• Jan 12th 2010, 11:51 AM
daniking
I've been needing some help with my algebra class.. thank you!

The ordered pair (2, 16) is a solution for which equation(s)? Check all that apply.
A. y = 2x2
B. y = (2x)2
C. y = 2x + 2
D. y = (x + 2)2

Here is my final problem I'm stuck at.. I read through my book several times but it does not mention what I do... My teacher said we whent over it in class.. But being absent has it's downs. =/
• Jan 12th 2010, 12:55 PM
pickslides
Quote:

Originally Posted by daniking

The ordered pair (2, 16) is a solution for which equation(s)? Check all that apply.
A. y = 2x2
B. y = (2x)2
C. y = 2x + 2
D. y = (x + 2)2

For each equation substitute in $\displaystyle x=2$, if the result gives 16 then it is a solution, i'll do the first one

$\displaystyle y=2x^2$

$\displaystyle x=2$

$\displaystyle y=2(2)^2= 2\times 4 = 8 \neq 16$
• Jan 12th 2010, 01:55 PM
daniking
Quote:

Originally Posted by pickslides
For each equation substitute in $\displaystyle x=2$, if the result gives 16 then it is a solution, i'll do the first one

$\displaystyle y=2x^2$

$\displaystyle x=2$

$\displaystyle y=2(2)^2= 2\times 4 = 8 \neq 16$

Ummmm Yeah that answer you gave me was wrong lol. (Giggle)

But it still helped me out so ty (:
• Jan 12th 2010, 06:00 PM
mr fantastic
Quote:

Originally Posted by daniking
Ummmm Yeah that answer you gave me was wrong lol. (Giggle)

But it still helped me out so ty (:

pickslides knows that this option is wrong. In fact, he clearly indicated that it was wrong. He started with option A and showed it was wrong as an example of how to follow his good advice:
Quote:

Originally Posted by pickslides
For each equation substitute in http://www.mathhelpforum.com/math-he...7a4b424e-1.gif, if the result gives 16 then it is a solution

The expectation was that you would test the remaining options in the same way until you found the option that worked.