Pre-Algebra Examples

$(x - 4) (x - 3) = 12$

Step 1

Simplify $(x - 4) (x - 3)$ .

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Step 1.1

Expand $(x - 4) (x - 3)$ using the FOIL Method.

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Step 1.1.1

Apply the distributive property.

$x (x - 3) - 4 (x - 3) = 12$

Step 1.1.2

Apply the distributive property.

$x \cdot x + x \cdot - 3 - 4 (x - 3) = 12$

Step 1.1.3

Apply the distributive property.

$x \cdot x + x \cdot - 3 - 4 x - 4 \cdot - 3 = 12$

Step 1.2

Simplify and combine like terms.

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Step 1.2.1

Simplify each term.

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Step 1.2.1.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 3 - 4 x - 4 \cdot - 3 = 12$

Step 1.2.1.2

Move $- 3$ to the left of $x$ .

$x^{2} - 3 \cdot x - 4 x - 4 \cdot - 3 = 12$

Step 1.2.1.3

Multiply $- 4$ by $- 3$ .

$x^{2} - 3 x - 4 x + 12 = 12$

Step 1.2.2

Subtract $4 x$ from $- 3 x$ .

$x^{2} - 7 x + 12 = 12$

Step 2

Subtract $12$ from both sides of the equation.

$x^{2} - 7 x + 12 - 12 = 0$

Step 3

Combine the opposite terms in $x^{2} - 7 x + 12 - 12$ .

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Step 3.1

Subtract $12$ from $12$ .

$x^{2} - 7 x + 0 = 0$

Step 3.2

Add $x^{2} - 7 x$ and $0$ .

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

Step 4

Factor $x$ out of $x^{2} - 7 x$ .

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Step 4.1

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

$x \cdot x - 7 x = 0$

Step 4.2

Factor $x$ out of $- 7 x$ .

$x \cdot x + x \cdot - 7 = 0$

Step 4.3

Factor $x$ out of $x \cdot x + x \cdot - 7$ .

$x (x - 7) = 0$

Step 5

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

$x = 0$

$x - 7 = 0$

Step 6

Set $x$ equal to $0$ .

$x = 0$

Step 7

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 7.2

Add $7$ to both sides of the equation.

$x = 7$

Step 8

The final solution is all the values that make $x (x - 7) = 0$ true.

$x = 0, 7$