Simplify x^2+16-6+7x+3x>3x

$x^{2} + 16 - 6 + 7 x + 3 x > 3 x$

Step 1

Simplify $x^{2} + 16 - 6 + 7 x + 3 x$ .

Subtract $6$ from $16$ .

$x^{2} + 10 + 7 x + 3 x > 3 x$

Add $7 x$ and $3 x$ .

$x^{2} + 10 + 10 x > 3 x$

Step 2

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

Subtract $3 x$ from both sides of the inequality.

$x^{2} + 10 + 10 x - 3 x > 0$

Subtract $3 x$ from $10 x$ .

$x^{2} + 10 + 7 x > 0$

Step 3

Convert the inequality to an equation.

$x^{2} + 10 + 7 x = 0$

Step 4

Factor $x^{2} + 10 + 7 x$ using the AC method.

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 $10$ and whose sum is $7$ .

$2, 5$

Write the factored form using these integers.

$(x + 2) (x + 5) = 0$

Step 5

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

$x + 2 = 0$

$x + 5 = 0$

Step 6

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 7

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

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 8

The final solution is all the values that make $(x + 2) (x + 5) = 0$ true.

$x = - 2, - 5$

Step 9

Use each root to create test intervals.

$x < - 5$

$- 5 < x < - 2$

$x > - 2$

Step 10

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

${(- 8)}^{2} + 16 - 6 + 7 (- 8) + 3 (- 8) > 3 (- 8)$

The left side $- 6$ is greater than the right side $- 24$ , which means that the given statement is always true.

True

Test a value on the interval $- 5 < x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

${(- 4)}^{2} + 16 - 6 + 7 (- 4) + 3 (- 4) > 3 (- 4)$

The left side $- 14$ is not greater than the right side $- 12$ , which means that the given statement is false.

False

Test a value on the interval $x > - 2$ to see if it makes the inequality true.

Choose a value on the interval $x > - 2$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

${(0)}^{2} + 16 - 6 + 7 (0) + 3 (0) > 3 (0)$

The left side $10$ is greater than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 5$ True

$- 5 < x < - 2$ False

$x > - 2$ True

$x < - 5$ True

$- 5 < x < - 2$ False

$x > - 2$ True

Step 11

The solution consists of all of the true intervals.

$x < - 5$ or $x > - 2$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$x < - 5 or x > - 2$

Interval Notation:

$(- \infty, - 5) \cup (- 2, \infty)$

Step 13

Simplify x^2+16-6+7x+3x>3x

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