Solve the system of equations

Solve the system of equations

Postby 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:


[tex]|x^2 - xy - 2y^2 = 0[/tex]
[tex]|3.2^x - 2^{y+2} = 4[/tex]
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Postby 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

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Re: Solve the system of equations

Postby Bayntonsmith » Tue Feb 08, 2011 4:16 am

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

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Re: Solve the system of equations

Postby 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!

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Re: Solve the system of equations

Postby 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 [tex]3*2^{-y}-2^{y+2}=4[/tex]
[tex]3-4*2^{2y}=4*2^{y}[/tex]. Now we take [tex]2^{y}[/tex] as u,
[tex]3-4u=4u^{2}\\
4u^{2}+4u-3=0\\
(2u+3)*(2u-1)=0[/tex]

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

b) For x=2y then [tex]3*2^{2y}-2^{y+2}=4[/tex]
[tex]3*[2^{y}]^{2}-4*2^{y}=4[/tex]. Now we take [tex]2^{y}[/tex]as u,
[tex]3u^2-4u-4=0\\
(u-2)*(3u+2)=0[/tex]
Now [tex]u_1=-2/3[/tex] and [tex]u_2=2[/tex]. But[tex]2^{y}=-2/3[/tex] has no solution.
Then, [tex]2^{y}=2[/tex] 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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