Solve (x^2)/15-(7x-1)/5≤(5-2x)/9

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve for x

\frac{- 3001 + 53}{6} \leq x \leq \frac{3001 + 53}{6}

Alternative Form

x \in [\frac{- 3001 + 53}{6}, \frac{3001 + 53}{6}]

Evaluate

\frac{x ^{2}}{15} - \frac{7 x - 1}{5} \leq \frac{5 - 2 x}{9}

Multiply both sides of the inequality by 45

(\frac{x ^{2}}{15} - \frac{7 x - 1}{5}) \times 45 \leq \frac{5 - 2 x}{9} \times 45

Multiply the terms

More Steps

3 x^{2} - 63 x + 9 \leq \frac{5 - 2 x}{9} \times 45

Multiply the terms

More Steps

3 x^{2} - 63 x + 9 \leq 25 - 10 x

Move the expression to the left side

3 x^{2} - 63 x + 9 - (25 - 10 x) \leq 0

Calculate the sum or difference

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3 x^{2} - 53 x - 16 \leq 0

Rewrite the expression

3 x^{2} - 53 x - 16 = 0

Add or subtract both sides

3 x^{2} - 53 x = 16

Divide both sides

\frac{3 x ^{2} - 53 x}{3} = \frac{16}{3}

Evaluate

x^{2} - \frac{53}{3} x = \frac{16}{3}

Add the same value to both sides

x^{2} - \frac{53}{3} x + \frac{2809}{36} = \frac{16}{3} + \frac{2809}{36}

Simplify the expression

(x - \frac{53}{6})^{2} = \frac{3001}{36}

Take the root of both sides of the equation and remember to use both positive and negative roots

x - \frac{53}{6} = \pm \frac{3001}{36}

Simplify the expression

x - \frac{53}{6} = \pm \frac{3001}{6}

Separate the equation into 2 possible cases

x - \frac{53}{6} = \frac{3001}{6} x - \frac{53}{6} = - \frac{3001}{6}

Solve the equation

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x = \frac{3001 + 53}{6} x - \frac{53}{6} = - \frac{3001}{6}

Solve the equation

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x = \frac{3001 + 53}{6} x = \frac{- 3001 + 53}{6}

Determine the test intervals using the critical values

x < \frac{- 3001 + 53}{6} \frac{- 3001 + 53}{6} < x < \frac{3001 + 53}{6} x > \frac{3001 + 53}{6}

Choose a value form each interval

x_{1} = - 1 x_{2} = 9 x_{3} = 19

To determine if x < \frac{- 3001 + 53}{6} is the solution to the inequality,test if the chosen value x = - 1 satisfies the initial inequality

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x < \frac{- 3001 + 53}{6} is not a solution x_{2} = 9 x_{3} = 19

To determine if \frac{- 3001 + 53}{6} < x < \frac{3001 + 53}{6} is the solution to the inequality,test if the chosen value x = 9 satisfies the initial inequality

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x < \frac{- 3001 + 53}{6} is not a solution \frac{- 3001 + 53}{6} < x < \frac{3001 + 53}{6} is the solution x_{3} = 19

To determine if x > \frac{3001 + 53}{6} is the solution to the inequality,test if the chosen value x = 19 satisfies the initial inequality

More Steps

x < \frac{- 3001 + 53}{6} is not a solution \frac{- 3001 + 53}{6} < x < \frac{3001 + 53}{6} is the solution x > \frac{3001 + 53}{6} is not a solution

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

\frac{- 3001 + 53}{6} \leq x \leq \frac{3001 + 53}{6} is the solution

Solution

\frac{- 3001 + 53}{6} \leq x \leq \frac{3001 + 53}{6}

Alternative Form

x \in [\frac{- 3001 + 53}{6}, \frac{3001 + 53}{6}]

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