Solve the System of Equations x^2+y^2=1 y=2x+2

$x^{2} + y^{2} = 1$ $y = 2 x + 2$

Step 1

Replace all occurrences of $y$ with $2 x + 2$ in each equation.

Replace all occurrences of $y$ in $x^{2} + y^{2} = 1$ with $2 x + 2$ .

$x^{2} + {(2 x + 2)}^{2} = 1$

$y = 2 x + 2$

Simplify the left side.

Simplify $x^{2} + {(2 x + 2)}^{2}$ .

Simplify each term.

Rewrite ${(2 x + 2)}^{2}$ as $(2 x + 2) (2 x + 2)$ .

$x^{2} + (2 x + 2) (2 x + 2) = 1$

$y = 2 x + 2$

Expand $(2 x + 2) (2 x + 2)$ using the FOIL Method.

Apply the distributive property.

$x^{2} + 2 x (2 x + 2) + 2 (2 x + 2) = 1$

$y = 2 x + 2$

Apply the distributive property.

$x^{2} + 2 x (2 x) + 2 x \cdot 2 + 2 (2 x + 2) = 1$

$y = 2 x + 2$

Apply the distributive property.

$x^{2} + 2 x (2 x) + 2 x \cdot 2 + 2 (2 x) + 2 \cdot 2 = 1$

$y = 2 x + 2$

$x^{2} + 2 x (2 x) + 2 x \cdot 2 + 2 (2 x) + 2 \cdot 2 = 1$

$y = 2 x + 2$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$x^{2} + 2 \cdot (2 x \cdot x) + 2 x \cdot 2 + 2 (2 x) + 2 \cdot 2 = 1$

$y = 2 x + 2$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$x^{2} + 2 \cdot (2 (x \cdot x)) + 2 x \cdot 2 + 2 (2 x) + 2 \cdot 2 = 1$

$y = 2 x + 2$

Multiply $x$ by $x$ .

$x^{2} + 2 \cdot (2 x^{2}) + 2 x \cdot 2 + 2 (2 x) + 2 \cdot 2 = 1$

$y = 2 x + 2$

$x^{2} + 2 \cdot (2 x^{2}) + 2 x \cdot 2 + 2 (2 x) + 2 \cdot 2 = 1$

$y = 2 x + 2$

Multiply $2$ by $2$ .

$x^{2} + 4 x^{2} + 2 x \cdot 2 + 2 (2 x) + 2 \cdot 2 = 1$

$y = 2 x + 2$

Multiply $2$ by $2$ .

$x^{2} + 4 x^{2} + 4 x + 2 (2 x) + 2 \cdot 2 = 1$

$y = 2 x + 2$

Multiply $2$ by $2$ .

$x^{2} + 4 x^{2} + 4 x + 4 x + 2 \cdot 2 = 1$

$y = 2 x + 2$

Multiply $2$ by $2$ .

$x^{2} + 4 x^{2} + 4 x + 4 x + 4 = 1$

$y = 2 x + 2$

$x^{2} + 4 x^{2} + 4 x + 4 x + 4 = 1$

$y = 2 x + 2$

Add $4 x$ and $4 x$ .

$x^{2} + 4 x^{2} + 8 x + 4 = 1$

$y = 2 x + 2$

$x^{2} + 4 x^{2} + 8 x + 4 = 1$

$y = 2 x + 2$

$x^{2} + 4 x^{2} + 8 x + 4 = 1$

$y = 2 x + 2$

Add $x^{2}$ and $4 x^{2}$ .

$5 x^{2} + 8 x + 4 = 1$

$y = 2 x + 2$

$5 x^{2} + 8 x + 4 = 1$

$y = 2 x + 2$

$5 x^{2} + 8 x + 4 = 1$

$y = 2 x + 2$

$5 x^{2} + 8 x + 4 = 1$

$y = 2 x + 2$

Step 2

Solve for $x$ in $5 x^{2} + 8 x + 4 = 1$ .

Subtract $1$ from both sides of the equation.

$5 x^{2} + 8 x + 4 - 1 = 0$

$y = 2 x + 2$

Subtract $1$ from $4$ .

$5 x^{2} + 8 x + 3 = 0$

$y = 2 x + 2$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 5 \cdot 3 = 15$ and whose sum is $b = 8$ .

Factor $8$ out of $8 x$ .

$5 x^{2} + 8 (x) + 3 = 0$

$y = 2 x + 2$

Rewrite $8$ as $3$ plus $5$

$5 x^{2} + (3 + 5) x + 3 = 0$

$y = 2 x + 2$

Apply the distributive property.

$5 x^{2} + 3 x + 5 x + 3 = 0$

$y = 2 x + 2$

$5 x^{2} + 3 x + 5 x + 3 = 0$

$y = 2 x + 2$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(5 x^{2} + 3 x) + 5 x + 3 = 0$

$y = 2 x + 2$

Factor out the greatest common factor (GCF) from each group.

