Algebra Examples

$x (2 x - 3) < 5 x < 3 x (x + 7)$

Step 1

Solve $x (2 x - 3) < 5 x$ for $x$ .

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Step 1.1

Simplify $x (2 x - 3)$ .

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Step 1.1.1

Rewrite.

$0 + 0 + x (2 x - 3) < 5 x$

Step 1.1.2

Simplify by multiplying through.

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Step 1.1.2.1

Apply the distributive property.

$x (2 x) + x \cdot - 3 < 5 x$

Step 1.1.2.2

Reorder.

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Step 1.1.2.2.1

Rewrite using the commutative property of multiplication.

$2 x \cdot x + x \cdot - 3 < 5 x$

Step 1.1.2.2.2

Move $- 3$ to the left of $x$ .

$2 x \cdot x - 3 \cdot x < 5 x$

Step 1.1.3

Multiply $x$ by $x$ by adding the exponents.

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Step 1.1.3.1

Move $x$ .

$2 (x \cdot x) - 3 \cdot x < 5 x$

Step 1.1.3.2

Multiply $x$ by $x$ .

$2 x^{2} - 3 \cdot x < 5 x$

$2 x^{2} - 3 x < 5 x$

Step 1.2

Move all terms containing $x$ to the left side of the inequality.

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Step 1.2.1

Subtract $5 x$ from both sides of the inequality.

$2 x^{2} - 3 x - 5 x < 0$

Step 1.2.2

Subtract $5 x$ from $- 3 x$ .

$2 x^{2} - 8 x < 0$

Step 1.3

Convert the inequality to an equation.

$2 x^{2} - 8 x = 0$

Step 1.4

Factor $2 x$ out of $2 x^{2} - 8 x$ .

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Step 1.4.1

Factor $2 x$ out of $2 x^{2}$ .

$2 x (x) - 8 x = 0$

Step 1.4.2

Factor $2 x$ out of $- 8 x$ .

$2 x (x) + 2 x (- 4) = 0$

Step 1.4.3

Factor $2 x$ out of $2 x (x) + 2 x (- 4)$ .

$2 x (x - 4) = 0$

Step 1.5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 4 = 0$

Step 1.6

Set $x$ equal to $0$ .

$x = 0$

Step 1.7

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 1.7.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 1.7.2

Add $4$ to both sides of the equation.

$x = 4$

Step 1.8

The final solution is all the values that make $2 x (x - 4) = 0$ true.

$x = 0, 4$

Step 1.9

Use each root to create test intervals.

$x < 0$

$0 < x < 4$

$x > 4$

Step 1.10

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 1.10.1

Test a value on the interval $x < 0$ to see if it makes the inequality true.

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Step 1.10.1.1

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 1.10.1.2

Replace $x$ with $- 2$ in the original inequality.

$(- 2) (2 (- 2) - 3) < 5 (- 2)$

Step 1.10.1.3

The left side $14$ is not less than the right side $- 10$ , which means that the given statement is false.

False

Step 1.10.2

Test a value on the interval $0 < x < 4$ to see if it makes the inequality true.

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Step 1.10.2.1

Choose a value on the interval $0 < x < 4$ and see if this value makes the original inequality true.

$x = 2$

Step 1.10.2.2

Replace $x$ with $2$ in the original inequality.

$(2) (2 (2) - 3) < 5 (2)$

Step 1.10.2.3

The left side $2$ is less than the right side $10$ , which means that the given statement is always true.

True

Step 1.10.3

Test a value on the interval $x > 4$ to see if it makes the inequality true.

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Step 1.10.3.1

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

Step 1.10.3.2

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

$(6) (2 (6) - 3) < 5 (6)$

Step 1.10.3.3

The left side $54$ is not less than the right side $30$ , which means that the given statement is false.

False

Step 1.10.4

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

$x < 0$ False

$0 < x < 4$ True

$x > 4$ False

$x < 0$ False

$0 < x < 4$ True

$x > 4$ False

Step 1.11

The solution consists of all of the true intervals.

$0 < x < 4$

Step 2

Solve $5 x < 3 x (x + 7)$ for $x$ .

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Step 2.1

Rewrite so $x$ is on the left side of the inequality.

$3 x (x + 7) > 5 x$

Step 2.2

Simplify $3 x (x + 7)$ .

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Step 2.2.1

Rewrite.

$0 + 0 + 3 x (x + 7) > 5 x$

Step 2.2.2

Simplify by adding zeros.

$3 x (x + 7) > 5 x$

Step 2.2.3

Apply the distributive property.

$3 x \cdot x + 3 x \cdot 7 > 5 x$

Step 2.2.4

Multiply $x$ by $x$ by adding the exponents.

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Step 2.2.4.1

Move $x$ .

