Precalculus Examples

$x + y = 1$ , $x^{2} + x y - y^{2} = - 11$

Step 1

Subtract $y$ from both sides of the equation.

$x = 1 - y$

$x^{2} + x y - y^{2} = - 11$

Step 2

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

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Step 2.1

Replace all occurrences of $x$ in $x^{2} + x y - y^{2} = - 11$ with $1 - y$ .

${(1 - y)}^{2} + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

Step 2.2

Simplify the left side.

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Step 2.2.1

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

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

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

$(1 - y) (1 - y) + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.1.2

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

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Step 2.2.1.1.2.1

Apply the distributive property.

$1 (1 - y) - y (1 - y) + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.1.2.2

Apply the distributive property.

$1 \cdot 1 + 1 (- y) - y (1 - y) + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.1.2.3

Apply the distributive property.

$1 \cdot 1 + 1 (- y) - y \cdot 1 - y (- y) + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

$1 \cdot 1 + 1 (- y) - y \cdot 1 - y (- y) + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

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

Multiply $1$ by $1$ .

$1 + 1 (- y) - y \cdot 1 - y (- y) + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.1.3.1.2

Multiply $- y$ by $1$ .

$1 - y - y \cdot 1 - y (- y) + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.1.3.1.3

Multiply $- 1$ by $1$ .

$1 - y - y - y (- y) + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.1.3.1.4

Rewrite using the commutative property of multiplication.

$1 - y - y - 1 \cdot (- 1 y \cdot y) + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.1.3.1.5

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

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Step 2.2.1.1.3.1.5.1

Move $y$ .

$1 - y - y - 1 \cdot (- 1 (y \cdot y)) + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.1.3.1.5.2

Multiply $y$ by $y$ .

$1 - y - y - 1 \cdot (- 1 y^{2}) + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

$1 - y - y - 1 \cdot (- 1 y^{2}) + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.1.3.1.6

Multiply $- 1$ by $- 1$ .

$1 - y - y + 1 y^{2} + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.1.3.1.7

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

$1 - y - y + y^{2} + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

$1 - y - y + y^{2} + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.1.3.2

Subtract $y$ from $- y$ .

$1 - 2 y + y^{2} + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

$1 - 2 y + y^{2} + (1 - y) \cdot y - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.1.4

Apply the distributive property.

$1 - 2 y + y^{2} + 1 y - y \cdot y - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.1.5

Multiply $y$ by $1$ .

$1 - 2 y + y^{2} + y - y \cdot y - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.1.6

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

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Step 2.2.1.1.6.1

Move $y$ .

$1 - 2 y + y^{2} + y - (y \cdot y) - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.1.6.2

Multiply $y$ by $y$ .

$1 - 2 y + y^{2} + y - y^{2} - y^{2} = - 11$

$x = 1 - y$

$1 - 2 y + y^{2} + y - y^{2} - y^{2} = - 11$

$x = 1 - y$

$1 - 2 y + y^{2} + y - y^{2} - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.2

Simplify by adding terms.

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Step 2.2.1.2.1

Combine the opposite terms in $1 - 2 y + y^{2} + y - y^{2} - y^{2}$ .

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Step 2.2.1.2.1.1

Subtract $y^{2}$ from $y^{2}$ .

$1 - 2 y + y + 0 - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.2.1.2

Add $1 - 2 y + y$ and $0$ .

$1 - 2 y + y - y^{2} = - 11$

$x = 1 - y$

$1 - 2 y + y - y^{2} = - 11$

$x = 1 - y$

Step 2.2.1.2.2

Add $- 2 y$ and $y$ .

$1 - y - y^{2} = - 11$

$x = 1 - y$

$1 - y - y^{2} = - 11$

$x = 1 - y$

$1 - y - y^{2} = - 11$

$x = 1 - y$

$1 - y - y^{2} = - 11$

$x = 1 - y$

$1 - y - y^{2} = - 11$

$x = 1 - y$

Step 3

Solve for $y$ in $1 - y - y^{2} = - 11$ .

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Step 3.1

Add $11$ to both sides of the equation.

$1 - y - y^{2} + 11 = 0$

$x = 1 - y$

Step 3.2

Add $1$ and $11$ .

$- y - y^{2} + 12 = 0$

$x = 1 - y$

Step 3.3

Factor the left side of the equation.

