Algebra Examples

$y = x^{2} + 10 x + 3$ $y = - x^{2} - 5$

Step 1

Eliminate the equal sides of each equation and combine.

$x^{2} + 10 x + 3 = - x^{2} - 5$

Step 2

Solve $x^{2} + 10 x + 3 = - x^{2} - 5$ for $x$ .

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Step 2.1

Move all terms containing $x$ to the left side of the equation.

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Step 2.1.1

Add $x^{2}$ to both sides of the equation.

$x^{2} + 10 x + 3 + x^{2} = - 5$

Step 2.1.2

Add $x^{2}$ and $x^{2}$ .

$2 x^{2} + 10 x + 3 = - 5$

Step 2.2

Add $5$ to both sides of the equation.

$2 x^{2} + 10 x + 3 + 5 = 0$

Step 2.3

Add $3$ and $5$ .

$2 x^{2} + 10 x + 8 = 0$

Step 2.4

Factor the left side of the equation.

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Step 2.4.1

Factor $2$ out of $2 x^{2} + 10 x + 8$ .

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Step 2.4.1.1

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

$2 (x^{2}) + 10 x + 8 = 0$

Step 2.4.1.2

Factor $2$ out of $10 x$ .

$2 (x^{2}) + 2 (5 x) + 8 = 0$

Step 2.4.1.3

Factor $2$ out of $8$ .

$2 x^{2} + 2 (5 x) + 2 \cdot 4 = 0$

Step 2.4.1.4

Factor $2$ out of $2 x^{2} + 2 (5 x)$ .

$2 (x^{2} + 5 x) + 2 \cdot 4 = 0$

Step 2.4.1.5

Factor $2$ out of $2 (x^{2} + 5 x) + 2 \cdot 4$ .

$2 (x^{2} + 5 x + 4) = 0$

Step 2.4.2

Factor.

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Step 2.4.2.1

Factor $x^{2} + 5 x + 4$ using the AC method.

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Step 2.4.2.1.1

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 $4$ and whose sum is $5$ .

$1, 4 + y = - x^{2} - 5$

Step 2.4.2.1.2

Write the factored form using these integers.

$2 ((x + 1) (x + 4)) = 0$

Step 2.4.2.2

Remove unnecessary parentheses.

$2 (x + 1) (x + 4) = 0$

Step 2.5

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

$x + 1 = 0$

$x + 4 = 0 + y = - x^{2} - 5$

Step 2.6

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

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Step 2.6.1

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

$x + 1 = 0$

Step 2.6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.7

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

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Step 2.7.1

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

$x + 4 = 0$

Step 2.7.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 2.8

The final solution is all the values that make $2 (x + 1) (x + 4) = 0$ true.

$x = - 1, - 4$

Step 3

Evaluate $y$ when $x = - 1$ .

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Step 3.1

Substitute $- 1$ for $x$ .

$y = - {(- 1)}^{2} - 5$

Step 3.2

Simplify $- {(- 1)}^{2} - 5$ .

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

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Step 3.2.1.1.1

Multiply $- 1$ by ${(- 1)}^{2}$ .

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Step 3.2.1.1.1.1

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

$y = {(- 1)}^{1} {(- 1)}^{2} - 5$

Step 3.2.1.1.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = {(- 1)}^{1 + 2} - 5$

Step 3.2.1.1.2

Add $1$ and $2$ .

$y = {(- 1)}^{3} - 5$

Step 3.2.1.2

Raise $- 1$ to the power of $3$ .

$y = - 1 - 5$

Step 3.2.2

Subtract $5$ from $- 1$ .

$y = - 6$

Step 4

Evaluate $y$ when $x = - 4$ .

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Step 4.1

Substitute $- 4$ for $x$ .

$y = - {(- 4)}^{2} - 5$

Step 4.2

Simplify $- {(- 4)}^{2} - 5$ .

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

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

$y = - 1 \cdot 16 - 5$

Step 4.2.1.2

Multiply $- 1$ by $16$ .

$y = - 16 - 5$

Step 4.2.2

Subtract $5$ from $- 16$ .

$y = - 21$

Step 5

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

$(- 1, - 6)$

$(- 4, - 21)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- 1, - 6), (- 4, - 21)$

Equation Form:

$x = - 1, y = - 6$

$x = - 4, y = - 21$

Step 7