Pre-Algebra Examples

$x + 4 + x + 9 + x^{2} - 3 x + 4 x - 4 = 49$

Step 1

Simplify $x + 4 + x + 9 + x^{2} - 3 x + 4 x - 4$ .

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Step 1.1

Combine the opposite terms in $x + 4 + x + 9 + x^{2} - 3 x + 4 x - 4$ .

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Step 1.1.1

Subtract $4$ from $4$ .

$x + x + 9 + x^{2} - 3 x + 4 x + 0 = 49$

Step 1.1.2

Add $x + x + 9 + x^{2} - 3 x + 4 x$ and $0$ .

$x + x + 9 + x^{2} - 3 x + 4 x = 49$

Step 1.2

Add $x$ and $x$ .

$2 x + 9 + x^{2} - 3 x + 4 x = 49$

Step 1.3

Subtract $3 x$ from $2 x$ .

$- x + 9 + x^{2} + 4 x = 49$

Step 1.4

Add $- x$ and $4 x$ .

$3 x + 9 + x^{2} = 49$

Step 2

Subtract $49$ from both sides of the equation.

$3 x + 9 + x^{2} - 49 = 0$

Step 3

Subtract $49$ from $9$ .

$3 x + x^{2} - 40 = 0$

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 $x$ .

$3 u + u^{2} - 40 = 0$

Step 4.2

Factor $3 u + u^{2} - 40$ 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 $- 40$ and whose sum is $3$ .

$- 5, 8$

Step 4.2.2

Write the factored form using these integers.

$(u - 5) (u + 8) = 0$

Step 4.3

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

$(x - 5) (x + 8) = 0$

Step 5

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

$x - 5 = 0$

$x + 8 = 0$

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 $x + 8$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Step 7.2

Subtract $8$ from both sides of the equation.

$x = - 8$

Step 8

The final solution is all the values that make $(x - 5) (x + 8) = 0$ true.

$x = 5, - 8$