Pre-Algebra Examples

$x (x - 3) = 7 x + 24$

Step 1

Simplify $x (x - 3)$ .

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Step 1.1

Rewrite.

$0 + 0 + x (x - 3) = 7 x + 24$

Step 1.2

Simplify by adding zeros.

$x (x - 3) = 7 x + 24$

Step 1.3

Apply the distributive property.

$x \cdot x + x \cdot - 3 = 7 x + 24$

Step 1.4

Simplify the expression.

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Step 1.4.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 3 = 7 x + 24$

Step 1.4.2

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

$x^{2} - 3 x = 7 x + 24$

Step 2

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

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Step 2.1

Subtract $7 x$ from both sides of the equation.

$x^{2} - 3 x - 7 x = 24$

Step 2.2

Subtract $7 x$ from $- 3 x$ .

$x^{2} - 10 x = 24$

Step 3

Subtract $24$ from both sides of the equation.

$x^{2} - 10 x - 24 = 0$

Step 4

Factor $x^{2} - 10 x - 24$ using the AC method.

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Step 4.1

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 $- 24$ and whose sum is $- 10$ .

$- 12, 2$

Step 4.2

Write the factored form using these integers.

$(x - 12) (x + 2) = 0$

Step 5

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

$x - 12 = 0$

$x + 2 = 0$

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

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

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Step 7.1

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

$x + 2 = 0$

Step 7.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 8

The final solution is all the values that make $(x - 12) (x + 2) = 0$ true.

$x = 12, - 2$