Solve x: -8x + 2 > 16 - 3x

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve for x

- 42 + 12 < x < 42 + 12

Alternative Form

x \in (- 42 + 12, 42 + 12)

Evaluate

x \div (- 8) x + 2 > 16 - 3 x

Simplify

More Steps

- \frac{x ^{2}}{8} + 2 > 16 - 3 x

Multiply both sides of the inequality by 8

(- \frac{x ^{2}}{8} + 2) \times 8 > (16 - 3 x) \times 8

Multiply the terms

More Steps

- x^{2} + 16 > (16 - 3 x) \times 8

Multiply the terms

More Steps

- x^{2} + 16 > 128 - 24 x

Move the expression to the left side

- x^{2} + 16 - (128 - 24 x) > 0

Calculate the sum or difference

More Steps

- x^{2} - 112 + 24 x > 0

Rewrite the expression

- x^{2} - 112 + 24 x = 0

Add or subtract both sides

- x^{2} + 24 x = 112

Divide both sides

\frac{- x ^{2} + 24 x}{- 1} = \frac{112}{- 1}

Evaluate

x^{2} - 24 x = - 112

Add the same value to both sides

x^{2} - 24 x + 144 = - 112 + 144

Simplify the expression

(x - 12)^{2} = 32

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

x - 12 = \pm 32

Simplify the expression

x - 12 = \pm 42

Separate the equation into 2 possible cases

x - 12 = 42 x - 12 = - 42

Move the constant to the right-hand side and change its sign

x = 42 + 12 x - 12 = - 42

Move the constant to the right-hand side and change its sign

x = 42 + 12 x = - 42 + 12

Determine the test intervals using the critical values

x < - 42 + 12 - 42 + 12 < x < 42 + 12 x > 42 + 12

Choose a value form each interval

x_{1} = 5 x_{2} = 12 x_{3} = 19

To determine if x < - 42 + 12 is the solution to the inequality,test if the chosen value x = 5 satisfies the initial inequality

More Steps

x < - 42 + 12 is not a solution x_{2} = 12 x_{3} = 19

To determine if - 42 + 12 < x < 42 + 12 is the solution to the inequality,test if the chosen value x = 12 satisfies the initial inequality

More Steps

x < - 42 + 12 is not a solution - 42 + 12 < x < 42 + 12 is the solution x_{3} = 19

To determine if x > 42 + 12 is the solution to the inequality,test if the chosen value x = 19 satisfies the initial inequality

More Steps

x < - 42 + 12 is not a solution - 42 + 12 < x < 42 + 12 is the solution x > 42 + 12 is not a solution

Solution

- 42 + 12 < x < 42 + 12

Alternative Form

x \in (- 42 + 12, 42 + 12)

Show Solution