Algebra Examples

$y = 5 x^{2} + 10$

Step 1

Interchange the variables.

$x = 5 y^{2} + 10$

Step 2

Solve for

$y$ .

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Step 2.1

Rewrite the equation as

$5 y^{2} + 10 = x$ .

$5 y^{2} + 10 = x$

Step 2.2

Subtract

$10$ from both sides of the equation.

$5 y^{2} = x - 10$

Step 2.3

Divide each term in

$5 y^{2} = x - 10$ by

$5$ and simplify.

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Step 2.3.1

Divide each term in

$5 y^{2} = x - 10$ by

$5$ .

$\frac{5 y^{2}}{5} = \frac{x}{5} + \frac{- 10}{5}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of

$5$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{5 y^{2}}{5} = \frac{x}{5} + \frac{- 10}{5}$

Step 2.3.2.1.2

Divide

$y^{2}$ by

$1$ .

$y^{2} = \frac{x}{5} + \frac{- 10}{5}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide

$- 10$ by

$5$ .

$y^{2} = \frac{x}{5} - 2$

Step 2.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{\frac{x}{5} - 2}$

Step 2.5

Simplify

$\pm \sqrt{\frac{x}{5} - 2}$ .

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Step 2.5.1

To write

$- 2$ as a fraction with a common denominator, multiply by

$\frac{5}{5}$ .

$y = \pm \sqrt{\frac{x}{5} - 2 \cdot \frac{5}{5}}$

Step 2.5.2

Combine

$- 2$ and

$\frac{5}{5}$ .

$y = \pm \sqrt{\frac{x}{5} + \frac{- 2 \cdot 5}{5}}$

Step 2.5.3

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{x - 2 \cdot 5}{5}}$

Step 2.5.4

Multiply

$- 2$ by

$5$ .

$y = \pm \sqrt{\frac{x - 10}{5}}$

Step 2.5.5

Rewrite

$\sqrt{\frac{x - 10}{5}}$ as

$\frac{\sqrt{x - 10}}{\sqrt{5}}$ .

$y = \pm \frac{\sqrt{x - 10}}{\sqrt{5}}$

Step 2.5.6

Multiply

$\frac{\sqrt{x - 10}}{\sqrt{5}}$ by

$\frac{\sqrt{5}}{\sqrt{5}}$ .

$y = \pm \frac{\sqrt{x - 10}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$

Step 2.5.7

Combine and simplify the denominator.

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Step 2.5.7.1

Multiply

$\frac{\sqrt{x - 10}}{\sqrt{5}}$ by

$\frac{\sqrt{5}}{\sqrt{5}}$ .

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{\sqrt{5} \sqrt{5}}$

Step 2.5.7.2

Raise

$\sqrt{5}$ to the power of

$1$ .

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{{\sqrt{5}}^{1} \sqrt{5}}$

Step 2.5.7.3

Raise

$\sqrt{5}$ to the power of

$1$ .

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{{\sqrt{5}}^{1} {\sqrt{5}}^{1}}$

Step 2.5.7.4

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{{\sqrt{5}}^{1 + 1}}$

Step 2.5.7.5

Add

$1$ and

$1$ .

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{{\sqrt{5}}^{2}}$

Step 2.5.7.6

Rewrite

${\sqrt{5}}^{2}$ as

$5$ .

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Step 2.5.7.6.1

Use

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

$\sqrt{5}$ as

$5^{\frac{1}{2}}$ .

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{{(5^{\frac{1}{2}})}^{2}}$

Step 2.5.7.6.2

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{5^{\frac{1}{2} \cdot 2}}$

Step 2.5.7.6.3

Combine

$\frac{1}{2}$ and

$2$ .

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{5^{\frac{2}{2}}}$

Step 2.5.7.6.4

Cancel the common factor of

$2$ .

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Step 2.5.7.6.4.1

Cancel the common factor.

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{5^{\frac{2}{2}}}$

Step 2.5.7.6.4.2

Rewrite the expression.

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{5^{1}}$

Step 2.5.7.6.5

Evaluate the exponent.

$y = \pm \frac{\sqrt{x - 10} \sqrt{5}}{5}$

Step 2.5.8

Combine using the product rule for radicals.

$y = \pm \frac{\sqrt{(x - 10) \cdot 5}}{5}$

Step 2.5.9

Reorder factors in

$\pm \frac{\sqrt{(x - 10) \cdot 5}}{5}$ .

