Precalculus Examples

$x y - x^{2} = - 20$ , $x - 2 y = 3$

Step 1

Add $2 y$ to both sides of the equation.

$x = 3 + 2 y$

$x y - x^{2} = - 20$

Step 2

Replace all occurrences of $x$ with $3 + 2 y$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $x y - x^{2} = - 20$ with $3 + 2 y$ .

$(3 + 2 y) \cdot y - {(3 + 2 y)}^{2} = - 20$

$x = 3 + 2 y$

Step 2.2

Simplify the left side.

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Step 2.2.1

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

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Apply the distributive property.

$3 y + 2 y \cdot y - {(3 + 2 y)}^{2} = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.2

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

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Step 2.2.1.1.2.1

Move $y$ .

$3 y + 2 (y \cdot y) - {(3 + 2 y)}^{2} = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.2.2

Multiply $y$ by $y$ .

$3 y + 2 y^{2} - {(3 + 2 y)}^{2} = - 20$

$x = 3 + 2 y$

$3 y + 2 y^{2} - {(3 + 2 y)}^{2} = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.3

Rewrite ${(3 + 2 y)}^{2}$ as $(3 + 2 y) (3 + 2 y)$ .

$3 y + 2 y^{2} - ((3 + 2 y) (3 + 2 y)) = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.4

Expand $(3 + 2 y) (3 + 2 y)$ using the FOIL Method.

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Step 2.2.1.1.4.1

Apply the distributive property.

$3 y + 2 y^{2} - (3 (3 + 2 y) + 2 y (3 + 2 y)) = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.4.2

Apply the distributive property.

$3 y + 2 y^{2} - (3 \cdot 3 + 3 (2 y) + 2 y (3 + 2 y)) = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.4.3

Apply the distributive property.

$3 y + 2 y^{2} - (3 \cdot 3 + 3 (2 y) + 2 y \cdot 3 + 2 y (2 y)) = - 20$

$x = 3 + 2 y$

$3 y + 2 y^{2} - (3 \cdot 3 + 3 (2 y) + 2 y \cdot 3 + 2 y (2 y)) = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.5

Simplify and combine like terms.

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Step 2.2.1.1.5.1

Simplify each term.

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Step 2.2.1.1.5.1.1

Multiply $3$ by $3$ .

$3 y + 2 y^{2} - (9 + 3 (2 y) + 2 y \cdot 3 + 2 y (2 y)) = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.5.1.2

Multiply $2$ by $3$ .

$3 y + 2 y^{2} - (9 + 6 y + 2 y \cdot 3 + 2 y (2 y)) = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.5.1.3

Multiply $3$ by $2$ .

$3 y + 2 y^{2} - (9 + 6 y + 6 y + 2 y (2 y)) = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.5.1.4

Rewrite using the commutative property of multiplication.

$3 y + 2 y^{2} - (9 + 6 y + 6 y + 2 \cdot (2 y \cdot y)) = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.5.1.5

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

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Step 2.2.1.1.5.1.5.1

Move $y$ .

$3 y + 2 y^{2} - (9 + 6 y + 6 y + 2 \cdot (2 (y \cdot y))) = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.5.1.5.2

Multiply $y$ by $y$ .

$3 y + 2 y^{2} - (9 + 6 y + 6 y + 2 \cdot (2 y^{2})) = - 20$

$x = 3 + 2 y$

$3 y + 2 y^{2} - (9 + 6 y + 6 y + 2 \cdot (2 y^{2})) = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.5.1.6

Multiply $2$ by $2$ .

$3 y + 2 y^{2} - (9 + 6 y + 6 y + 4 y^{2}) = - 20$

$x = 3 + 2 y$

$3 y + 2 y^{2} - (9 + 6 y + 6 y + 4 y^{2}) = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.5.2

Add $6 y$ and $6 y$ .

$3 y + 2 y^{2} - (9 + 12 y + 4 y^{2}) = - 20$

$x = 3 + 2 y$

$3 y + 2 y^{2} - (9 + 12 y + 4 y^{2}) = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.6

Apply the distributive property.

$3 y + 2 y^{2} - 1 \cdot 9 - (12 y) - (4 y^{2}) = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.7

Simplify.

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Step 2.2.1.1.7.1

Multiply $- 1$ by $9$ .

$3 y + 2 y^{2} - 9 - (12 y) - (4 y^{2}) = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.7.2

Multiply $12$ by $- 1$ .

$3 y + 2 y^{2} - 9 - 12 y - (4 y^{2}) = - 20$

$x = 3 + 2 y$

Step 2.2.1.1.7.3

Multiply $4$ by $- 1$ .