$x (5 x + 3) + 1 (5 x + 3) = 0$

$y = 2 x + 2$

$x (5 x + 3) + 1 (5 x + 3) = 0$

$y = 2 x + 2$

Factor the polynomial by factoring out the greatest common factor, $5 x + 3$ .

$(5 x + 3) (x + 1) = 0$

$y = 2 x + 2$

$(5 x + 3) (x + 1) = 0$

$y = 2 x + 2$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$5 x + 3 = 0$

$x + 1 = 0$

$y = 2 x + 2$

Set $5 x + 3$ equal to $0$ and solve for $x$ .

Set $5 x + 3$ equal to $0$ .

$5 x + 3 = 0$

$y = 2 x + 2$

Solve $5 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$5 x = - 3$

$y = 2 x + 2$

Divide each term in $5 x = - 3$ by $5$ and simplify.

Divide each term in $5 x = - 3$ by $5$ .

$\frac{5 x}{5} = \frac{- 3}{5}$

$y = 2 x + 2$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{- 3}{5}$

$y = 2 x + 2$

Divide $x$ by $1$ .

$x = \frac{- 3}{5}$

$y = 2 x + 2$

$x = \frac{- 3}{5}$

$y = 2 x + 2$

$x = \frac{- 3}{5}$

$y = 2 x + 2$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{5}$

$y = 2 x + 2$

$x = - \frac{3}{5}$

$y = 2 x + 2$

$x = - \frac{3}{5}$

$y = 2 x + 2$

$x = - \frac{3}{5}$

$y = 2 x + 2$

$x = - \frac{3}{5}$

$y = 2 x + 2$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

$y = 2 x + 2$

Subtract $1$ from both sides of the equation.

$x = - 1$

$y = 2 x + 2$

$x = - 1$

$y = 2 x + 2$

The final solution is all the values that make $(5 x + 3) (x + 1) = 0$ true.

$x = - \frac{3}{5}, - 1$

$y = 2 x + 2$

$x = - \frac{3}{5}, - 1$

$y = 2 x + 2$

Step 3

Replace all occurrences of $x$ with $- \frac{3}{5}$ in each equation.

Replace all occurrences of $x$ in $y = 2 x + 2$ with $- \frac{3}{5}$ .

$y = 2 (- \frac{3}{5}) + 2$

$x = - \frac{3}{5}$

Simplify the right side.

Simplify $2 (- \frac{3}{5}) + 2$ .

Simplify each term.

Multiply $2 (- \frac{3}{5})$ .

Multiply $- 1$ by $2$ .

$y = - 2 (\frac{3}{5}) + 2$

$x = - \frac{3}{5}$

Combine $- 2$ and $\frac{3}{5}$ .

$y = \frac{- 2 \cdot 3}{5} + 2$

$x = - \frac{3}{5}$

Multiply $- 2$ by $3$ .

$y = \frac{- 6}{5} + 2$

$x = - \frac{3}{5}$

$y = \frac{- 6}{5} + 2$

$x = - \frac{3}{5}$

Move the negative in front of the fraction.

$y = - \frac{6}{5} + 2$

$x = - \frac{3}{5}$

$y = - \frac{6}{5} + 2$

$x = - \frac{3}{5}$

To write $2$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$y = - \frac{6}{5} + 2 \cdot \frac{5}{5}$

$x = - \frac{3}{5}$

Combine $2$ and $\frac{5}{5}$ .

$y = - \frac{6}{5} + \frac{2 \cdot 5}{5}$

$x = - \frac{3}{5}$

Combine the numerators over the common denominator.

$y = \frac{- 6 + 2 \cdot 5}{5}$

$x = - \frac{3}{5}$

Simplify the numerator.

Multiply $2$ by $5$ .

$y = \frac{- 6 + 10}{5}$

$x = - \frac{3}{5}$

Add $- 6$ and $10$ .

$y = \frac{4}{5}$

$x = - \frac{3}{5}$

$y = \frac{4}{5}$

$x = - \frac{3}{5}$

$y = \frac{4}{5}$

$x = - \frac{3}{5}$

$y = \frac{4}{5}$

$x = - \frac{3}{5}$

$y = \frac{4}{5}$

$x = - \frac{3}{5}$

Step 4

Replace all occurrences of $x$ with $- 1$ in each equation.

Replace all occurrences of $x$ in $y = 2 x + 2$ with $- 1$ .

$y = 2 (- 1) + 2$

$x = - 1$

Simplify the right side.

Simplify $2 (- 1) + 2$ .

Multiply $2$ by $- 1$ .

$y = - 2 + 2$

$x = - 1$

Add $- 2$ and $2$ .

$y = 0$

$x = - 1$

$y = 0$

$x = - 1$

$y = 0$

$x = - 1$

$y = 0$

$x = - 1$

Step 5

The solution to the system is the complete set of ordered pairs that are valid solutions.

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

$(- 1, 0)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- \frac{3}{5}, \frac{4}{5}), (- 1, 0)$

Equation Form:

$x = - \frac{3}{5}, y = \frac{4}{5}$

$x = - 1, y = 0$

Step 7

Solve the System of Equations x^2+y^2=1 y=2x+2

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