Algebra Examples

$\frac{2}{x^{2} - 3 x + 2} \leq \frac{6}{x^{2} - 4}$

Step 1

Solve $x < - 2 or 1 < x < 2 or x \geq \frac{5}{2}$ .

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Step 1.1

Multiply both sides by $x^{2} - 3 x + 2$ .

$\frac{2}{x^{2} - 3 x + 2} (x^{2} - 3 x + 2) = \frac{6}{x^{2} - 4} (x^{2} - 3 x + 2)$

Step 1.2

Simplify.

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Step 1.2.1

Simplify the left side.

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Step 1.2.1.1

Cancel the common factor of $x^{2} - 3 x + 2$ .

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Step 1.2.1.1.1

Cancel the common factor.

$\frac{2}{x^{2} - 3 x + 2} (x^{2} - 3 x + 2) = \frac{6}{x^{2} - 4} (x^{2} - 3 x + 2)$

Step 1.2.1.1.2

Rewrite the expression.

$2 = \frac{6}{x^{2} - 4} (x^{2} - 3 x + 2)$

Step 1.2.2

Simplify the right side.

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Step 1.2.2.1

Simplify $\frac{6}{x^{2} - 4} (x^{2} - 3 x + 2)$ .

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Step 1.2.2.1.1

Simplify the denominator.

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Step 1.2.2.1.1.1

Rewrite $4$ as $2^{2}$ .

$2 = \frac{6}{x^{2} - 2^{2}} (x^{2} - 3 x + 2)$

Step 1.2.2.1.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$2 = \frac{6}{(x + 2) (x - 2)} (x^{2} - 3 x + 2)$

Step 1.2.2.1.2

Multiply $\frac{6}{(x + 2) (x - 2)}$ by $x^{2} - 3 x + 2$ .

$2 = \frac{6 (x^{2} - 3 x + 2)}{(x + 2) (x - 2)}$

Step 1.2.2.1.3

Factor $x^{2} - 3 x + 2$ using the AC method.

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Step 1.2.2.1.3.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $2$ and whose sum is $- 3$ .

$- 2, - 1$

Step 1.2.2.1.3.2

Write the factored form using these integers.

$2 = \frac{6 ((x - 2) (x - 1))}{(x + 2) (x - 2)}$

$2 = \frac{6 (x - 2) (x - 1)}{(x + 2) (x - 2)}$

Step 1.2.2.1.4

Cancel the common factor of $x - 2$ .

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Step 1.2.2.1.4.1

Cancel the common factor.

$2 = \frac{6 (x - 2) (x - 1)}{(x + 2) (x - 2)}$

Step 1.2.2.1.4.2

Rewrite the expression.

$2 = \frac{6 (x - 1)}{x + 2}$

Step 1.3

Solve for $x$ .

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Step 1.3.1

Rewrite the equation as $\frac{6 (x - 1)}{x + 2} = 2$ .

$\frac{6 (x - 1)}{x + 2} = 2$

Step 1.3.2

Find the LCD of the terms in the equation.

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Step 1.3.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x + 2, 1$

Step 1.3.2.2

Remove parentheses.

$x + 2, 1$

Step 1.3.2.3

The LCM of one and any expression is the expression.

$x + 2$

Step 1.3.3

Multiply each term in $\frac{6 (x - 1)}{x + 2} = 2$ by $x + 2$ to eliminate the fractions.

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Step 1.3.3.1

Multiply each term in $\frac{6 (x - 1)}{x + 2} = 2$ by $x + 2$ .

$\frac{6 (x - 1)}{x + 2} (x + 2) = 2 (x + 2)$

Step 1.3.3.2

Simplify the left side.

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Step 1.3.3.2.1

Cancel the common factor of $x + 2$ .

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Step 1.3.3.2.1.1

Cancel the common factor.

$\frac{6 (x - 1)}{x + 2} (x + 2) = 2 (x + 2)$

Step 1.3.3.2.1.2

Rewrite the expression.

