Algebra Examples

$y = x^{2} - 2 x - 19$ $y + 4 x = 5$

Replace all occurrences of $y$ with $x^{2} - 2 x - 19$ in each equation.

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Replace all occurrences of $y$ in $y + 4 x = 5$ with $x^{2} - 2 x - 19$ .

$(x^{2} - 2 x - 19) + 4 x = 5$

$y = x^{2} - 2 x - 19$

Simplify the left side.

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Add $- 2 x$ and $4 x$ .

$x^{2} + 2 x - 19 = 5$

$y = x^{2} - 2 x - 19$

$x^{2} + 2 x - 19 = 5$

$y = x^{2} - 2 x - 19$

$x^{2} + 2 x - 19 = 5$

$y = x^{2} - 2 x - 19$

Solve for $x$ in $x^{2} + 2 x - 19 = 5$ .

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Subtract $5$ from both sides of the equation.

$x^{2} + 2 x - 19 - 5 = 0$

$y = x^{2} - 2 x - 19$

Subtract $5$ from $- 19$ .

$x^{2} + 2 x - 24 = 0$

$y = x^{2} - 2 x - 19$

Factor $x^{2} + 2 x - 24$ using the AC method.

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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 $- 24$ and whose sum is $2$ .

$- 4, 6$

$y = x^{2} - 2 x - 19$

Write the factored form using these integers.

$(x - 4) (x + 6) = 0$

$y = x^{2} - 2 x - 19$

$(x - 4) (x + 6) = 0$

$y = x^{2} - 2 x - 19$

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

$x - 4 = 0$

$x + 6 = 0$

$y = x^{2} - 2 x - 19$

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

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

$x - 4 = 0$

$y = x^{2} - 2 x - 19$

Add $4$ to both sides of the equation.

$x = 4$

$y = x^{2} - 2 x - 19$

$x = 4$

$y = x^{2} - 2 x - 19$

Set $x + 6$ equal to $0$ and solve for $x$ .

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Set $x + 6$ equal to $0$ .

$x + 6 = 0$

$y = x^{2} - 2 x - 19$

Subtract $6$ from both sides of the equation.

$x = - 6$

$y = x^{2} - 2 x - 19$

$x = - 6$

$y = x^{2} - 2 x - 19$

The final solution is all the values that make $(x - 4) (x + 6) = 0$ true.

$x = 4, - 6$

$y = x^{2} - 2 x - 19$

$x = 4, - 6$

$y = x^{2} - 2 x - 19$

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

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Replace all occurrences of $x$ in $y = x^{2} - 2 x - 19$ with $4$ .

$y = {(4)}^{2} - 2 \cdot 4 - 19$

$x = 4$

Simplify the right side.

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Simplify ${(4)}^{2} - 2 \cdot 4 - 19$ .

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Simplify each term.

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

$y = 16 - 2 \cdot 4 - 19$

$x = 4$

Multiply $- 2$ by $4$ .

$y = 16 - 8 - 19$

$x = 4$

$y = 16 - 8 - 19$

$x = 4$

Simplify by subtracting numbers.

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Subtract $8$ from $16$ .

$y = 8 - 19$

$x = 4$

Subtract $19$ from $8$ .

$y = - 11$

$x = 4$

$y = - 11$

$x = 4$

$y = - 11$

$x = 4$

$y = - 11$

$x = 4$

$y = - 11$

$x = 4$

Replace all occurrences of $x$ with $- 6$ in each equation.

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Replace all occurrences of $x$ in $y = x^{2} - 2 x - 19$ with $- 6$ .

$y = {(- 6)}^{2} - 2 \cdot - 6 - 19$

$x = - 6$

Simplify the right side.

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Simplify ${(- 6)}^{2} - 2 \cdot - 6 - 19$ .

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Simplify each term.

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Raise $- 6$ to the power of $2$ .

$y = 36 - 2 \cdot - 6 - 19$

$x = - 6$

Multiply $- 2$ by $- 6$ .

$y = 36 + 12 - 19$

$x = - 6$

$y = 36 + 12 - 19$

$x = - 6$

Simplify by adding and subtracting.

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Add $36$ and $12$ .

$y = 48 - 19$

$x = - 6$

Subtract $19$ from $48$ .

$y = 29$

$x = - 6$

$y = 29$

$x = - 6$

$y = 29$

$x = - 6$

$y = 29$

$x = - 6$

$y = 29$

$x = - 6$

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

$(4, - 11)$

$(- 6, 29)$

The result can be shown in multiple forms.

Point Form:

$(4, - 11), (- 6, 29)$

Equation Form:

$x = 4, y = - 11$

$x = - 6, y = 29$