Pre-Algebra Examples

$9 x^{2} - 49 < 0$

Step 1

Add $49$ to both sides of the inequality.

$9 x^{2} < 49$

Step 2

Divide each term in $9 x^{2} < 49$ by $9$ and simplify.

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Step 2.1

Divide each term in $9 x^{2} < 49$ by $9$ .

$\frac{9 x^{2}}{9} < \frac{49}{9}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $9$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{9 x^{2}}{9} < \frac{49}{9}$

Step 2.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} < \frac{49}{9}$

Step 3

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

$\sqrt{x^{2}} < \sqrt{\frac{49}{9}}$

Step 4

Simplify the equation.

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Step 4.1

Simplify the left side.

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Step 4.1.1

Pull terms out from under the radical.

$| x | < \sqrt{\frac{49}{9}}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $\sqrt{\frac{49}{9}}$ .

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Step 4.2.1.1

Rewrite $\sqrt{\frac{49}{9}}$ as $\frac{\sqrt{49}}{\sqrt{9}}$ .

$| x | < \frac{\sqrt{49}}{\sqrt{9}}$

Step 4.2.1.2

Simplify the numerator.

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Step 4.2.1.2.1

Rewrite $49$ as $7^{2}$ .

$| x | < \frac{\sqrt{7^{2}}}{\sqrt{9}}$

Step 4.2.1.2.2

Pull terms out from under the radical.

$| x | < \frac{| 7 |}{\sqrt{9}}$

Step 4.2.1.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $7$ is $7$ .

$| x | < \frac{7}{\sqrt{9}}$

Step 4.2.1.3

Simplify the denominator.

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Step 4.2.1.3.1

Rewrite $9$ as $3^{2}$ .

$| x | < \frac{7}{\sqrt{3^{2}}}$

Step 4.2.1.3.2

Pull terms out from under the radical.

$| x | < \frac{7}{| 3 |}$

Step 4.2.1.3.3

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$| x | < \frac{7}{3}$

Step 5

Write $| x | < \frac{7}{3}$ as a piecewise.

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Step 5.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 5.2

In the piece where $x$ is non-negative, remove the absolute value.

$x < \frac{7}{3}$

Step 5.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 5.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x < \frac{7}{3}$

Step 5.5

Write as a piecewise.

${\begin{cases} x < \frac{7}{3} & x \geq 0 \\ - x < \frac{7}{3} & x < 0 \end{cases}$

Step 6

Find the intersection of $x < \frac{7}{3}$ and $x \geq 0$ .

$0 \leq x < \frac{7}{3}$

Step 7

Solve $- x < \frac{7}{3}$ when $x < 0$ .

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Step 7.1

Divide each term in $- x < \frac{7}{3}$ by $- 1$ and simplify.

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Step 7.1.1

Divide each term in $- x < \frac{7}{3}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} > \frac{\frac{7}{3}}{- 1}$

Step 7.1.2

Simplify the left side.

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Step 7.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} > \frac{\frac{7}{3}}{- 1}$

Step 7.1.2.2

Divide $x$ by $1$ .

$x > \frac{\frac{7}{3}}{- 1}$

Step 7.1.3

Simplify the right side.

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Step 7.1.3.1

Move the negative one from the denominator of $\frac{\frac{7}{3}}{- 1}$ .

$x > - 1 \cdot \frac{7}{3}$

Step 7.1.3.2

Rewrite $- 1 \cdot \frac{7}{3}$ as $- \frac{7}{3}$ .

$x > - \frac{7}{3}$

Step 7.2

Find the intersection of $x > - \frac{7}{3}$ and $x < 0$ .

$- \frac{7}{3} < x < 0$

Step 8

Find the union of the solutions.

$- \frac{7}{3} < x < \frac{7}{3}$

Step 9