Precalculus Examples

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

Step 1

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

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Step 1.1

Subtract $x^{2}$ from both sides of the equation.

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

$2 x - y = - 1$

Step 1.2

Divide each term in $- y = 2 - x^{2}$ by $- 1$ and simplify.

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Step 1.2.1

Divide each term in $- y = 2 - x^{2}$ by $- 1$ .

$\frac{- y}{- 1} = \frac{2}{- 1} + \frac{- x^{2}}{- 1}$

$2 x - y = - 1$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{2}{- 1} + \frac{- x^{2}}{- 1}$

$2 x - y = - 1$

Step 1.2.2.2

Divide $y$ by $1$ .

$y = \frac{2}{- 1} + \frac{- x^{2}}{- 1}$

$2 x - y = - 1$

$y = \frac{2}{- 1} + \frac{- x^{2}}{- 1}$

$2 x - y = - 1$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Divide $2$ by $- 1$ .

$y = - 2 + \frac{- x^{2}}{- 1}$

$2 x - y = - 1$

Step 1.2.3.1.2

Dividing two negative values results in a positive value.

$y = - 2 + \frac{x^{2}}{1}$

$2 x - y = - 1$

Step 1.2.3.1.3

Divide $x^{2}$ by $1$ .

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

$2 x - y = - 1$

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

$2 x - y = - 1$

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

$2 x - y = - 1$

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

$2 x - y = - 1$

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

$2 x - y = - 1$

Step 2

Replace all occurrences of $y$ with $- 2 + x^{2}$ in each equation.

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Step 2.1

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

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

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

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Apply the distributive property.

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

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

Step 2.2.1.2

Multiply $- 1$ by $- 2$ .

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

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

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

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

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

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

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

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

Step 3

Solve for $x$ in $2 x + 2 - x^{2} = - 1$ .

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Step 3.1

Add $1$ to both sides of the equation.

$2 x + 2 - x^{2} + 1 = 0$

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

Step 3.2

Add $2$ and $1$ .

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

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

Step 3.3

Factor the left side of the equation.

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Step 3.3.1

Factor $- 1$ out of $2 x - x^{2} + 3$ .

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Step 3.3.1.1

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

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

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

Step 3.3.1.2

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

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

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

Step 3.3.1.3

Factor $- 1$ out of $2 x$ .

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

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

Step 3.3.1.4

Rewrite $3$ as $- 1 (- 3)$ .

$- (x^{2}) - (- 2 x) - 1 \cdot - 3 = 0$

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

Step 3.3.1.5

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

$- (x^{2} - 2 x) - 1 \cdot - 3 = 0$

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

Step 3.3.1.6

Factor $- 1$ out of $- (x^{2} - 2 x) - 1 (- 3)$ .

$- (x^{2} - 2 x - 3) = 0$

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

$- (x^{2} - 2 x - 3) = 0$

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

Step 3.3.2

Factor.

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Step 3.3.2.1

Factor $x^{2} - 2 x - 3$ 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 $- 3$ and whose sum is $- 2$ .

$- 3, 1$

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

Step 3.3.2.1.2

Write the factored form using these integers.

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

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

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

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

Step 3.3.2.2

Remove unnecessary parentheses.

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

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

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

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

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

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

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 $0$ .

$x - 3 = 0$

$x + 1 = 0$

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

Step 3.5

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

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Step 3.5.1

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

$x - 3 = 0$

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

Step 3.5.2

Add $3$ to both sides of the equation.

$x = 3$

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

$x = 3$

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

Step 3.6

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

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Step 3.6.1

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

$x + 1 = 0$

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

Step 3.6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

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

$x = - 1$

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

Step 3.7

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

$x = 3, - 1$

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

$x = 3, - 1$

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

Step 4

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

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Step 4.1

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

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

$x = 3$

Step 4.2

Simplify the right side.

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Step 4.2.1

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

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Step 4.2.1.1

Raise $3$ to the power of $2$ .

$y = - 2 + 9$

$x = 3$

Step 4.2.1.2

Add $- 2$ and $9$ .

$y = 7$

$x = 3$

$y = 7$

$x = 3$

$y = 7$

$x = 3$

$y = 7$

$x = 3$

Step 5

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

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Step 5.1

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

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

$x = - 1$

Step 5.2

Simplify the right side.

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Step 5.2.1

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

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Step 5.2.1.1

Raise $- 1$ to the power of $2$ .

$y = - 2 + 1$

$x = - 1$

Step 5.2.1.2

Add $- 2$ and $1$ .

$y = - 1$

$x = - 1$

$y = - 1$

$x = - 1$

$y = - 1$

$x = - 1$

$y = - 1$

$x = - 1$

Step 6

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

$(3, 7)$

$(- 1, - 1)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(3, 7), (- 1, - 1)$

Equation Form:

$x = 3, y = 7$

$x = - 1, y = - 1$

Step 8

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