Pre-Algebra Examples

$8 x (x - 1) = 7 x^{2} + 4 x$

Step 1

Simplify $8 x (x - 1)$ .

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Step 1.1

Rewrite.

$0 + 0 + 8 x (x - 1) = 7 x^{2} + 4 x$

Step 1.2

Simplify by adding zeros.

$8 x (x - 1) = 7 x^{2} + 4 x$

Step 1.3

Apply the distributive property.

$8 x \cdot x + 8 x \cdot - 1 = 7 x^{2} + 4 x$

Step 1.4

Multiply $x$ by $x$ by adding the exponents.

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Step 1.4.1

Move $x$ .

$8 (x \cdot x) + 8 x \cdot - 1 = 7 x^{2} + 4 x$

Step 1.4.2

Multiply $x$ by $x$ .

$8 x^{2} + 8 x \cdot - 1 = 7 x^{2} + 4 x$

Step 1.5

Multiply $- 1$ by $8$ .

$8 x^{2} - 8 x = 7 x^{2} + 4 x$

Step 2

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

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Step 2.1

Subtract $7 x^{2}$ from both sides of the equation.

$8 x^{2} - 8 x - 7 x^{2} = 4 x$

Step 2.2

Subtract $4 x$ from both sides of the equation.

$8 x^{2} - 8 x - 7 x^{2} - 4 x = 0$

Step 2.3

Subtract $7 x^{2}$ from $8 x^{2}$ .

$x^{2} - 8 x - 4 x = 0$

Step 2.4

Subtract $4 x$ from $- 8 x$ .

$x^{2} - 12 x = 0$

Step 3

Factor $x$ out of $x^{2} - 12 x$ .

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Step 3.1

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

$x \cdot x - 12 x = 0$

Step 3.2

Factor $x$ out of $- 12 x$ .

$x \cdot x + x \cdot - 12 = 0$

Step 3.3

Factor $x$ out of $x \cdot x + x \cdot - 12$ .

$x (x - 12) = 0$

Step 4

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

$x = 0$

$x - 12 = 0$

Step 5

Set $x$ equal to $0$ .

$x = 0$

Step 6

Set $x - 12$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x - 12$ equal to $0$ .

$x - 12 = 0$

Step 6.2

Add $12$ to both sides of the equation.

$x = 12$

Step 7

The final solution is all the values that make $x (x - 12) = 0$ true.

$x = 0, 12$