Answer:
3y[tex]\sqrt{2}[/tex]
Step-by-step explanation:
Solving the first term in the expression
y[tex]\sqrt{72}[/tex] is divisible by 2 and 36.
y[tex]\sqrt{2*36}[/tex] 36 is a perfect square of 6, and 2 is not a perfect square
6y[tex]\sqrt{2}[/tex] is the simplified version of y[tex]\sqrt{72}[/tex]
Solving the second term in the expression
[tex]\sqrt{18y^{2} }[/tex] 18 is divisible by 2 and 9. y^2 is already a perfect square
[tex]\sqrt{(2*9)y^2}[/tex] 2 is not a perfect square so it stays on the inside
[tex]3y\sqrt{2}[/tex] is the simplified version of[tex]\sqrt{18y^{2} }[/tex]
Solving for the entire expression
Now taking the two terms and solving for the original expression
6y[tex]\sqrt{2}[/tex] - [tex]3y\sqrt{2}[/tex] = 3y[tex]\sqrt{2}[/tex]
I hope this helped!
