Question

When x=3 and y=5 by how much does the value of 3x^2-2y exceed the value of 2x^2-3/

Answer 1

x = 3

y = 5

3x^2 – 2y

= 3(3)^2 - 2(5)

=> 3(9) - 10 = 17

And 2x^2– 3y

=> 2(3)^2 - 3(5)

=> 2(9) - 15 = 3

17 - 3 = 14

This gives 3x^2 – 2y exceeding 2x^2– 3y by 17 - 3 = 14

Answer 2

14 is the correct answer.

When you use x = 3 and

 y = 5 in the given expressions, 3x2 – 2y = 3(3)2– 2(5) = 27 – 10 = 17 and

2x2 – 3y = 2(3)2 – 3(5) = 18 – 15 = 3.

Then subtract 3 from 17....17-3 = 14.

14 is your answer.

Hope I helped ;]