y^-1 = (3x^2y)/y#

=#3x^2#