Respuesta :
Answer:
(-3/5, 4/5)
Explanation:
Given the system of equations:
[tex]\begin{gathered} x^2+y^2=1 \\ y=2x+2 \end{gathered}[/tex]First, we substitute y=2x+2 into the first equation to obtain:
[tex]\begin{gathered} x^2+(2x+2)^2=1 \\ x^2+(2x+2)(2x+2)=1 \\ x^2+4x^2+4x+4x+4=1 \\ 5x^2+8x+4-1=0 \\ 5x^2+8x+3=0 \end{gathered}[/tex]We solve the derived quadratic equation for x,
[tex]\begin{gathered} 5x^2+8x+3=0 \\ 5x^2+5x+3x+3=0 \\ 5x(x+1)+3(x+1)=0 \\ (5x+3)(x+1)=0 \\ 5x+3=0\text{ or }x+1=0 \\ x=-\frac{3}{5}\text{ or -1} \end{gathered}[/tex]We then solve for the corresponding values of y using any of the equations.
[tex]\begin{gathered} \text{When x=-1} \\ y=2x+2 \\ y=2(-1)+2 \\ y=0 \\ When\text{ }x=-\frac{3}{5} \\ y=2(-\frac{3}{5})+2 \\ =\frac{4}{5} \end{gathered}[/tex]Therefore, the solutions o this system are:
(-1,0) and (-3/5, 4/5).
The other solution is (-3/5, 4/5).
