Algebra Examples

$x (x - 1) - 5 (x - 2) = 2$

Step 1

Simplify $x (x - 1) - 5 (x - 2)$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

Apply the distributive property.

$x \cdot x + x \cdot - 1 - 5 (x - 2) = 2$

Step 1.1.2

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 1 - 5 (x - 2) = 2$

Step 1.1.3

Move $- 1$ to the left of $x$ .

$x^{2} - 1 \cdot x - 5 (x - 2) = 2$

Step 1.1.4

Rewrite $- 1 x$ as $- x$ .

$x^{2} - x - 5 (x - 2) = 2$

Step 1.1.5

Apply the distributive property.

$x^{2} - x - 5 x - 5 \cdot - 2 = 2$

Step 1.1.6

Multiply $- 5$ by $- 2$ .

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

Step 1.2

Subtract $5 x$ from $- x$ .

$x^{2} - 6 x + 10 = 2$

Step 2

Subtract $2$ from both sides of the equation.

$x^{2} - 6 x + 10 - 2 = 0$

Step 3

Subtract $2$ from $10$ .

$x^{2} - 6 x + 8 = 0$

Step 4

Factor $x^{2} - 6 x + 8$ 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 $8$ and whose sum is $- 6$ .

$- 4, - 2$

Step 4.2

Write the factored form using these integers.

$(x - 4) (x - 2) = 0$

Step 5

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

$x - 4 = 0$

$x - 2 = 0$

Step 6

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

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Step 6.1

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

$x - 4 = 0$

Step 6.2

Add $4$ to both sides of the equation.

$x = 4$

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

Add $2$ to both sides of the equation.

$x = 2$

Step 8

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

$x = 4, 2$