Solve the Inequality for x 6x^2+18x>7x-3

$6 x^{2} + 18 x > 7 x - 3$

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

Subtract $7 x$ from both sides of the inequality.

$6 x^{2} + 18 x - 7 x > - 3$

Subtract $7 x$ from $18 x$ .

$6 x^{2} + 11 x > - 3$

Convert the inequality to an equation.

$6 x^{2} + 11 x = - 3$

Add $3$ to both sides of the equation.

$6 x^{2} + 11 x + 3 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 6 \cdot 3 = 18$ and whose sum is $b = 11$ .

Factor $11$ out of $11 x$ .

$6 x^{2} + 11 (x) + 3 = 0$

Rewrite $11$ as $2$ plus $9$

$6 x^{2} + (2 + 9) x + 3 = 0$

Apply the distributive property.

$6 x^{2} + 2 x + 9 x + 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(6 x^{2} + 2 x) + 9 x + 3 = 0$

Factor out the greatest common factor (GCF) from each group.

$2 x (3 x + 1) + 3 (3 x + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x + 1$ .

$(3 x + 1) (2 x + 3) = 0$

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

$3 x + 1 = 0$

$2 x + 3 = 0$

Set $3 x + 1$ equal to $0$ and solve for $x$ .

Set $3 x + 1$ equal to $0$ .

$3 x + 1 = 0$

Solve $3 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Divide each term in $3 x = - 1$ by $3$ and simplify.

Divide each term in $3 x = - 1$ by $3$ .

$\frac{3 x}{3} = \frac{- 1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 1}{3}$

Divide $x$ by $1$ .

$x = \frac{- 1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{3}$

Set $2 x + 3$ equal to $0$ and solve for $x$ .

Set $2 x + 3$ equal to $0$ .

$2 x + 3 = 0$

Solve $2 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Divide each term in $2 x = - 3$ by $2$ and simplify.

Divide each term in $2 x = - 3$ by $2$ .

$\frac{2 x}{2} = \frac{- 3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 3}{2}$

Divide $x$ by $1$ .

$x = \frac{- 3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{2}$

The final solution is all the values that make $(3 x + 1) (2 x + 3) = 0$ true.

$x = - \frac{1}{3}, - \frac{3}{2}$

Use each root to create test intervals.

$x < - \frac{3}{2}$

$- \frac{3}{2} < x < - \frac{1}{3}$

$x > - \frac{1}{3}$

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 $x < - \frac{3}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x < - \frac{3}{2}$ and see if this value makes the original inequality true.

$x = - 4$

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

$6 {(- 4)}^{2} + 18 (- 4) > 7 (- 4) - 3$

The left side $24$ is greater than the right side $- 31$ , which means that the given statement is always true.

True

Test a value on the interval $- \frac{3}{2} < x < - \frac{1}{3}$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{3}{2} < x < - \frac{1}{3}$ and see if this value makes the original inequality true.

$x = - 1$

Replace $x$ with $- 1$ in the original inequality.

$6 {(- 1)}^{2} + 18 (- 1) > 7 (- 1) - 3$

The left side $- 12$ is not greater than the right side $- 10$ , which means that the given statement is false.

False

Test a value on the interval $x > - \frac{1}{3}$ to see if it makes the inequality true.

Choose a value on the interval $x > - \frac{1}{3}$ and see if this value makes the original inequality true.

$x = 0$

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

$6 {(0)}^{2} + 18 (0) > 7 (0) - 3$

The left side $0$ is greater than the right side $- 3$ , which means that the given statement is always true.

True

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

$x < - \frac{3}{2}$ True

$- \frac{3}{2} < x < - \frac{1}{3}$ False

$x > - \frac{1}{3}$ True

$x < - \frac{3}{2}$ True

$- \frac{3}{2} < x < - \frac{1}{3}$ False

$x > - \frac{1}{3}$ True

The solution consists of all of the true intervals.

$x < - \frac{3}{2}$ or $x > - \frac{1}{3}$

The result can be shown in multiple forms.

Inequality Form:

$x < - \frac{3}{2} or x > - \frac{1}{3}$

Interval Notation:

$(- \infty, - \frac{3}{2}) \cup (- \frac{1}{3}, \infty)$

Solve the Inequality for x 6x^2+18x>7x-3

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