Solve (2x^2)/4 - (x 1)/3 <= 3x - 33/6

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

Solve for x

3 \leq x \leq \frac{11}{3}

Alternative Form

x \in [3, \frac{11}{3}]

Evaluate

\frac{2 x ^{2}}{4} - \frac{x \times 1}{3} \leq 3 x - \frac{33}{6}

Simplify

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\frac{x ^{2}}{2} - \frac{x}{3} \leq 3 x - \frac{33}{6}

Cancel out the common factor 3

\frac{x ^{2}}{2} - \frac{x}{3} \leq 3 x - \frac{11}{2}

Multiply both sides of the inequality by 6

(\frac{x ^{2}}{2} - \frac{x}{3}) \times 6 \leq (3 x - \frac{11}{2}) \times 6

Multiply the terms

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3 x^{2} - 2 x \leq (3 x - \frac{11}{2}) \times 6

Multiply the terms

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3 x^{2} - 2 x \leq 18 x - 33

Move the expression to the left side

3 x^{2} - 2 x - (18 x - 33) \leq 0

Subtract the terms

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3 x^{2} - 20 x + 33 \leq 0

Rewrite the expression

3 x^{2} - 20 x + 33 = 0

Factor the expression

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(x - 3) (3 x - 11) = 0

When the product of factors equals 0,at least one factor is 0

x - 3 = 0 3 x - 11 = 0

Solve the equation for x

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x = 3 3 x - 11 = 0

Solve the equation for x

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x = 3 x = \frac{11}{3}

Determine the test intervals using the critical values

x < 3 3 < x < \frac{11}{3} x > \frac{11}{3}

Choose a value form each interval

x_{1} = 2 x_{2} = \frac{10}{3} x_{3} = 5

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

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x < 3 is not a solution x_{2} = \frac{10}{3} x_{3} = 5

To determine if 3 < x < \frac{11}{3} is the solution to the inequality,test if the chosen value x = \frac{10}{3} satisfies the initial inequality

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x < 3 is not a solution 3 < x < \frac{11}{3} is the solution x_{3} = 5

To determine if x > \frac{11}{3} is the solution to the inequality,test if the chosen value x = 5 satisfies the initial inequality

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x < 3 is not a solution 3 < x < \frac{11}{3} is the solution x > \frac{11}{3} is not a solution

The original inequality is a nonstrict inequality,so include the critical value in the solution

3 \leq x \leq \frac{11}{3} is the solution

Solution

3 \leq x \leq \frac{11}{3}

Alternative Form

x \in [3, \frac{11}{3}]

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