Pre-Algebra Examples

$x - 1 + x^{2} - 3 x = 2 x - 4$

Step 1

Subtract $3 x$ from $x$ .

$- 2 x - 1 + x^{2} = 2 x - 4$

Step 2

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

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Step 2.1

Subtract $2 x$ from both sides of the equation.

$- 2 x - 1 + x^{2} - 2 x = - 4$

Step 2.2

Subtract $2 x$ from $- 2 x$ .

$- 4 x - 1 + x^{2} = - 4$

Step 3

Add $4$ to both sides of the equation.

$- 4 x - 1 + x^{2} + 4 = 0$

Step 4

Add $- 1$ and $4$ .

$- 4 x + x^{2} + 3 = 0$

Step 5

Factor the left side of the equation.

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Step 5.1

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$- 4 u + u^{2} + 3 = 0$

Step 5.2

Factor $- 4 u + u^{2} + 3$ using the AC method.

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Step 5.2.1

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 $3$ and whose sum is $- 4$ .

$- 3, - 1$

Step 5.2.2

Write the factored form using these integers.

$(u - 3) (u - 1) = 0$

Step 5.3

Replace all occurrences of $u$ with $x$ .

$(x - 3) (x - 1) = 0$

Step 6

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

$x - 3 = 0$

$x - 1 = 0$

Step 7

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 7.2

Add $3$ to both sides of the equation.

$x = 3$

Step 8

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 8.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 8.2

Add $1$ to both sides of the equation.

$x = 1$

Step 9

The final solution is all the values that make $(x - 3) (x - 1) = 0$ true.

$x = 3, 1$