Precalculus Examples

$x + y + 1 = 0$ , $x^{2} + y^{2} + 2 y - 3 x = - 1$

Step 1

Move all terms not containing $x$ to the right side of the equation.

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Step 1.1

Subtract $y$ from both sides of the equation.

$x + 1 = - y$

$x^{2} + y^{2} + 2 y - 3 x = - 1$

Step 1.2

Subtract $1$ from both sides of the equation.

$x = - y - 1$

$x^{2} + y^{2} + 2 y - 3 x = - 1$

$x = - y - 1$

$x^{2} + y^{2} + 2 y - 3 x = - 1$

Step 2

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

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Step 2.1

Replace all occurrences of $x$ in $x^{2} + y^{2} + 2 y - 3 x = - 1$ with $- y - 1$ .

${(- y - 1)}^{2} + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify ${(- y - 1)}^{2} + y^{2} + 2 y - 3 (- y - 1)$ .

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Rewrite ${(- y - 1)}^{2}$ as $(- y - 1) (- y - 1)$ .

$(- y - 1) (- y - 1) + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

Step 2.2.1.1.2

Expand $(- y - 1) (- y - 1)$ using the FOIL Method.

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Step 2.2.1.1.2.1

Apply the distributive property.

$- y (- y - 1) - 1 (- y - 1) + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

Step 2.2.1.1.2.2

Apply the distributive property.

$- y (- y) - y \cdot - 1 - 1 (- y - 1) + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

Step 2.2.1.1.2.3

Apply the distributive property.

$- y (- y) - y \cdot - 1 - 1 (- y) - 1 \cdot - 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

$- y (- y) - y \cdot - 1 - 1 (- y) - 1 \cdot - 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

Step 2.2.1.1.3

Simplify and combine like terms.

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Step 2.2.1.1.3.1

Simplify each term.

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Step 2.2.1.1.3.1.1

Rewrite using the commutative property of multiplication.

$- 1 \cdot (- 1 y \cdot y) - y \cdot - 1 - 1 (- y) - 1 \cdot - 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

Step 2.2.1.1.3.1.2

Multiply $y$ by $y$ by adding the exponents.

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Step 2.2.1.1.3.1.2.1

Move $y$ .

$- 1 \cdot (- 1 (y \cdot y)) - y \cdot - 1 - 1 (- y) - 1 \cdot - 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

Step 2.2.1.1.3.1.2.2

Multiply $y$ by $y$ .

$- 1 \cdot (- 1 y^{2}) - y \cdot - 1 - 1 (- y) - 1 \cdot - 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

$- 1 \cdot (- 1 y^{2}) - y \cdot - 1 - 1 (- y) - 1 \cdot - 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

Step 2.2.1.1.3.1.3

Multiply $- 1$ by $- 1$ .

$1 y^{2} - y \cdot - 1 - 1 (- y) - 1 \cdot - 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

Step 2.2.1.1.3.1.4

Multiply $y^{2}$ by $1$ .

$y^{2} - y \cdot - 1 - 1 (- y) - 1 \cdot - 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

Step 2.2.1.1.3.1.5

Multiply $- y \cdot - 1$ .

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Step 2.2.1.1.3.1.5.1

Multiply $- 1$ by $- 1$ .

$y^{2} + 1 y - 1 (- y) - 1 \cdot - 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

Step 2.2.1.1.3.1.5.2

Multiply $y$ by $1$ .

$y^{2} + y - 1 (- y) - 1 \cdot - 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

$y^{2} + y - 1 (- y) - 1 \cdot - 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

Step 2.2.1.1.3.1.6

Multiply $- 1 (- y)$ .

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Step 2.2.1.1.3.1.6.1

Multiply $- 1$ by $- 1$ .

$y^{2} + y + 1 y - 1 \cdot - 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

Step 2.2.1.1.3.1.6.2

Multiply $y$ by $1$ .

$y^{2} + y + y - 1 \cdot - 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

$y^{2} + y + y - 1 \cdot - 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

Step 2.2.1.1.3.1.7

Multiply $- 1$ by $- 1$ .

