Solve the Inequality for x -6x^2+27x-13>=8x-10

$- 6 x^{2} + 27 x - 13 \geq 8 x - 10$

Step 1

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

Subtract $8 x$ from both sides of the inequality.

$- 6 x^{2} + 27 x - 13 - 8 x \geq - 10$

Subtract $8 x$ from $27 x$ .

$- 6 x^{2} + 19 x - 13 \geq - 10$

Step 2

Convert the inequality to an equation.

$- 6 x^{2} + 19 x - 13 = - 10$

Step 3

Add $10$ to both sides of the equation.

$- 6 x^{2} + 19 x - 13 + 10 = 0$

Step 4

Add $- 13$ and $10$ .

$- 6 x^{2} + 19 x - 3 = 0$

Step 5

Factor the left side of the equation.

Factor $- 1$ out of $- 6 x^{2} + 19 x - 3$ .

Factor $- 1$ out of $- 6 x^{2}$ .

$- (6 x^{2}) + 19 x - 3 = 0$

Factor $- 1$ out of $19 x$ .

$- (6 x^{2}) - (- 19 x) - 3 = 0$

Rewrite $- 3$ as $- 1 (3)$ .

$- (6 x^{2}) - (- 19 x) - 1 \cdot 3 = 0$

Factor $- 1$ out of $- (6 x^{2}) - (- 19 x)$ .

$- (6 x^{2} - 19 x) - 1 \cdot 3 = 0$

Factor $- 1$ out of $- (6 x^{2} - 19 x) - 1 (3)$ .

$- (6 x^{2} - 19 x + 3) = 0$

Factor.

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 $b = - 19$ .

Factor $- 19$ out of $- 19 x$ .

$- (6 x^{2} - 19 x + 3) = 0$

Rewrite $- 19$ as $- 1$ plus $- 18$

$- (6 x^{2} + (- 1 - 18) x + 3) = 0$

Apply the distributive property.

$- (6 x^{2} - 1 x - 18 x + 3) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((6 x^{2} - 1 x) - 18 x + 3) = 0$

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

$- (x (6 x - 1) - 3 (6 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $6 x - 1$ .

$- ((6 x - 1) (x - 3)) = 0$

Remove unnecessary parentheses.

$- (6 x - 1) (x - 3) = 0$

Step 6

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

$6 x - 1 = 0$

$x - 3 = 0$

Step 7

Set $6 x - 1$ equal to $0$ and solve for $x$ .

Set $6 x - 1$ equal to $0$ .

$6 x - 1 = 0$

Solve $6 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$6 x = 1$

Divide each term in $6 x = 1$ by $6$ and simplify.

Divide each term in $6 x = 1$ by $6$ .

$\frac{6 x}{6} = \frac{1}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{1}{6}$

Divide $x$ by $1$ .

$x = \frac{1}{6}$

Step 8

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 9

The final solution is all the values that make $- (6 x - 1) (x - 3) = 0$ true.

$x = \frac{1}{6}, 3$

Step 10

Use each root to create test intervals.

$x < \frac{1}{6}$

$\frac{1}{6} < x < 3$

$x > 3$

Step 11

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 $x < \frac{1}{6}$ to see if it makes the inequality true.

Choose a value on the interval $x < \frac{1}{6}$ and see if this value makes the original inequality true.

$x = 0$

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

$- 6 {(0)}^{2} + 27 (0) - 13 \geq 8 (0) - 10$

The left side $- 13$ is less than the right side $- 10$ , which means that the given statement is false.

False

Test a value on the interval $\frac{1}{6} < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $\frac{1}{6} < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

$- 6 {(2)}^{2} + 27 (2) - 13 \geq 8 (2) - 10$

The left side $17$ is greater than the right side $6$ , which means that the given statement is always true.

True

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$- 6 {(6)}^{2} + 27 (6) - 13 \geq 8 (6) - 10$

The left side $- 67$ is less than the right side $38$ , which means that the given statement is false.

False

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

$x < \frac{1}{6}$ False

$\frac{1}{6} < x < 3$ True

$x > 3$ False

$x < \frac{1}{6}$ False

$\frac{1}{6} < x < 3$ True

$x > 3$ False

Step 12

The solution consists of all of the true intervals.

$\frac{1}{6} \leq x \leq 3$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$\frac{1}{6} \leq x \leq 3$

Interval Notation:

$[\frac{1}{6}, 3]$

Step 14

Solve the Inequality for x -6x^2+27x-13>=8x-10

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