Solve 2(9x-9) 2≥4(8x-6) 9x

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve for x

\frac{- 17 + 7}{16} \leq x \leq \frac{17 + 7}{16}

Alternative Form

x \in [\frac{- 17 + 7}{16}, \frac{17 + 7}{16}]

Evaluate

2 (9 x - 9) \times 2 \geq 4 (8 x - 6) \times 9 x

Multiply the terms

4 (9 x - 9) \geq 4 (8 x - 6) \times 9 x

Multiply

More Steps

4 (9 x - 9) \geq 36 x (8 x - 6)

Move the expression to the left side

4 (9 x - 9) - 36 x (8 x - 6) \geq 0

Subtract the terms

More Steps

252 x - 36 - 288 x^{2} \geq 0

Rewrite the expression

252 x - 36 - 288 x^{2} = 0

Add or subtract both sides

252 x - 288 x^{2} = 36

Divide both sides

\frac{252 x - 288 x ^{2}}{- 288} = \frac{36}{- 288}

Evaluate

- \frac{7}{8} x + x^{2} = - \frac{1}{8}

Add the same value to both sides

- \frac{7}{8} x + x^{2} + \frac{49}{256} = - \frac{1}{8} + \frac{49}{256}

Simplify the expression

(x - \frac{7}{16})^{2} = \frac{17}{256}

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

x - \frac{7}{16} = \pm \frac{17}{256}

Simplify the expression

x - \frac{7}{16} = \pm \frac{17}{16}

Separate the equation into 2 possible cases

x - \frac{7}{16} = \frac{17}{16} x - \frac{7}{16} = - \frac{17}{16}

Solve the equation

More Steps

x = \frac{17 + 7}{16} x - \frac{7}{16} = - \frac{17}{16}

Solve the equation

More Steps

x = \frac{17 + 7}{16} x = \frac{- 17 + 7}{16}

Determine the test intervals using the critical values

x < \frac{- 17 + 7}{16} \frac{- 17 + 7}{16} < x < \frac{17 + 7}{16} x > \frac{17 + 7}{16}

Choose a value form each interval

x_{1} = - 1 x_{2} = \frac{7}{16} x_{3} = 2

To determine if x < \frac{- 17 + 7}{16} is the solution to the inequality,test if the chosen value x = - 1 satisfies the initial inequality

More Steps

x < \frac{- 17 + 7}{16} is not a solution x_{2} = \frac{7}{16} x_{3} = 2

To determine if \frac{- 17 + 7}{16} < x < \frac{17 + 7}{16} is the solution to the inequality,test if the chosen value x = \frac{7}{16} satisfies the initial inequality

More Steps

x < \frac{- 17 + 7}{16} is not a solution \frac{- 17 + 7}{16} < x < \frac{17 + 7}{16} is the solution x_{3} = 2

To determine if x > \frac{17 + 7}{16} is the solution to the inequality,test if the chosen value x = 2 satisfies the initial inequality

More Steps

x < \frac{- 17 + 7}{16} is not a solution \frac{- 17 + 7}{16} < x < \frac{17 + 7}{16} is the solution x > \frac{17 + 7}{16} is not a solution

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

\frac{- 17 + 7}{16} \leq x \leq \frac{17 + 7}{16} is the solution

Solution

\frac{- 17 + 7}{16} \leq x \leq \frac{17 + 7}{16}

Alternative Form

x \in [\frac{- 17 + 7}{16}, \frac{17 + 7}{16}]

Show Solution