$3 y + 2 y^{2} - 9 - 12 y - 4 y^{2} = - 20$

$x = 3 + 2 y$

$3 y + 2 y^{2} - 9 - 12 y - 4 y^{2} = - 20$

$x = 3 + 2 y$

$3 y + 2 y^{2} - 9 - 12 y - 4 y^{2} = - 20$

$x = 3 + 2 y$

Step 2.2.1.2

Simplify by adding terms.

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Step 2.2.1.2.1

Subtract $12 y$ from $3 y$ .

$- 9 y + 2 y^{2} - 9 - 4 y^{2} = - 20$

$x = 3 + 2 y$

Step 2.2.1.2.2

Subtract $4 y^{2}$ from $2 y^{2}$ .

$- 9 y - 2 y^{2} - 9 = - 20$

$x = 3 + 2 y$

$- 9 y - 2 y^{2} - 9 = - 20$

$x = 3 + 2 y$

$- 9 y - 2 y^{2} - 9 = - 20$

$x = 3 + 2 y$

$- 9 y - 2 y^{2} - 9 = - 20$

$x = 3 + 2 y$

$- 9 y - 2 y^{2} - 9 = - 20$

$x = 3 + 2 y$

Step 3

Solve for $y$ in $- 9 y - 2 y^{2} - 9 = - 20$ .

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Step 3.1

Add $20$ to both sides of the equation.

$- 9 y - 2 y^{2} - 9 + 20 = 0$

$x = 3 + 2 y$

Step 3.2

Add $- 9$ and $20$ .

$- 9 y - 2 y^{2} + 11 = 0$

$x = 3 + 2 y$

Step 3.3

Factor the left side of the equation.

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Step 3.3.1

Factor $- 1$ out of $- 9 y - 2 y^{2} + 11$ .

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Step 3.3.1.1

Reorder $- 9 y$ and $- 2 y^{2}$ .

$- 2 y^{2} - 9 y + 11 = 0$

$x = 3 + 2 y$

Step 3.3.1.2

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

$- (2 y^{2}) - 9 y + 11 = 0$

$x = 3 + 2 y$

Step 3.3.1.3

Factor $- 1$ out of $- 9 y$ .

$- (2 y^{2}) - (9 y) + 11 = 0$

$x = 3 + 2 y$

Step 3.3.1.4

Rewrite $11$ as $- 1 (- 11)$ .

$- (2 y^{2}) - (9 y) - 1 \cdot - 11 = 0$

$x = 3 + 2 y$

Step 3.3.1.5

Factor $- 1$ out of $- (2 y^{2}) - (9 y)$ .

$- (2 y^{2} + 9 y) - 1 \cdot - 11 = 0$

$x = 3 + 2 y$

Step 3.3.1.6

Factor $- 1$ out of $- (2 y^{2} + 9 y) - 1 (- 11)$ .

$- (2 y^{2} + 9 y - 11) = 0$

$x = 3 + 2 y$

$- (2 y^{2} + 9 y - 11) = 0$

$x = 3 + 2 y$

Step 3.3.2

Factor.

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Step 3.3.2.1

Factor by grouping.

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Step 3.3.2.1.1

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 $a \cdot c = 2 \cdot - 11 = - 22$ and whose sum is $b = 9$ .

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Step 3.3.2.1.1.1

Factor $9$ out of $9 y$ .

$- (2 y^{2} + 9 (y) - 11) = 0$

$x = 3 + 2 y$

Step 3.3.2.1.1.2

Rewrite $9$ as $- 2$ plus $11$

$- (2 y^{2} + (- 2 + 11) y - 11) = 0$

$x = 3 + 2 y$

Step 3.3.2.1.1.3

Apply the distributive property.

$- (2 y^{2} - 2 y + 11 y - 11) = 0$

$x = 3 + 2 y$

$- (2 y^{2} - 2 y + 11 y - 11) = 0$

$x = 3 + 2 y$

Step 3.3.2.1.2

Factor out the greatest common factor from each group.

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Step 3.3.2.1.2.1

Group the first two terms and the last two terms.

$- ((2 y^{2} - 2 y) + 11 y - 11) = 0$

$x = 3 + 2 y$

Step 3.3.2.1.2.2

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

$- (2 y (y - 1) + 11 (y - 1)) = 0$

$x = 3 + 2 y$

$- (2 y (y - 1) + 11 (y - 1)) = 0$

$x = 3 + 2 y$

Step 3.3.2.1.3

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

$- ((y - 1) (2 y + 11)) = 0$

$x = 3 + 2 y$

$- ((y - 1) (2 y + 11)) = 0$

$x = 3 + 2 y$

Step 3.3.2.2

Remove unnecessary parentheses.

