Precalculus Examples

$y = - x^{2} + 15$ , $2 x + y = 0$

Step 1

Replace all occurrences of $y$ with $- x^{2} + 15$ in each equation.

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Step 1.1

Replace all occurrences of $y$ in $2 x + y = 0$ with $- x^{2} + 15$ .

$2 x - x^{2} + 15 = 0$

$y = - x^{2} + 15$

Step 1.2

Simplify the left side.

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Step 1.2.1

Remove parentheses.

$2 x - x^{2} + 15 = 0$

$y = - x^{2} + 15$

$2 x - x^{2} + 15 = 0$

$y = - x^{2} + 15$

$2 x - x^{2} + 15 = 0$

$y = - x^{2} + 15$

Step 2

Solve for $x$ in $2 x - x^{2} + 15 = 0$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$2 u - u^{2} + 15 = 0$

$y = - x^{2} + 15$

Step 2.1.2

Factor by grouping.

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Step 2.1.2.1

Reorder terms.

$- u^{2} + 2 u + 15 = 0$

$y = - x^{2} + 15$

Step 2.1.2.2

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 $a \cdot c = - 1 \cdot 15 = - 15$ and whose sum is $b = 2$ .

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Step 2.1.2.2.1

Factor $2$ out of $2 u$ .

$- u^{2} + 2 (u) + 15 = 0$

$y = - x^{2} + 15$

Step 2.1.2.2.2

Rewrite $2$ as $- 3$ plus $5$

$- u^{2} + (- 3 + 5) u + 15 = 0$

$y = - x^{2} + 15$

Step 2.1.2.2.3

Apply the distributive property.

$- u^{2} - 3 u + 5 u + 15 = 0$

$y = - x^{2} + 15$

$- u^{2} - 3 u + 5 u + 15 = 0$

$y = - x^{2} + 15$

Step 2.1.2.3

Factor out the greatest common factor from each group.

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Step 2.1.2.3.1

Group the first two terms and the last two terms.

$(- u^{2} - 3 u) + 5 u + 15 = 0$

$y = - x^{2} + 15$

Step 2.1.2.3.2

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

$u (- u - 3) - 5 (- u - 3) = 0$

$y = - x^{2} + 15$

$u (- u - 3) - 5 (- u - 3) = 0$

$y = - x^{2} + 15$

Step 2.1.2.4

Factor the polynomial by factoring out the greatest common factor, $- u - 3$ .

$(- u - 3) (u - 5) = 0$

$y = - x^{2} + 15$

$(- u - 3) (u - 5) = 0$

$y = - x^{2} + 15$

Step 2.1.3

Replace all occurrences of $u$ with $x$ .

$(- x - 3) (x - 5) = 0$

$y = - x^{2} + 15$

$(- x - 3) (x - 5) = 0$

$y = - x^{2} + 15$

Step 2.2

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

$- x - 3 = 0$

$x - 5 = 0$

$y = - x^{2} + 15$

Step 2.3

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

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Step 2.3.1

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

$- x - 3 = 0$

$y = - x^{2} + 15$

Step 2.3.2

Solve $- x - 3 = 0$ for $x$ .

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Step 2.3.2.1

Add $3$ to both sides of the equation.

$- x = 3$

$y = - x^{2} + 15$

Step 2.3.2.2

Divide each term in $- x = 3$ by $- 1$ and simplify.

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Step 2.3.2.2.1

Divide each term in $- x = 3$ by $- 1$ .

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

$y = - x^{2} + 15$

Step 2.3.2.2.2

Simplify the left side.

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Step 2.3.2.2.2.1

Dividing two negative values results in a positive value.

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

$y = - x^{2} + 15$

Step 2.3.2.2.2.2

Divide $x$ by $1$ .

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

$y = - x^{2} + 15$

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

$y = - x^{2} + 15$

Step 2.3.2.2.3

Simplify the right side.

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Step 2.3.2.2.3.1

Divide $3$ by $- 1$ .

$x = - 3$

$y = - x^{2} + 15$

$x = - 3$

$y = - x^{2} + 15$

$x = - 3$

$y = - x^{2} + 15$

$x = - 3$

$y = - x^{2} + 15$

$x = - 3$

$y = - x^{2} + 15$

Step 2.4

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

$y = - x^{2} + 15$

Step 2.4.2

Add $5$ to both sides of the equation.

$x = 5$

$y = - x^{2} + 15$

$x = 5$

$y = - x^{2} + 15$

Step 2.5

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

$x = - 3, 5$

$y = - x^{2} + 15$

$x = - 3, 5$

$y = - x^{2} + 15$

Step 3

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

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Step 3.1

Replace all occurrences of $x$ in $y = - x^{2} + 15$ with $- 3$ .

$y = - {(- 3)}^{2} + 15$

$x = - 3$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify $- {(- 3)}^{2} + 15$ .

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Step 3.2.1.1

Simplify each term.

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Step 3.2.1.1.1

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

$y = - 1 \cdot 9 + 15$

$x = - 3$

Step 3.2.1.1.2

Multiply $- 1$ by $9$ .

$y = - 9 + 15$

$x = - 3$

$y = - 9 + 15$

$x = - 3$

Step 3.2.1.2

Add $- 9$ and $15$ .

$y = 6$

$x = - 3$

$y = 6$

$x = - 3$

$y = 6$

$x = - 3$

$y = 6$

$x = - 3$

Step 4

Replace all occurrences of $x$ with $5$ in each equation.

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Step 4.1

Replace all occurrences of $x$ in $y = - x^{2} + 15$ with $5$ .

$y = - {(5)}^{2} + 15$

$x = 5$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $- {(5)}^{2} + 15$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Raise $5$ to the power of $2$ .

$y = - 1 \cdot 25 + 15$

$x = 5$

Step 4.2.1.1.2

Multiply $- 1$ by $25$ .

$y = - 25 + 15$

$x = 5$

$y = - 25 + 15$

$x = 5$

Step 4.2.1.2

Add $- 25$ and $15$ .

$y = - 10$

$x = 5$

$y = - 10$

$x = 5$

$y = - 10$

$x = 5$

$y = - 10$

$x = 5$

Step 5

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

$(- 3, 6)$

$(5, - 10)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- 3, 6), (5, - 10)$

Equation Form:

$x = - 3, y = 6$

$x = 5, y = - 10$

Step 7