$y^{2} + y + y + 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

$y^{2} + y + y + 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

Step 2.2.1.1.3.2

Add $y$ and $y$ .

$y^{2} + 2 y + 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

$y^{2} + 2 y + 1 + y^{2} + 2 y - 3 (- y - 1) = - 1$

$x = - y - 1$

Step 2.2.1.1.4

Apply the distributive property.

$y^{2} + 2 y + 1 + y^{2} + 2 y - 3 (- y) - 3 \cdot - 1 = - 1$

$x = - y - 1$

Step 2.2.1.1.5

Multiply $- 1$ by $- 3$ .

$y^{2} + 2 y + 1 + y^{2} + 2 y + 3 y - 3 \cdot - 1 = - 1$

$x = - y - 1$

Step 2.2.1.1.6

Multiply $- 3$ by $- 1$ .

$y^{2} + 2 y + 1 + y^{2} + 2 y + 3 y + 3 = - 1$

$x = - y - 1$

$y^{2} + 2 y + 1 + y^{2} + 2 y + 3 y + 3 = - 1$

$x = - y - 1$

Step 2.2.1.2

Simplify by adding terms.

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Step 2.2.1.2.1

Add $y^{2}$ and $y^{2}$ .

$2 y^{2} + 2 y + 1 + 2 y + 3 y + 3 = - 1$

$x = - y - 1$

Step 2.2.1.2.2

Add $2 y$ and $2 y$ .

$2 y^{2} + 4 y + 1 + 3 y + 3 = - 1$

$x = - y - 1$

Step 2.2.1.2.3

Add $4 y$ and $3 y$ .

$2 y^{2} + 7 y + 1 + 3 = - 1$

$x = - y - 1$

Step 2.2.1.2.4

Add $1$ and $3$ .

$2 y^{2} + 7 y + 4 = - 1$

$x = - y - 1$

$2 y^{2} + 7 y + 4 = - 1$

$x = - y - 1$

$2 y^{2} + 7 y + 4 = - 1$

$x = - y - 1$

$2 y^{2} + 7 y + 4 = - 1$

$x = - y - 1$

$2 y^{2} + 7 y + 4 = - 1$

$x = - y - 1$

Step 3

Solve for $y$ in $2 y^{2} + 7 y + 4 = - 1$ .

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Step 3.1

Add $1$ to both sides of the equation.

$2 y^{2} + 7 y + 4 + 1 = 0$

$x = - y - 1$

Step 3.2

Add $4$ and $1$ .

$2 y^{2} + 7 y + 5 = 0$

$x = - y - 1$

Step 3.3

Factor by grouping.

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Step 3.3.1

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 $a \cdot c = 2 \cdot 5 = 10$ and whose sum is $b = 7$ .

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Step 3.3.1.1

Factor $7$ out of $7 y$ .

$2 y^{2} + 7 (y) + 5 = 0$

$x = - y - 1$

Step 3.3.1.2

Rewrite $7$ as $2$ plus $5$

$2 y^{2} + (2 + 5) y + 5 = 0$

$x = - y - 1$

Step 3.3.1.3

Apply the distributive property.

$2 y^{2} + 2 y + 5 y + 5 = 0$

$x = - y - 1$

$2 y^{2} + 2 y + 5 y + 5 = 0$

$x = - y - 1$

Step 3.3.2

Factor out the greatest common factor from each group.

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Step 3.3.2.1

Group the first two terms and the last two terms.

$(2 y^{2} + 2 y) + 5 y + 5 = 0$

$x = - y - 1$

Step 3.3.2.2

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

$2 y (y + 1) + 5 (y + 1) = 0$

$x = - y - 1$

$2 y (y + 1) + 5 (y + 1) = 0$

$x = - y - 1$

Step 3.3.3

Factor the polynomial by factoring out the greatest common factor, $y + 1$ .

$(y + 1) (2 y + 5) = 0$

$x = - y - 1$

$(y + 1) (2 y + 5) = 0$

$x = - y - 1$

Step 3.4

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

$y + 1 = 0$

$2 y + 5 = 0$

$x = - y - 1$

Step 3.5

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

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Step 3.5.1

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

$y + 1 = 0$

$x = - y - 1$

Step 3.5.2

Subtract $1$ from both sides of the equation.

