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$We want to solve the following system of equations. {[x^(2)+y^(2)=1],[y=2x+2]:} One of the solutions to this system is (-1,0). Find the other solution. Your answer must be exact.$

We want to solve the following system of equations. $\newline$ $\left\{\begin{array}{l} x^{2}+y^{2}=1 \\ y=2 x+2 \end{array}\right.$ $\newline$ One of the solutions to this system is $(-1,0)$ . $\newline$ Find the other solution. Your answer must be exact.

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Grade 8

Find what percent one number is of another: word problems

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Q. We want to solve the following system of equations.

\newline

\left\{\begin{array}{l} x^{2}+y^{2}=1 \\ y=2 x+2 \end{array}\right.

\newline

One of the solutions to this system is

(-1,0)

\newline

Find the other solution. Your answer must be exact.

Given Equations: We are given the system of equations: $\newline$ $1$ . $x^2 + y^2 = 1$ $\newline$ $2$ . $y = 2x + 2$ $\newline$ We need to find the solution to this system other than $(-1,0)$ .
Substituting $y$ in the first equation: Substitute the expression for $y$ from the second equation into the first equation to eliminate $y$ and solve for $x$ . $\newline$ So, we replace $y$ in the first equation with $(2x + 2)$ to get: $\newline$ $x^2 + (2x + 2)^2 = 1$
Expanding the squared term: Expand the squared term $(2x + 2)^2$ to simplify the equation: $\newline$ $x^2 + (4x^2 + 8x + 4) = 1$
Combining like terms: Combine like terms to form a quadratic equation: $\newline$ $x^2 + 4x^2 + 8x + 4 = 1$ $\newline$ $5x^2 + 8x + 4 = 1$
Setting the quadratic equation to zero: Subtract $1$ from both sides to set the quadratic equation to zero: $\newline$ $5x^2 + 8x + 4 - 1 = 0$ $\newline$ $5x^2 + 8x + 3 = 0$
Factoring the quadratic equation: Factor the quadratic equation to find the values of $x$ : $(5x + 3)(x + 1) = 0$
Solving for x: Set each factor equal to zero and solve for x: $\newline$ $5x + 3 = 0$ or $x + 1 = 0$ $\newline$ $x = -\frac{3}{5}$ or $x = -1$ $\newline$ Since we already know that $x = -1$ is part of the solution $(-1,0)$ , we will use $x = -\frac{3}{5}$ for the other solution.
Substituting $x$ into the second equation: Substitute $x = -\frac{3}{5}$ into the second equation $y = 2x + 2$ to find the corresponding value of $y$ :
$y = 2\left(-\frac{3}{5}\right) + 2$
$y = -\frac{6}{5} + 2$
$y = -\frac{6}{5} + \frac{10}{5}$
$y = \frac{4}{5}$
Final Solution: The other solution to the system of equations is $(-\frac{3}{5}, \frac{4}{5})$ .