# Two equations and 3 variables

• March 24th 2010, 09:04 AM
artabro
Two equations and 3 variables
Another problem from the GRE program. Determine which option is greater:

x = v + 2
y = 2v

Option A: x^2 + y^2
Option B: (x+y)^2

I decided that B is greater because it results in: x^2 + y^2 + 2xy. So I guess that this 2xy makes the possitive difference. However, I have tried it out with v=1 and v=-1 and both of them are ok with my result, B is greater. However, in the corrections it says that it can´t be determined which one is greater. Could anyone help me to know why??

Thank you so much!
• March 24th 2010, 10:00 AM

If v = -1.

x = 1,
y = -2

So 2xy = -4

Hence option A gives
$1^2 + (-2)^2 = 1 + 4 = 5$
Option B gives
$1^2 + (-2)^2 + (-4) = 5 - 4 = 1$

What happens when $-2 < v < 0$?
• March 24th 2010, 01:47 PM
Soroban
Hello, artabro!

Quote:

Determine which option is greater: . $\begin{array}{ccc}x \;=\;v+2 \\ y \:=\:2v \end{array}$

. . $\begin{array}{cccc}\text{Option A:} & x^2+y^2 \\ \text{Option B:} & (x+y)^2 \end{array}$

I decided that B is greater because it results in: $x^2 + y^2 + 2xy$
So I guess that this $2xy$ makes the positive difference.
How do we know this is a positive difference?

Option A: . $x^2 + y^2 \;=\;(v+2)^2 + (2v)^2 \;=\;5v^2 + 4v + 4$

Option B: . $(x+y)^2 \;=\;(v+2+2v)^2 \-=\;(3v+2)^2 \;=\;9v^2 + 12v + 4$

How do they compare?

. . . . . $\text{Option B} \;\;\begin{array}{c}> \\[-2mm] < \end{array}\;\;\text{Option A}$

. . $9v^2 + 12v + 4 \;\;\begin{array}{c}> \\[-2mm]<\end{array}\;\;5v^2 + 4v + 4$

. . . . . $4v^2 + 8v \;\;\begin{array}{c}> \\[-2mm]<\end{array}\;\;0$

. . . . . $4v(v+2)\;\;\begin{array}{c}>\\[-2mm] <\end{array}\;\;0$

$\text{Option B }\: > \:\text{ Option A }\:\text{ if }\,v < \text{-}2\,\text{ or }\,v > 0$

$\text{Option B }\:<\:\text{ Option A }\:\text{ if }\,\text{-}2 < v < 0$