$y = - 1$

$x = - y - 1$

$y = - 1$

$x = - y - 1$

Step 3.6

Set $2 y + 5$ equal to $0$ and solve for $y$ .

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Step 3.6.1

Set $2 y + 5$ equal to $0$ .

$2 y + 5 = 0$

$x = - y - 1$

Step 3.6.2

Solve $2 y + 5 = 0$ for $y$ .

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Step 3.6.2.1

Subtract $5$ from both sides of the equation.

$2 y = - 5$

$x = - y - 1$

Step 3.6.2.2

Divide each term in $2 y = - 5$ by $2$ and simplify.

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Step 3.6.2.2.1

Divide each term in $2 y = - 5$ by $2$ .

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

$x = - y - 1$

Step 3.6.2.2.2

Simplify the left side.

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Step 3.6.2.2.2.1

Cancel the common factor of $2$ .

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Step 3.6.2.2.2.1.1

Cancel the common factor.

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

$x = - y - 1$

Step 3.6.2.2.2.1.2

Divide $y$ by $1$ .

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

$x = - y - 1$

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

$x = - y - 1$

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

$x = - y - 1$

Step 3.6.2.2.3

Simplify the right side.

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Step 3.6.2.2.3.1

Move the negative in front of the fraction.

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

$x = - y - 1$

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

$x = - y - 1$

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

$x = - y - 1$

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

$x = - y - 1$

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

$x = - y - 1$

Step 3.7

The final solution is all the values that make $(y + 1) (2 y + 5) = 0$ true.

$y = - 1, - \frac{5}{2}$

$x = - y - 1$

$y = - 1, - \frac{5}{2}$

$x = - y - 1$

Step 4

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

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Step 4.1

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

$x = - (- 1) - 1$

$y = - 1$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $- (- 1) - 1$ .

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Step 4.2.1.1

Multiply $- 1$ by $- 1$ .

$x = 1 - 1$

$y = - 1$

Step 4.2.1.2

Subtract $1$ from $1$ .

$x = 0$

$y = - 1$

$x = 0$

$y = - 1$

$x = 0$

$y = - 1$

$x = 0$

$y = - 1$

Step 5

Replace all occurrences of $y$ with $- \frac{5}{2}$ in each equation.

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Step 5.1

Replace all occurrences of $y$ in $x = - y - 1$ with $- \frac{5}{2}$ .

$x = - (- \frac{5}{2}) - 1$

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

Step 5.2

Simplify the right side.

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Step 5.2.1

Simplify $- (- \frac{5}{2}) - 1$ .

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Step 5.2.1.1

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

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Step 5.2.1.1.1

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{5}{2}) - 1$

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

Step 5.2.1.1.2

Multiply $\frac{5}{2}$ by $1$ .

$x = \frac{5}{2} - 1$

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

$x = \frac{5}{2} - 1$

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

Step 5.2.1.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = \frac{5}{2} - 1 \cdot \frac{2}{2}$

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

Step 5.2.1.3

Combine $- 1$ and $\frac{2}{2}$ .

$x = \frac{5}{2} + \frac{- 1 \cdot 2}{2}$

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

Step 5.2.1.4

Combine the numerators over the common denominator.

$x = \frac{5 - 1 \cdot 2}{2}$

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

Step 5.2.1.5

Simplify the numerator.

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Step 5.2.1.5.1

Multiply $- 1$ by $2$ .

$x = \frac{5 - 2}{2}$

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

Step 5.2.1.5.2

Subtract $2$ from $5$ .

$x = \frac{3}{2}$

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

$x = \frac{3}{2}$

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

$x = \frac{3}{2}$

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

$x = \frac{3}{2}$

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

$x = \frac{3}{2}$

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

Step 6

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

$(0, - 1)$

$(\frac{3}{2}, - \frac{5}{2})$

Step 7

The result can be shown in multiple forms.

Point Form:

$(0, - 1), (\frac{3}{2}, - \frac{5}{2})$

Equation Form:

$x = 0, y = - 1$

$x = \frac{3}{2}, y = - \frac{5}{2}$

Step 8