# When x = 3 and y = 5, by how much does the value of 3x^2 – 2y exceed the value of 2x^2– 3y ?

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### 2 Answers

When the value of x is equal to 3 and the y is equal to 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

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

Given the equations:

3x^2 - 2y

2x^2 - 3y

Also we know that x= 3 and y= 5

We need to know the difference in values between both equations.

First we will substitute x and y values in each equation.

==> 3x^2 -2y = 3*(3^2) - 2*5 = 27 -10= 17

==> 2x^2 -3y = 2*(3^2) - 3*5 = 18-15 = 3

Now we will calculate the difference between both values.

==> 17 -3 = 14.

**Then the values of 3x^2 -2y exceeds the values of 2x^2 -3y by 14.**