# 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

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Joined: Sun Oct 09, 2005 11:37 am
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x≈2.14
y≈1.08

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

ahmed bakoush

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Joined: Sat Oct 16, 2010 7:00 am
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### Re: Solve the system of equations

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

Bayntonsmith

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Joined: Wed Dec 15, 2010 11:53 pm
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### 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
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