Solve the inequality by separating into cases of x+12/x<7

x \in \left(-\infty,0\right)\cup \left(3,4\right)

Question :

x+12/x<7

x \in \left(-\infty,0\right)\cup \left(3,4\right)

x \in \left(-\infty,0\right)\cup \left(3,4\right)

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

x \in (- \infty, 0) \cup (3, 4)

Evaluate

x + \frac{12}{x} < 7

Find the domain

x + \frac{12}{x} < 7, x \neq = 0

Move the expression to the left side

x + \frac{12}{x} - 7 < 0

Calculate the sum or difference

More Steps

\frac{x ^{2} + 12 - 7 x}{x} < 0

Set the numerator and denominator of \frac{x ^{2} + 12 - 7 x}{x} equal to 0 to find the values of x where sign changes may occur

x^{2} + 12 - 7 x = 0 x = 0

Calculate

More Steps

x = 4 x = 3 x = 0

Determine the test intervals using the critical values

x < 0 0 < x < 3 3 < x < 4 x > 4

Choose a value form each interval

x_{1} = - 1 x_{2} = 2 x_{3} = \frac{7}{2} x_{4} = 5

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

More Steps

x < 0 is the solution x_{2} = 2 x_{3} = \frac{7}{2} x_{4} = 5

To determine if 0 < x < 3 is the solution to the inequality,test if the chosen value x = 2 satisfies the initial inequality

More Steps

x < 0 is the solution 0 < x < 3 is not a solution x_{3} = \frac{7}{2} x_{4} = 5

To determine if 3 < x < 4 is the solution to the inequality,test if the chosen value x = \frac{7}{2} satisfies the initial inequality

More Steps

x < 0 is the solution 0 < x < 3 is not a solution 3 < x < 4 is the solution x_{4} = 5

To determine if x > 4 is the solution to the inequality,test if the chosen value x = 5 satisfies the initial inequality

More Steps

x < 0 is the solution 0 < x < 3 is not a solution 3 < x < 4 is the solution x > 4 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 (3, 4)

x \in (- \infty, 0) \cup (3, 4)

Check if the solution is in the defined range

x \in (- \infty, 0) \cup (3, 4), x \neq = 0

Solution

x \in (- \infty, 0) \cup (3, 4)

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