Solve the system of equations

by **Math Tutor** » Thu Jul 29, 2010 4:14 am

Solve the system of equations of the entrance examination to the most prestigious bulgarian university - Sofia University:

$|x^2 - xy - 2y^2 = 0$
$|3.2^x - 2^{y+2} = 4$

by **ahmed bakoush** » Sat Oct 16, 2010 7:30 am

x≈2.14
y≈1.08

and it has anther solution but i will try to get it

by **Bayntonsmith** » Tue Feb 08, 2011 4:16 am

Take common both x and Y and then solve them..........

by **the system** » Tue Feb 08, 2011 4:30 am

What do you mean Bayntonsmith?
I do not get your idea. The system is really tricky!

by **Guest** » Fri Jan 11, 2013 4:01 pm

From 1st equation (x+y)*(x-2y)=0. We have 2 conditions:

a) For x=-y then $3*2^{-y}-2^{y+2}=4$
$3-4*2^{2y}=4*2^{y}$ . Now we take $2^{y}$ as u,
$3-4u=4u^{2}\\<br />4u^{2}+4u-3=0\\<br />(2u+3)*(2u-1)=0$

Now $u1=-3/2$ and u2=1/2. But $2^{y}=-3/2$ has no solution.
Then, $2^{y}=\frac{1}{2}$ has a solution as y=-1. Thus, x must be 1.

b) For x=2y then $3*2^{2y}-2^{y+2}=4$
$3*[2^{y}]^{2}-4*2^{y}=4$ . Now we take $2^{y}$ as u,
$3u^2-4u-4=0\\<br />(u-2)*(3u+2)=0$
Now $u_1=-2/3$ and $u_2=2$ . But $2^{y}=-2/3$ has no solution.
Then, $2^{y}=2$ has a solution as y=1.
Thus, x must be 2. Consequently this equation system has 2 solutions for x and y: (1,-1) and (2, 1).

Solver: Cem Shentin

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