Pre-Algebra Examples

$(x + 5) + (x^{2} - 2 x) + (x + 2) + (2 x - 4) = 38$

Step 1

Simplify $x + 5 + x^{2} - 2 x + x + 2 + 2 x - 4$ .

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Step 1.1

Combine the opposite terms in $x + 5 + x^{2} - 2 x + x + 2 + 2 x - 4$ .

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Step 1.1.1

Add $- 2 x$ and $2 x$ .

$x + 5 + x^{2} + x + 2 + 0 - 4 = 38$

Step 1.1.2

Add $x + 5 + x^{2} + x + 2$ and $0$ .

$x + 5 + x^{2} + x + 2 - 4 = 38$

Step 1.2

Add $x$ and $x$ .

$2 x + 5 + x^{2} + 2 - 4 = 38$

Step 1.3

Simplify by adding and subtracting.

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Step 1.3.1

Add $5$ and $2$ .

$2 x + x^{2} + 7 - 4 = 38$

Step 1.3.2

Subtract $4$ from $7$ .

$2 x + x^{2} + 3 = 38$

Step 2

Subtract $38$ from both sides of the equation.

$2 x + x^{2} + 3 - 38 = 0$

Step 3

Subtract $38$ from $3$ .

$2 x + x^{2} - 35 = 0$

Step 4

Factor the left side of the equation.

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Step 4.1

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

$2 u + u^{2} - 35 = 0$

Step 4.2

Factor $2 u + u^{2} - 35$ using the AC method.

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Step 4.2.1

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 $- 35$ and whose sum is $2$ .

$- 5, 7$

Step 4.2.2

Write the factored form using these integers.

$(u - 5) (u + 7) = 0$

Step 4.3

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

$(x - 5) (x + 7) = 0$

Step 5

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

$x - 5 = 0$

$x + 7 = 0$

Step 6

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 6.2

Add $5$ to both sides of the equation.

$x = 5$

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

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 8

The final solution is all the values that make $(x - 5) (x + 7) = 0$ true.

$x = 5, - 7$