$6 (x - 1) = 2 (x + 2)$

Step 1.3.3.2.2

Apply the distributive property.

$6 x + 6 \cdot - 1 = 2 (x + 2)$

Step 1.3.3.2.3

Multiply $6$ by $- 1$ .

$6 x - 6 = 2 (x + 2)$

Step 1.3.3.3

Simplify the right side.

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Step 1.3.3.3.1

Apply the distributive property.

$6 x - 6 = 2 x + 2 \cdot 2$

Step 1.3.3.3.2

Multiply $2$ by $2$ .

$6 x - 6 = 2 x + 4$

Step 1.3.4

Solve the equation.

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Step 1.3.4.1

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

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Step 1.3.4.1.1

Subtract $2 x$ from both sides of the equation.

$6 x - 6 - 2 x = 4$

Step 1.3.4.1.2

Subtract $2 x$ from $6 x$ .

$4 x - 6 = 4$

Step 1.3.4.2

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

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Step 1.3.4.2.1

Add $6$ to both sides of the equation.

$4 x = 4 + 6$

Step 1.3.4.2.2

Add $4$ and $6$ .

$4 x = 10$

Step 1.3.4.3

Divide each term in $4 x = 10$ by $4$ and simplify.

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Step 1.3.4.3.1

Divide each term in $4 x = 10$ by $4$ .

$\frac{4 x}{4} = \frac{10}{4}$

Step 1.3.4.3.2

Simplify the left side.

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Step 1.3.4.3.2.1

Cancel the common factor of $4$ .

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Step 1.3.4.3.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{10}{4}$

Step 1.3.4.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{10}{4}$

Step 1.3.4.3.3

Simplify the right side.

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Step 1.3.4.3.3.1

Cancel the common factor of $10$ and $4$ .

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Step 1.3.4.3.3.1.1

Factor $2$ out of $10$ .

$x = \frac{2 (5)}{4}$

Step 1.3.4.3.3.1.2

Cancel the common factors.

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Step 1.3.4.3.3.1.2.1

Factor $2$ out of $4$ .

$x = \frac{2 \cdot 5}{2 \cdot 2}$

Step 1.3.4.3.3.1.2.2

Cancel the common factor.

$x = \frac{2 \cdot 5}{2 \cdot 2}$

Step 1.3.4.3.3.1.2.3

Rewrite the expression.

$x = \frac{5}{2}$

Step 1.4

Find the domain of $\frac{2}{(x - 2) (x - 1)} - \frac{6}{(x + 2) (x - 2)}$ .

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Step 1.4.1

Set the denominator in $\frac{2}{(x - 2) (x - 1)}$ equal to $0$ to find where the expression is undefined.

$(x - 2) (x - 1) = 0$

Step 1.4.2

Solve for $x$ .

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Step 1.4.2.1

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

$x - 2 = 0$

$x - 1 = 0$

Step 1.4.2.2

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 1.4.2.2.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 1.4.2.2.2

Add $2$ to both sides of the equation.

$x = 2$

Step 1.4.2.3

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 1.4.2.3.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 1.4.2.3.2

Add $1$ to both sides of the equation.

$x = 1$

Step 1.4.2.4

The final solution is all the values that make $(x - 2) (x - 1) = 0$ true.

$x = 2, 1$

Step 1.4.3

Set the denominator in $\frac{6}{(x + 2) (x - 2)}$ equal to $0$ to find where the expression is undefined.

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

Step 1.4.4

Solve for $x$ .

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Step 1.4.4.1

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

$x + 2 = 0$

$x - 2 = 0$

Step 1.4.4.2

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 1.4.4.2.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 1.4.4.2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 1.4.4.3

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 1.4.4.3.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 1.4.4.3.2

Add $2$ to both sides of the equation.

