Precalculus Examples

$x^{2} + y = 6$ , $13 x + y = 42$

Step 1

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

$y = 6 - x^{2}$

$13 x + y = 42$

Step 2

Replace all occurrences of $y$ with $6 - x^{2}$ in each equation.

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Step 2.1

Replace all occurrences of $y$ in $13 x + y = 42$ with $6 - x^{2}$ .

$13 x + 6 - x^{2} = 42$

$y = 6 - x^{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Remove parentheses.

$13 x + 6 - x^{2} = 42$

$y = 6 - x^{2}$

$13 x + 6 - x^{2} = 42$

$y = 6 - x^{2}$

$13 x + 6 - x^{2} = 42$

$y = 6 - x^{2}$

Step 3

Solve for $x$ in $13 x + 6 - x^{2} = 42$ .

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Step 3.1

Subtract $42$ from both sides of the equation.

$13 x + 6 - x^{2} - 42 = 0$

$y = 6 - x^{2}$

Step 3.2

Subtract $42$ from $6$ .

$13 x - x^{2} - 36 = 0$

$y = 6 - x^{2}$

Step 3.3

Factor the left side of the equation.

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Step 3.3.1

Factor $- 1$ out of $13 x - x^{2} - 36$ .

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Step 3.3.1.1

Reorder $13 x$ and $- x^{2}$ .

$- x^{2} + 13 x - 36 = 0$

$y = 6 - x^{2}$

Step 3.3.1.2

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

$- (x^{2}) + 13 x - 36 = 0$

$y = 6 - x^{2}$

Step 3.3.1.3

Factor $- 1$ out of $13 x$ .

$- (x^{2}) - (- 13 x) - 36 = 0$

$y = 6 - x^{2}$

Step 3.3.1.4

Rewrite $- 36$ as $- 1 (36)$ .

$- (x^{2}) - (- 13 x) - 1 \cdot 36 = 0$

$y = 6 - x^{2}$

Step 3.3.1.5

Factor $- 1$ out of $- (x^{2}) - (- 13 x)$ .

$- (x^{2} - 13 x) - 1 \cdot 36 = 0$

$y = 6 - x^{2}$

Step 3.3.1.6

Factor $- 1$ out of $- (x^{2} - 13 x) - 1 (36)$ .

$- (x^{2} - 13 x + 36) = 0$

$y = 6 - x^{2}$

$- (x^{2} - 13 x + 36) = 0$

$y = 6 - x^{2}$

Step 3.3.2

Factor.

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Step 3.3.2.1

Factor $x^{2} - 13 x + 36$ 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 $36$ and whose sum is $- 13$ .

$- 9, - 4$

$y = 6 - x^{2}$

Step 3.3.2.1.2

Write the factored form using these integers.

$- ((x - 9) (x - 4)) = 0$

$y = 6 - x^{2}$

$- ((x - 9) (x - 4)) = 0$

$y = 6 - x^{2}$

Step 3.3.2.2

Remove unnecessary parentheses.

$- (x - 9) (x - 4) = 0$

$y = 6 - x^{2}$

$- (x - 9) (x - 4) = 0$

$y = 6 - x^{2}$

$- (x - 9) (x - 4) = 0$

$y = 6 - x^{2}$

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 $0$ .

$x - 9 = 0$

$x - 4 = 0$

$y = 6 - x^{2}$

Step 3.5

Set $x - 9$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

$y = 6 - x^{2}$

Step 3.5.2

Add $9$ to both sides of the equation.

$x = 9$

$y = 6 - x^{2}$

$x = 9$

$y = 6 - x^{2}$

Step 3.6

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

$y = 6 - x^{2}$

Step 3.6.2

Add $4$ to both sides of the equation.

$x = 4$

$y = 6 - x^{2}$

$x = 4$

$y = 6 - x^{2}$

Step 3.7

The final solution is all the values that make $- (x - 9) (x - 4) = 0$ true.

$x = 9, 4$

$y = 6 - x^{2}$

$x = 9, 4$

$y = 6 - x^{2}$

Step 4

Replace all occurrences of $x$ with $9$ in each equation.

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Step 4.1

Replace all occurrences of $x$ in $y = 6 - x^{2}$ with $9$ .

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

$x = 9$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $6 - {(9)}^{2}$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Raise $9$ to the power of $2$ .

$y = 6 - 1 \cdot 81$

$x = 9$

Step 4.2.1.1.2

Multiply $- 1$ by $81$ .

$y = 6 - 81$

$x = 9$

$y = 6 - 81$

$x = 9$

Step 4.2.1.2

Subtract $81$ from $6$ .

$y = - 75$

$x = 9$

$y = - 75$

$x = 9$

$y = - 75$

$x = 9$

$y = - 75$

$x = 9$

Step 5

Replace all occurrences of $x$ with $4$ in each equation.

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Step 5.1

Replace all occurrences of $x$ in $y = 6 - x^{2}$ with $4$ .

$y = 6 - {(4)}^{2}$

$x = 4$

Step 5.2

Simplify the right side.

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Step 5.2.1

Simplify $6 - {(4)}^{2}$ .

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Step 5.2.1.1

Simplify each term.

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Step 5.2.1.1.1

Raise $4$ to the power of $2$ .

$y = 6 - 1 \cdot 16$

$x = 4$

Step 5.2.1.1.2

Multiply $- 1$ by $16$ .

$y = 6 - 16$

$x = 4$

$y = 6 - 16$

$x = 4$

Step 5.2.1.2

Subtract $16$ from $6$ .

$y = - 10$

$x = 4$

$y = - 10$

$x = 4$

$y = - 10$

$x = 4$

$y = - 10$

$x = 4$

Step 6

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

$(9, - 75)$

$(4, - 10)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(9, - 75), (4, - 10)$

Equation Form:

$x = 9, y = - 75$

$x = 4, y = - 10$

Step 8