# Solve the system of equations

### Solve the system of equations

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

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

Math Tutor

Posts: 312
Joined: Sun Oct 09, 2005 11:37 am

x≈2.14
y≈1.08

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

Posts: 2
Joined: Sat Oct 16, 2010 7:00 am

### Re: Solve the system of equations

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

Posts: 1
Joined: Wed Dec 15, 2010 11:53 pm

### Re: Solve the system of equations

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

### Re: Solve the system of equations

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,
$3-4u=4u^{2}\\
4u^{2}+4u-3=0\\
(2u+3)*(2u-1)=0$

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,
$3u^2-4u-4=0\\
(u-2)*(3u+2)=0$

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
Guest