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

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clara2 | Student, Undergraduate | eNoter

Posted March 20, 2011 at 12:15 PM via web

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

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted March 20, 2011 at 12:19 PM (Answer #1)

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Given that x= 3 and y= 5

First we will find the value for each expression then compare the results.

The first expression is 3x^2 - 2y

We will substitute with x- 3 and y- 5

==> 3*3^2 - 2*5 = 27 - 10 = 17..........(1)

Now we will calculate the second expression 2x^2 - 3y

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

Now the difference between the values is 17-3 = 14.

Then we conclude that the value of 3x^2 - 2y exceeds the value of 2x^2 - 3y by 14.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted March 20, 2011 at 12:26 PM (Answer #2)

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We are given that x = 3 and y = 5. We have to determine by how much the value of 3x^2 - 2y exceeds that of 2x^2 - 3y

3x^2 - 2y - (2x^2 - 3y)

=> 3x^2 - 2y - 2x^2 + 3y

=> x^2 + y

=> 3^2 + 5

=> 9 + 5

=> 14

The required value by which 3x^2 - 2y exceeds 2x^2 - 3y is 14

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