Solve (x 7)/(x-9) > x/2

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

x \in (- \infty, 0) \cup (9, 23)

Evaluate

\frac{x \times 7}{x - 9} > \frac{x}{2}

Find the domain

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\frac{x \times 7}{x - 9} > \frac{x}{2}, x \neq = 9

Use the commutative property to reorder the terms

\frac{7 x}{x - 9} > \frac{x}{2}

Move the expression to the left side

\frac{7 x}{x - 9} - \frac{x}{2} > 0

Subtract the terms

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\frac{23 x - x ^{2}}{2 ( x - 9 )} > 0

Set the numerator and denominator of \frac{23 x - x ^{2}}{2 ( x - 9 )} equal to 0 to find the values of x where sign changes may occur

23 x - x^{2} = 0 2 (x - 9) = 0

Calculate

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x = 0 x = 23 2 (x - 9) = 0

Calculate

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x = 0 x = 23 x = 9

Determine the test intervals using the critical values

x < 0 0 < x < 9 9 < x < 23 x > 23

Choose a value form each interval

x_{1} = - 1 x_{2} = 5 x_{3} = 16 x_{4} = 24

To determine if x < 0 is the solution to the inequality,test if the chosen value x = - 1 satisfies the initial inequality

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x < 0 is the solution x_{2} = 5 x_{3} = 16 x_{4} = 24

To determine if 0 < x < 9 is the solution to the inequality,test if the chosen value x = 5 satisfies the initial inequality

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x < 0 is the solution 0 < x < 9 is not a solution x_{3} = 16 x_{4} = 24

To determine if 9 < x < 23 is the solution to the inequality,test if the chosen value x = 16 satisfies the initial inequality

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x < 0 is the solution 0 < x < 9 is not a solution 9 < x < 23 is the solution x_{4} = 24

To determine if x > 23 is the solution to the inequality,test if the chosen value x = 24 satisfies the initial inequality

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x < 0 is the solution 0 < x < 9 is not a solution 9 < x < 23 is the solution x > 23 is not a solution

The original inequality is a strict inequality,so does not include the critical value ,the final solution is x \in (- \infty, 0) \cup (9, 23)

x \in (- \infty, 0) \cup (9, 23)

Check if the solution is in the defined range

x \in (- \infty, 0) \cup (9, 23), x \neq = 9

Solution

x \in (- \infty, 0) \cup (9, 23)

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