$- (y - 1) (2 y + 11) = 0$

$x = 3 + 2 y$

$- (y - 1) (2 y + 11) = 0$

$x = 3 + 2 y$

$- (y - 1) (2 y + 11) = 0$

$x = 3 + 2 y$

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 $0$ .

$y - 1 = 0$

$2 y + 11 = 0$

$x = 3 + 2 y$

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 $0$ .

$y - 1 = 0$

$x = 3 + 2 y$

Step 3.5.2

Add $1$ to both sides of the equation.

$y = 1$

$x = 3 + 2 y$

$y = 1$

$x = 3 + 2 y$

Step 3.6

Set $2 y + 11$ equal to $0$ and solve for $y$ .

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Step 3.6.1

Set $2 y + 11$ equal to $0$ .

$2 y + 11 = 0$

$x = 3 + 2 y$

Step 3.6.2

Solve $2 y + 11 = 0$ for $y$ .

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Step 3.6.2.1

Subtract $11$ from both sides of the equation.

$2 y = - 11$

$x = 3 + 2 y$

Step 3.6.2.2

Divide each term in $2 y = - 11$ by $2$ and simplify.

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Step 3.6.2.2.1

Divide each term in $2 y = - 11$ by $2$ .

$\frac{2 y}{2} = \frac{- 11}{2}$

$x = 3 + 2 y$

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.

$\frac{2 y}{2} = \frac{- 11}{2}$

$x = 3 + 2 y$

Step 3.6.2.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 11}{2}$

$x = 3 + 2 y$

$y = \frac{- 11}{2}$

$x = 3 + 2 y$

$y = \frac{- 11}{2}$

$x = 3 + 2 y$

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.

$y = - \frac{11}{2}$

$x = 3 + 2 y$

$y = - \frac{11}{2}$

$x = 3 + 2 y$

$y = - \frac{11}{2}$

$x = 3 + 2 y$

$y = - \frac{11}{2}$

$x = 3 + 2 y$

$y = - \frac{11}{2}$

$x = 3 + 2 y$

Step 3.7

The final solution is all the values that make $- (y - 1) (2 y + 11) = 0$ true.

$y = 1, - \frac{11}{2}$

$x = 3 + 2 y$

$y = 1, - \frac{11}{2}$

$x = 3 + 2 y$

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 $x = 3 + 2 y$ with $1$ .

$x = 3 + 2 (1)$

$y = 1$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $3 + 2 (1)$ .

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Step 4.2.1.1

Multiply $2$ by $1$ .

$x = 3 + 2$

$y = 1$

Step 4.2.1.2

Add $3$ and $2$ .

$x = 5$

$y = 1$

$x = 5$

$y = 1$

$x = 5$

$y = 1$

$x = 5$

$y = 1$

Step 5

Replace all occurrences of $y$ with $- \frac{11}{2}$ in each equation.

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Step 5.1

Replace all occurrences of $y$ in $x = 3 + 2 y$ with $- \frac{11}{2}$ .

$x = 3 + 2 (- \frac{11}{2})$

$y = - \frac{11}{2}$

Step 5.2

Simplify the right side.

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Step 5.2.1

Simplify $3 + 2 (- \frac{11}{2})$ .

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Step 5.2.1.1

Cancel the common factor of $2$ .

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Step 5.2.1.1.1

Move the leading negative in $- \frac{11}{2}$ into the numerator.

$x = 3 + 2 (\frac{- 11}{2})$

$y = - \frac{11}{2}$

Step 5.2.1.1.2

Cancel the common factor.

$x = 3 + 2 (\frac{- 11}{2})$

$y = - \frac{11}{2}$

Step 5.2.1.1.3

Rewrite the expression.

$x = 3 - 11$

$y = - \frac{11}{2}$

$x = 3 - 11$

$y = - \frac{11}{2}$

Step 5.2.1.2

Subtract $11$ from $3$ .

$x = - 8$

$y = - \frac{11}{2}$

$x = - 8$

$y = - \frac{11}{2}$

$x = - 8$

$y = - \frac{11}{2}$

$x = - 8$

$y = - \frac{11}{2}$

Step 6

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

$(5, 1)$

$(- 8, - \frac{11}{2})$

Step 7

The result can be shown in multiple forms.

Point Form:

$(5, 1), (- 8, - \frac{11}{2})$

Equation Form:

$x = 5, y = 1$

$x = - 8, y = - \frac{11}{2}$

Step 8