$y = \pm \frac{\sqrt{5 (x - 10)}}{5}$

Step 2.6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.6.1

First, use the positive value of the

$\pm$ to find the first solution.

$y = \frac{\sqrt{5 (x - 10)}}{5}$

Step 2.6.2

Next, use the negative value of the

$\pm$ to find the second solution.

$y = - \frac{\sqrt{5 (x - 10)}}{5}$

Step 2.6.3

The complete solution is the result of both the positive and negative portions of the solution.

$y = \frac{\sqrt{5 (x - 10)}}{5}$

$y = - \frac{\sqrt{5 (x - 10)}}{5}$

$y = \frac{\sqrt{5 (x - 10)}}{5}$

$y = - \frac{\sqrt{5 (x - 10)}}{5}$

$y = \frac{\sqrt{5 (x - 10)}}{5}$

$y = - \frac{\sqrt{5 (x - 10)}}{5}$

Step 3

Replace

$y$ with

$f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = \frac{\sqrt{5 (x - 10)}}{5}, - \frac{\sqrt{5 (x - 10)}}{5}$

Step 4

Verify if

$f^{- 1} (x) = \frac{\sqrt{5 (x - 10)}}{5}, - \frac{\sqrt{5 (x - 10)}}{5}$ is the inverse of

$f (x) = 5 x^{2} + 10$ .

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Step 4.1

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of

$f (x) = 5 x^{2} + 10$ and

$f^{- 1} (x) = \frac{\sqrt{5 (x - 10)}}{5}, - \frac{\sqrt{5 (x - 10)}}{5}$ and compare them.

Step 4.2

Find the range of

$f (x) = 5 x^{2} + 10$ .

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Step 4.2.1

The range is the set of all valid

$y$ values. Use the graph to find the range.

Interval Notation:

$[10, \infty)$

Step 4.3

Find the domain of

$\frac{\sqrt{5 (x - 10)}}{5}$ .

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Step 4.3.1

Set the radicand in

$\sqrt{5 (x - 10)}$ greater than or equal to

$0$ to find where the expression is defined.

$5 (x - 10) \geq 0$

Step 4.3.2

Solve for

$x$ .

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Step 4.3.2.1

Divide each term in

$5 (x - 10) \geq 0$ by

$5$ and simplify.

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Step 4.3.2.1.1

Divide each term in

$5 (x - 10) \geq 0$ by

$5$ .

$\frac{5 (x - 10)}{5} \geq \frac{0}{5}$

Step 4.3.2.1.2

Simplify the left side.

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Step 4.3.2.1.2.1

Cancel the common factor of

$5$ .

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Step 4.3.2.1.2.1.1

Cancel the common factor.

$\frac{5 (x - 10)}{5} \geq \frac{0}{5}$

Step 4.3.2.1.2.1.2

Divide

$x - 10$ by

$1$ .

$x - 10 \geq \frac{0}{5}$

Step 4.3.2.1.3

Simplify the right side.

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Step 4.3.2.1.3.1

Divide

$0$ by

$5$ .

$x - 10 \geq 0$

Step 4.3.2.2

Add

$10$ to both sides of the inequality.

$x \geq 10$

Step 4.3.3

The domain is all values of

$x$ that make the expression defined.

$[10, \infty)$

Step 4.4

Find the domain of

$f (x) = 5 x^{2} + 10$ .

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Step 4.4.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

$(- \infty, \infty)$

Step 4.5

Since the domain of

$f^{- 1} (x) = \frac{\sqrt{5 (x - 10)}}{5}, - \frac{\sqrt{5 (x - 10)}}{5}$ is the range of

$f (x) = 5 x^{2} + 10$ and the range of

$f^{- 1} (x) = \frac{\sqrt{5 (x - 10)}}{5}, - \frac{\sqrt{5 (x - 10)}}{5}$ is the domain of

$f (x) = 5 x^{2} + 10$ , then

$f^{- 1} (x) = \frac{\sqrt{5 (x - 10)}}{5}, - \frac{\sqrt{5 (x - 10)}}{5}$ is the inverse of

$f (x) = 5 x^{2} + 10$ .

$f^{- 1} (x) = \frac{\sqrt{5 (x - 10)}}{5}, - \frac{\sqrt{5 (x - 10)}}{5}$

Step 5