Algebra Examples

$y = x^{3} - x$ , $y = 3 x$

Step 1

Eliminate the equal sides of each equation and combine.

$x^{3} - x = 3 x$

Step 2

Solve $x^{3} - x = 3 x$ 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

Subtract $3 x$ from both sides of the equation.

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

Step 2.1.2

Subtract $3 x$ from $- x$ .

$x^{3} - 4 x = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $x$ out of $x^{3} - 4 x$ .

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Step 2.2.1.1

Factor $x$ out of $x^{3}$ .

$x \cdot x^{2} - 4 x = 0$

Step 2.2.1.2

Factor $x$ out of $- 4 x$ .

$x \cdot x^{2} + x \cdot - 4 = 0$

Step 2.2.1.3

Factor $x$ out of $x \cdot x^{2} + x \cdot - 4$ .

$x (x^{2} - 4) = 0$

Step 2.2.2

Rewrite $4$ as $2^{2}$ .

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

Step 2.2.3

Factor.

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Step 2.2.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$x ((x + 2) (x - 2)) = 0$

Step 2.2.3.2

Remove unnecessary parentheses.

$x (x + 2) (x - 2) = 0$

Step 2.3

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

$x = 0$

$x + 2 = 0$

$x - 2 = 0 + y = 3 x$

Step 2.4

Set $x$ equal to $0$ .

$x = 0$

Step 2.5

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 2.5.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.6

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 2.6.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.7

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

$x = 0, - 2, 2$

Step 3

Evaluate $y$ when $x = 0$ .

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Step 3.1

Substitute $0$ for $x$ .

$y = 3 (0)$

Step 3.2

Multiply $3$ by $0$ .

$y = 0$

Step 4

Evaluate $y$ when $x = - 2$ .

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Step 4.1

Substitute $- 2$ for $x$ .

$y = 3 (- 2)$

Step 4.2

Multiply $3$ by $- 2$ .

$y = - 6$

Step 5

Evaluate $y$ when $x = 2$ .

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Step 5.1

Substitute $2$ for $x$ .

$y = 3 (2)$

Step 5.2

Multiply $3$ by $2$ .

$y = 6$

Step 6

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

$(0, 0)$

$(- 2, - 6)$

$(2, 6)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(0, 0), (- 2, - 6), (2, 6)$

Equation Form:

$x = 0, y = 0$

$x = - 2, y = - 6$

$x = 2, y = 6$

Step 8