Solve for x square root of 6x-18+6<=15

$\sqrt{6 x - 18} + 6 \leq 15$

Step 1

Move all terms not containing $\sqrt{6 x - 18}$ to the right side of the inequality.

Subtract $6$ from both sides of the inequality.

$\sqrt{6 x - 18} \leq 15 - 6$

Subtract $6$ from $15$ .

$\sqrt{6 x - 18} \leq 9$

Step 2

To remove the radical on the left side of the inequality, square both sides of the inequality.

${\sqrt{6 x - 18}}^{2} \leq 9^{2}$

Step 3

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{6 x - 18}$ as ${(6 x - 18)}^{\frac{1}{2}}$ .

${({(6 x - 18)}^{\frac{1}{2}})}^{2} \leq 9^{2}$

Simplify the left side.

Simplify ${({(6 x - 18)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(6 x - 18)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(6 x - 18)}^{\frac{1}{2} \cdot 2} \leq 9^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(6 x - 18)}^{\frac{1}{2} \cdot 2} \leq 9^{2}$

Rewrite the expression.

${(6 x - 18)}^{1} \leq 9^{2}$

Simplify.

$6 x - 18 \leq 9^{2}$

Simplify the right side.

Raise $9$ to the power of $2$ .

$6 x - 18 \leq 81$

Step 4

Solve for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Add $18$ to both sides of the inequality.

$6 x \leq 81 + 18$

Add $81$ and $18$ .

$6 x \leq 99$

Divide each term in $6 x \leq 99$ by $6$ and simplify.

Divide each term in $6 x \leq 99$ by $6$ .

$\frac{6 x}{6} \leq \frac{99}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} \leq \frac{99}{6}$

Divide $x$ by $1$ .

$x \leq \frac{99}{6}$

Simplify the right side.

Cancel the common factor of $99$ and $6$ .

Factor $3$ out of $99$ .

$x \leq \frac{3 (33)}{6}$

Cancel the common factors.

Factor $3$ out of $6$ .

$x \leq \frac{3 \cdot 33}{3 \cdot 2}$

Cancel the common factor.

$x \leq \frac{3 \cdot 33}{3 \cdot 2}$

Rewrite the expression.

$x \leq \frac{33}{2}$

Step 5

Find the domain of $\sqrt{6 (x - 3)} - 9$ .

Set the radicand in $\sqrt{6 (x - 3)}$ greater than or equal to $0$ to find where the expression is defined.

$6 (x - 3) \geq 0$

Solve for $x$ .

Divide each term in $6 (x - 3) \geq 0$ by $6$ and simplify.

Divide each term in $6 (x - 3) \geq 0$ by $6$ .

$\frac{6 (x - 3)}{6} \geq \frac{0}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 (x - 3)}{6} \geq \frac{0}{6}$

Divide $x - 3$ by $1$ .

$x - 3 \geq \frac{0}{6}$

Simplify the right side.

Divide $0$ by $6$ .

$x - 3 \geq 0$

Add $3$ to both sides of the inequality.

$x \geq 3$

The domain is all values of $x$ that make the expression defined.

$[3, \infty)$

Step 6

Use each root to create test intervals.

$x < 3$

$3 < x < \frac{33}{2}$

$x > \frac{33}{2}$

Step 7

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 $x < 3$ 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.

$x = 0$

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

$\sqrt{6 (0) - 18} + 6 \leq 15$

The left side is not equal to the right side, which means that the given statement is false.

False

Test a value on the interval $3 < x < \frac{33}{2}$ to see if it makes the inequality true.

Choose a value on the interval $3 < x < \frac{33}{2}$ and see if this value makes the original inequality true.

$x = 6$

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

$\sqrt{6 (6) - 18} + 6 \leq 15$

The left side $10.24264068$ is less than the right side $15$ , which means that the given statement is always true.

True

Test a value on the interval $x > \frac{33}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{33}{2}$ and see if this value makes the original inequality true.

$x = 19$

Replace $x$ with $19$ in the original inequality.

$\sqrt{6 (19) - 18} + 6 \leq 15$

The left side $15.79795897$ is greater than the right side $15$ , which means that the given statement is false.

False

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

$x < 3$ False

$3 < x < \frac{33}{2}$ True

$x > \frac{33}{2}$ False

$x < 3$ False

$3 < x < \frac{33}{2}$ True

$x > \frac{33}{2}$ False

Step 8

The solution consists of all of the true intervals.

$3 \leq x \leq \frac{33}{2}$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$3 \leq x \leq \frac{33}{2}$

Interval Notation:

$[3, \frac{33}{2}]$

Step 10

Solve for x square root of 6x-18+6<=15

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