Answer:
The solution sets are ([tex]\frac{1}{3},\frac{5}{3}[/tex]) and (5,-3)
Explanation:
We are given two equations and we are supposed to solve these two equations and find their solutions.
x + y = 2
4[tex]y^{2}-x^{2}=11[/tex]
Substitute the first equation in the second and make it a quadratic equation in one variable.
4[tex]y^{2}-(2-y)^{2}= 11[/tex]
4[tex]y^{2}-(y^{2}-4y+4)=11[/tex]
[tex]3y^{2}+4y-4=11[/tex]
[tex]3y^{2}+4y-15=[/tex]
[tex]3y^{2}+9y-5y-15=0[/tex]
(3y-5)(y+3) = 0
y = [tex]\frac{5}{3}[/tex] or y = -3
x = [tex]\frac{1}{3}[/tex] or x = 5
The solution sets are ([tex]\frac{1}{3},\frac{5}{3}[/tex]) and (5,-3)
