Algebra Examples

$2 x + 5 > 11$ or $- \frac{1}{3} (x + 15) < - 10$

Step 1

Simplify the first inequality.

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Step 1.1

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

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Step 1.1.1

Subtract $5$ from both sides of the inequality.

$2 x > 11 - 5$ or $- \frac{1}{3} \cdot (x + 15) < - 10$

Step 1.1.2

Subtract $5$ from $11$ .

$2 x > 6$ or $- \frac{1}{3} \cdot (x + 15) < - 10$

Step 1.2

Divide each term in $2 x > 6$ by $2$ and simplify.

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Step 1.2.1

Divide each term in $2 x > 6$ by $2$ .

$\frac{2 x}{2} > \frac{6}{2}$ or $- \frac{1}{3} \cdot (x + 15) < - 10$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Cancel the common factor of $2$ .

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Step 1.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} > \frac{6}{2}$ or $- \frac{1}{3} \cdot (x + 15) < - 10$

Step 1.2.2.1.2

Divide $x$ by $1$ .

$x > \frac{6}{2}$ or $- \frac{1}{3} \cdot (x + 15) < - 10$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

Divide $6$ by $2$ .

$x > 3$ or $- \frac{1}{3} \cdot (x + 15) < - 10$

Step 2

Simplify the second inequality.

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Step 2.1

Multiply each term in $- \frac{1}{3} \cdot (x + 15) < - 10$ by $- 3$ to eliminate the fractions.

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Step 2.1.1

Multiply each term in $- \frac{1}{3} \cdot (x + 15) < - 10$ by $- 3$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$x > 3$ or $- \frac{1}{3} \cdot (x + 15) \cdot - 3 > - 10 \cdot - 3$

Step 2.1.2

Simplify the left side.

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Step 2.1.2.1

Apply the distributive property.

$x > 3$ or $(- \frac{1}{3} x - \frac{1}{3} \cdot 15) \cdot - 3 > - 10 \cdot - 3$

Step 2.1.2.2

Combine $x$ and $\frac{1}{3}$ .

$x > 3$ or $(- \frac{x}{3} - \frac{1}{3} \cdot 15) \cdot - 3 > - 10 \cdot - 3$

Step 2.1.2.3

Cancel the common factor of $3$ .

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Step 2.1.2.3.1

Move the leading negative in $- \frac{1}{3}$ into the numerator.

$x > 3$ or $(- \frac{x}{3} + \frac{- 1}{3} \cdot 15) \cdot - 3 > - 10 \cdot - 3$

Step 2.1.2.3.2

Factor $3$ out of $15$ .

$x > 3$ or $(- \frac{x}{3} + \frac{- 1}{3} \cdot (3 (5))) \cdot - 3 > - 10 \cdot - 3$

Step 2.1.2.3.3

Cancel the common factor.

$x > 3$ or $(- \frac{x}{3} + \frac{- 1}{3} \cdot (3 \cdot 5)) \cdot - 3 > - 10 \cdot - 3$

Step 2.1.2.3.4

Rewrite the expression.

$x > 3$ or $(- \frac{x}{3} - 1 \cdot 5) \cdot - 3 > - 10 \cdot - 3$

Step 2.1.2.4

Multiply $- 1$ by $5$ .

$x > 3$ or $(- \frac{x}{3} - 5) \cdot - 3 > - 10 \cdot - 3$

Step 2.1.2.5

Apply the distributive property.

$x > 3$ or $- \frac{x}{3} \cdot - 3 - 5 \cdot - 3 > - 10 \cdot - 3$

Step 2.1.2.6

Cancel the common factor of $3$ .

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Step 2.1.2.6.1

Move the leading negative in $- \frac{x}{3}$ into the numerator.

$x > 3$ or $\frac{- x}{3} \cdot - 3 - 5 \cdot - 3 > - 10 \cdot - 3$

Step 2.1.2.6.2

Factor $3$ out of $- 3$ .

$x > 3$ or $\frac{- x}{3} \cdot (3 (- 1)) - 5 \cdot - 3 > - 10 \cdot - 3$

Step 2.1.2.6.3

Cancel the common factor.

$x > 3$ or $\frac{- x}{3} \cdot (3 \cdot - 1) - 5 \cdot - 3 > - 10 \cdot - 3$

Step 2.1.2.6.4

Rewrite the expression.

$x > 3$ or $- x \cdot - 1 - 5 \cdot - 3 > - 10 \cdot - 3$

Step 2.1.2.7

Multiply.

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Step 2.1.2.7.1

Multiply $- 1$ by $- 1$ .

$x > 3$ or $1 x - 5 \cdot - 3 > - 10 \cdot - 3$

Step 2.1.2.7.2

Multiply $x$ by $1$ .

$x > 3$ or $x - 5 \cdot - 3 > - 10 \cdot - 3$

Step 2.1.2.7.3

Multiply $- 5$ by $- 3$ .

$x > 3$ or $x + 15 > - 10 \cdot - 3$

Step 2.1.3

Simplify the right side.

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Step 2.1.3.1

Multiply $- 10$ by $- 3$ .

$x > 3$ or $x + 15 > 30$

Step 2.2

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

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Step 2.2.1

Subtract $15$ from both sides of the inequality.

$x > 3$ or $x > 30 - 15$

Step 2.2.2

Subtract $15$ from $30$ .

$x > 3$ or $x > 15$

Step 3

The union consists of all of the elements that are contained in each interval.

$x > 3$

Step 4

The result can be shown in multiple forms.

Inequality Form:

$x > 3$

Interval Notation:

$(3, \infty)$

Step 5