$3 (x \cdot x) + 3 x \cdot 7 > 5 x$

Step 2.2.4.2

Multiply $x$ by $x$ .

$3 x^{2} + 3 x \cdot 7 > 5 x$

Step 2.2.5

Multiply $7$ by $3$ .

$3 x^{2} + 21 x > 5 x$

Step 2.3

Move all terms containing $x$ to the left side of the inequality.

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Step 2.3.1

Subtract $5 x$ from both sides of the inequality.

$3 x^{2} + 21 x - 5 x > 0$

Step 2.3.2

Subtract $5 x$ from $21 x$ .

$3 x^{2} + 16 x > 0$

Step 2.4

Convert the inequality to an equation.

$3 x^{2} + 16 x = 0$

Step 2.5

Factor $x$ out of $3 x^{2} + 16 x$ .

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Step 2.5.1

Factor $x$ out of $3 x^{2}$ .

$x (3 x) + 16 x = 0$

Step 2.5.2

Factor $x$ out of $16 x$ .

$x (3 x) + x \cdot 16 = 0$

Step 2.5.3

Factor $x$ out of $x (3 x) + x \cdot 16$ .

$x (3 x + 16) = 0$

Step 2.6

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$3 x + 16 = 0$

Step 2.7

Set $x$ equal to $0$ .

$x = 0$

Step 2.8

Set $3 x + 16$ equal to $0$ and solve for $x$ .

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Step 2.8.1

Set $3 x + 16$ equal to $0$ .

$3 x + 16 = 0$

Step 2.8.2

Solve $3 x + 16 = 0$ for $x$ .

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Step 2.8.2.1

Subtract $16$ from both sides of the equation.

$3 x = - 16$

Step 2.8.2.2

Divide each term in $3 x = - 16$ by $3$ and simplify.

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Step 2.8.2.2.1

Divide each term in $3 x = - 16$ by $3$ .

$\frac{3 x}{3} = \frac{- 16}{3}$

Step 2.8.2.2.2

Simplify the left side.

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Step 2.8.2.2.2.1

Cancel the common factor of $3$ .

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Step 2.8.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 16}{3}$

Step 2.8.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 16}{3}$

Step 2.8.2.2.3

Simplify the right side.

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Step 2.8.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{16}{3}$

Step 2.9

The final solution is all the values that make $x (3 x + 16) = 0$ true.

$x = 0, - \frac{16}{3}$

Step 2.10

Use each root to create test intervals.

$x < - \frac{16}{3}$

$- \frac{16}{3} < x < 0$

$x > 0$

Step 2.11

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 2.11.1

Test a value on the interval $x < - \frac{16}{3}$ to see if it makes the inequality true.

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Step 2.11.1.1

Choose a value on the interval $x < - \frac{16}{3}$ and see if this value makes the original inequality true.

$x = - 8$

Step 2.11.1.2

Replace $x$ with $- 8$ in the original inequality.

$5 (- 8) < 3 (- 8) ((- 8) + 7)$

Step 2.11.1.3

The left side $- 40$ is less than the right side $24$ , which means that the given statement is always true.

True

Step 2.11.2

Test a value on the interval $- \frac{16}{3} < x < 0$ to see if it makes the inequality true.

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Step 2.11.2.1

Choose a value on the interval $- \frac{16}{3} < x < 0$ and see if this value makes the original inequality true.

$x = - 3$

Step 2.11.2.2

Replace $x$ with $- 3$ in the original inequality.

$5 (- 3) < 3 (- 3) ((- 3) + 7)$

Step 2.11.2.3

The left side $- 15$ is not less than the right side $- 36$ , which means that the given statement is false.

False

Step 2.11.3

Test a value on the interval $x > 0$ to see if it makes the inequality true.

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Step 2.11.3.1

Choose a value on the interval $x > 0$ and see if this value makes the original inequality true.

$x = 2$

Step 2.11.3.2

Replace $x$ with $2$ in the original inequality.

$5 (2) < 3 (2) ((2) + 7)$

Step 2.11.3.3

The left side $10$ is less than the right side $54$ , which means that the given statement is always true.

True

Step 2.11.4

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

$x < - \frac{16}{3}$ True

$- \frac{16}{3} < x < 0$ False

$x > 0$ True

$x < - \frac{16}{3}$ True

$- \frac{16}{3} < x < 0$ False

$x > 0$ True

Step 2.12

The solution consists of all of the true intervals.

$x < - \frac{16}{3}$ or $x > 0$

Step 3

Find the intersection of $0 < x < 4$ and $x < - \frac{16}{3} or x > 0$ .

$0 < x < 4$

Step 4

The result can be shown in multiple forms.

Inequality Form:

$0 < x < 4$

Interval Notation:

$(0, 4)$

Step 5