Solve the Inequality for x 10- square root of 2x+7<=3

$10 - \sqrt{2 x + 7} \leq 3$

Move all terms not containing $- \sqrt{2 x + 7}$ to the right side of the inequality.

Subtract $10$ from both sides of the inequality.

$- \sqrt{2 x + 7} \leq 3 - 10$

Subtract $10$ from $3$ .

$- \sqrt{2 x + 7} \leq - 7$

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

${(- \sqrt{2 x + 7})}^{2} \leq {(- 7)}^{2}$

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x + 7}$ as ${(2 x + 7)}^{\frac{1}{2}}$ .

${(- {(2 x + 7)}^{\frac{1}{2}})}^{2} \leq {(- 7)}^{2}$

Simplify the left side.

Simplify ${(- {(2 x + 7)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- {(2 x + 7)}^{\frac{1}{2}}$ .

${(- 1)}^{2} {({(2 x + 7)}^{\frac{1}{2}})}^{2} \leq {(- 7)}^{2}$

Raise $- 1$ to the power of $2$ .

$1 {({(2 x + 7)}^{\frac{1}{2}})}^{2} \leq {(- 7)}^{2}$

Multiply ${({(2 x + 7)}^{\frac{1}{2}})}^{2}$ by $1$ .

${({(2 x + 7)}^{\frac{1}{2}})}^{2} \leq {(- 7)}^{2}$

Multiply the exponents in ${({(2 x + 7)}^{\frac{1}{2}})}^{2}$ .

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

${(2 x + 7)}^{\frac{1}{2} \cdot 2} \leq {(- 7)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 x + 7)}^{\frac{1}{2} \cdot 2} \leq {(- 7)}^{2}$

Rewrite the expression.

${(2 x + 7)}^{1} \leq {(- 7)}^{2}$

Simplify.

$2 x + 7 \leq {(- 7)}^{2}$

Simplify the right side.

Raise $- 7$ to the power of $2$ .

$2 x + 7 \leq 49$

Solve for $x$ .

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

Subtract $7$ from both sides of the inequality.

$2 x \leq 49 - 7$

Subtract $7$ from $49$ .

$2 x \leq 42$

Divide each term in $2 x \leq 42$ by $2$ and simplify.

Divide each term in $2 x \leq 42$ by $2$ .

$\frac{2 x}{2} \leq \frac{42}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \leq \frac{42}{2}$

Divide $x$ by $1$ .

$x \leq \frac{42}{2}$

Simplify the right side.

Divide $42$ by $2$ .

$x \leq 21$

Find the domain of $7 - \sqrt{2 x + 7}$ .

Set the radicand in $\sqrt{2 x + 7}$ greater than or equal to $0$ to find where the expression is defined.

$2 x + 7 \geq 0$

Solve for $x$ .

Subtract $7$ from both sides of the inequality.

$2 x \geq - 7$

Divide each term in $2 x \geq - 7$ by $2$ and simplify.

Divide each term in $2 x \geq - 7$ by $2$ .

$\frac{2 x}{2} \geq \frac{- 7}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \geq \frac{- 7}{2}$

Divide $x$ by $1$ .

$x \geq \frac{- 7}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x \geq - \frac{7}{2}$

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

$[- \frac{7}{2}, \infty)$

The solution consists of all of the true intervals.

$x \geq 21$

The result can be shown in multiple forms.

Inequality Form:

$x \geq 21$

Interval Notation:

$[21, \infty)$

Solve the Inequality for x 10- square root of 2x+7<=3

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