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Step 3.3.1

Factor $- 1$ out of $- y - y^{2} + 12$ .

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Step 3.3.1.1

Reorder $- y$ and $- y^{2}$ .

$- y^{2} - y + 12 = 0$

$x = 1 - y$

Step 3.3.1.2

Factor $- 1$ out of $- y^{2}$ .

$- (y^{2}) - y + 12 = 0$

$x = 1 - y$

Step 3.3.1.3

Factor $- 1$ out of $- y$ .

$- (y^{2}) - (y) + 12 = 0$

$x = 1 - y$

Step 3.3.1.4

Rewrite $12$ as $- 1 (- 12)$ .

$- (y^{2}) - (y) - 1 \cdot - 12 = 0$

$x = 1 - y$

Step 3.3.1.5

Factor $- 1$ out of $- (y^{2}) - (y)$ .

$- (y^{2} + y) - 1 \cdot - 12 = 0$

$x = 1 - y$

Step 3.3.1.6

Factor $- 1$ out of $- (y^{2} + y) - 1 (- 12)$ .

$- (y^{2} + y - 12) = 0$

$x = 1 - y$

$- (y^{2} + y - 12) = 0$

$x = 1 - y$

Step 3.3.2

Factor.

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Step 3.3.2.1

Factor $y^{2} + y - 12$ using the AC method.

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Step 3.3.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $1$ .

$- 3, 4$

$x = 1 - y$

Step 3.3.2.1.2

Write the factored form using these integers.

$- ((y - 3) (y + 4)) = 0$

$x = 1 - y$

$- ((y - 3) (y + 4)) = 0$

$x = 1 - y$

Step 3.3.2.2

Remove unnecessary parentheses.

$- (y - 3) (y + 4) = 0$

$x = 1 - y$

$- (y - 3) (y + 4) = 0$

$x = 1 - y$

$- (y - 3) (y + 4) = 0$

$x = 1 - y$

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 $0$ .

$y - 3 = 0$

$y + 4 = 0$

$x = 1 - y$

Step 3.5

Set $y - 3$ equal to $0$ and solve for $y$ .

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Step 3.5.1

Set $y - 3$ equal to $0$ .

$y - 3 = 0$

$x = 1 - y$

Step 3.5.2

Add $3$ to both sides of the equation.

$y = 3$

$x = 1 - y$

$y = 3$

$x = 1 - y$

Step 3.6

Set $y + 4$ equal to $0$ and solve for $y$ .

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Step 3.6.1

Set $y + 4$ equal to $0$ .

$y + 4 = 0$

$x = 1 - y$

Step 3.6.2

Subtract $4$ from both sides of the equation.

$y = - 4$

$x = 1 - y$

$y = - 4$

$x = 1 - y$

Step 3.7

The final solution is all the values that make $- (y - 3) (y + 4) = 0$ true.

$y = 3, - 4$

$x = 1 - y$

$y = 3, - 4$

$x = 1 - y$

Step 4

Replace all occurrences of $y$ with $3$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = 1 - y$ with $3$ .

$x = 1 - (3)$

$y = 3$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $1 - (3)$ .

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Step 4.2.1.1

Multiply $- 1$ by $3$ .

$x = 1 - 3$

$y = 3$

Step 4.2.1.2

Subtract $3$ from $1$ .

$x = - 2$

$y = 3$

$x = - 2$

$y = 3$

$x = - 2$

$y = 3$

$x = - 2$

$y = 3$

Step 5

Replace all occurrences of $y$ with $- 4$ in each equation.

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Step 5.1

Replace all occurrences of $y$ in $x = 1 - y$ with $- 4$ .

$x = 1 - (- 4)$

$y = - 4$

Step 5.2

Simplify the right side.

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Step 5.2.1

Simplify $1 - (- 4)$ .

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Step 5.2.1.1

Multiply $- 1$ by $- 4$ .

$x = 1 + 4$

$y = - 4$

Step 5.2.1.2

Add $1$ and $4$ .

$x = 5$

$y = - 4$

$x = 5$

$y = - 4$

$x = 5$

$y = - 4$

$x = 5$

$y = - 4$

Step 6

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

$(- 2, 3)$

$(5, - 4)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(- 2, 3), (5, - 4)$

Equation Form:

$x = - 2, y = 3$

$x = 5, y = - 4$

Step 8