we want to solve the following system of equations.\n$\begin{cases}x^{2}+y^{2}=1\\y = 2x + 2end{cases}$\none…

Explanation:

Step1: Substitute $y$ into the first - equation

Substitute $y = 2x + 2$ into $x^{2}+y^{2}=1$. We get $x^{2}+(2x + 2)^{2}=1$. Expand $(2x + 2)^{2}$ using the formula $(a + b)^{2}=a^{2}+2ab + b^{2}$, where $a = 2x$ and $b = 2$. So $(2x + 2)^{2}=4x^{2}+8x + 4$. Then the equation becomes $x^{2}+4x^{2}+8x + 4 = 1$. Combine like - terms: $5x^{2}+8x+3 = 0$.

Step2: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 5$, $b = 8$, $c = 3$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor the equation. Factor $5x^{2}+8x + 3$: $5x^{2}+8x + 3=5x^{2}+5x+3x + 3=5x(x + 1)+3(x + 1)=(5x + 3)(x + 1)=0$. Set each factor equal to zero: $x+1 = 0$ gives $x=-1$ (which is the known solution). $5x+3 = 0$ gives $x=-\frac{3}{5}$.

Step3: Find the corresponding $y$ - value

Substitute $x = -\frac{3}{5}$ into $y = 2x+2$. $y=2\times(-\frac{3}{5})+2=\frac{-6 + 10}{5}=\frac{4}{5}$.

Answer:

$(-\frac{3}{5},\frac{4}{5})$