$x = 2$

Step 1.4.4.4

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

$x = - 2, 2$

Step 1.4.5

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

$(- \infty, - 2) \cup (- 2, 1) \cup (1, 2) \cup (2, \infty)$

Step 1.5

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 1$

$1 < x < 2$

$2 < x < \frac{5}{2}$

$x > \frac{5}{2}$

Step 1.6

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 1.6.1

Test a value on the interval $x < - 2$ to see if it makes the inequality true.

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Step 1.6.1.1

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Step 1.6.1.2

Replace $x$ with $- 4$ in the original inequality.

$\frac{2}{{(- 4)}^{2} - 3 \cdot - 4 + 2} \leq \frac{6}{{(- 4)}^{2} - 4}$

Step 1.6.1.3

The left side $0.0\overline{6}$ is less than the right side $0.5$ , which means that the given statement is always true.

True

Step 1.6.2

Test a value on the interval $- 2 < x < 1$ to see if it makes the inequality true.

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Step 1.6.2.1

Choose a value on the interval $- 2 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 1.6.2.2

Replace $x$ with $0$ in the original inequality.

$\frac{2}{{(0)}^{2} - 3 \cdot 0 + 2} \leq \frac{6}{{(0)}^{2} - 4}$

Step 1.6.2.3

The left side $1$ is greater than the right side $- 1.5$ , which means that the given statement is false.

False

Step 1.6.3

Test a value on the interval $1 < x < 2$ to see if it makes the inequality true.

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Step 1.6.3.1

Choose a value on the interval $1 < x < 2$ and see if this value makes the original inequality true.

$x = 1.5$

Step 1.6.3.2

Replace $x$ with $1.5$ in the original inequality.

$\frac{2}{{(1.5)}^{2} - 3 \cdot 1.5 + 2} \leq \frac{6}{{(1.5)}^{2} - 4}$

Step 1.6.3.3

The left side $- 8$ is less than the right side $- 3.\overline{428571}$ , which means that the given statement is always true.

True

Step 1.6.4

Test a value on the interval $2 < x < \frac{5}{2}$ to see if it makes the inequality true.

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Step 1.6.4.1

Choose a value on the interval $2 < x < \frac{5}{2}$ and see if this value makes the original inequality true.

$x = 2.25$

Step 1.6.4.2

Replace $x$ with $2.25$ in the original inequality.

$\frac{2}{{(2.25)}^{2} - 3 \cdot 2.25 + 2} \leq \frac{6}{{(2.25)}^{2} - 4}$

Step 1.6.4.3

The left side $6.4$ is greater than the right side $5.64705882$ , which means that the given statement is false.

False

Step 1.6.5

Test a value on the interval $x > \frac{5}{2}$ to see if it makes the inequality true.

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Step 1.6.5.1

Choose a value on the interval $x > \frac{5}{2}$ and see if this value makes the original inequality true.

$x = 5$

Step 1.6.5.2

Replace $x$ with $5$ in the original inequality.

$\frac{2}{{(5)}^{2} - 3 \cdot 5 + 2} \leq \frac{6}{{(5)}^{2} - 4}$

Step 1.6.5.3

The left side $0.1\overline{6}$ is less than the right side $0.\overline{285714}$ , which means that the given statement is always true.

True

Step 1.6.6

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

$x < - 2$ True

$- 2 < x < 1$ False

$1 < x < 2$ True

$2 < x < \frac{5}{2}$ False

$x > \frac{5}{2}$ True

$x < - 2$ True

$- 2 < x < 1$ False

$1 < x < 2$ True

$2 < x < \frac{5}{2}$ False

$x > \frac{5}{2}$ True

Step 1.7

The solution consists of all of the true intervals.

$x < - 2$ or $1 < x < 2$ or $x \geq \frac{5}{2}$

Step 2

Use the inequality $x < - 2 or 1 < x < 2 or x \geq \frac{5}{2}$ to build the set notation.

${x | x < - 2 or 1 < x < 2 or x \geq \frac{5}{2}}$

Step 3