Question 378901
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Hi,
solution(s) for following system
x^2-y^2=4
y^2=3x 
x^2 - 3x - 4 = 0   
Yes! on the substitution resulting in x = -1 or x = 4 and y^2 = 12 Or y = {{{2*sqrt(3)}}} 
(x = -1 is an extraneous solution as then Y^2 = -3 and y would then be an imaginary number.
Pt(4,{{{2*sqrt(3)}}}) is the solution for this system of equations
{{{drawing(300,300, -6, 6, -6, 6,
circle(4, 3.464,0.2),
graph( 300, 300, -6, 6, -6, 6, (3x)^.5,1*sqrt(x^2-4),-1*sqrt(x^2-4)))}}}