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 [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