Solve a greatest integer function inequality with operations - High school Algebra I Exercises

Solve the following inequality using operations.

\lfloor x-\dfrac{2}{7} \rfloor ≤ 3

x \lt \dfrac{30}{7}

x \gt \dfrac {-30} {7}

x \gt \dfrac {30} {7}

x \lt \dfrac {-30} {7}

\lfloor x-\dfrac{2}{7} \rfloor ≤ 3 can be rewritten as x-\dfrac{2}{7} \lt 4 since the greatest integer of all numbers less than 4 will be less than or equal to 3.

Solve x-\dfrac{2}{7} \lt 4 by adding \dfrac{2}{7} to both sides:

x- \dfrac {2} {7} + \dfrac {2} {7} \lt 4+ \dfrac {2} {7}

x is a solution of the equation if and only if:

x \lt \dfrac{30}{7}

\lfloor x-\dfrac{1}{4} \rfloor ≤ 2

x \lt \dfrac{4}{3}

x \lt \dfrac{13}{4}

x \lt \dfrac{9}{2}

x \lt \dfrac{-15}{7}

\lfloor x-\dfrac{1}{4} \rfloor ≤ 2 can be rewritten as x-\dfrac{1}{4} \lt 3 since the greatest integer of all numbers less than 3 will be less than or equal to 2.

Solve x-\dfrac{1}{4} \lt 3 by adding \dfrac{1}{4} to both sides:

x-\dfrac{1}{4}+\dfrac{1}{4} \lt 3+\dfrac{1}{4}

x is a solution of the equation if and only if:

x \lt \dfrac{13}{4}

\lfloor x-\dfrac{3}{10} \rfloor ≤ 4

x \lt \dfrac{57}{4}

x \lt \dfrac{42}{5}

x \lt \dfrac{63}{5}

x \lt \dfrac{53}{3}

\lfloor x-\dfrac{3}{10} \rfloor ≤ 4 can be rewritten as x-\dfrac{3}{10} \lt 5 since the greatest integer of all numbers less than 5 will be less than or equal to 4.

Solve x-\dfrac{3}{10} \lt 5 by adding \dfrac{3}{10} to both sides:

x-\dfrac{3}{10}+\dfrac{3}{10} \lt 5+\dfrac{3}{10}

x \lt \dfrac{53}{3}

x is a solution of the equation if and only if:

x \lt \dfrac{53}{3}

\lfloor 2x-\dfrac{1}{10} \rfloor ≤ 3

x \lt \dfrac{33}{25}

x \lt \dfrac{41}{20}

x \lt \dfrac{23}{15}

x \lt \dfrac{26}{17}

\lfloor 2x-\dfrac{1}{10} \rfloor ≤ 3 can be rewritten as 2x-\dfrac{1}{10} \lt 4 since the greatest integer of all numbers less than 4 will be less than or equal to 3.

Solve 2x-\dfrac{1}{10} \lt 4 by adding \dfrac{1}{10} to both sides:

2x-\dfrac{1}{10}+\dfrac{1}{10} \lt 4+\dfrac{1}{10}

Therefore:

2x \lt \dfrac{41}{10}

Multiply both sides by \dfrac{1}{2} :

\dfrac{1}{2}\cdot2x \lt \dfrac{1}{2}\cdot\dfrac{41}{10}

x is a solution of the equation if and only if:

x \lt \dfrac{41}{20}

\lfloor 2x-1 \rfloor \geq 5

x \geq 6

x \geq 12

x \geq 3

x \geq 5

\lfloor 2x-1 \rfloor \geq 5 can be rewritten as 2x-1 \geq 5 since the greatest integer of all numbers greater than or equal to 5 will be greater than or equal to 5.

Solve 2x-1 \geq 5 by adding 1 to both sides:

2x-1+1 \geq 5+1

Therefore:

2x \geq 6

Multiply both sides by \dfrac{1}{2} :

\dfrac{1}{2}\cdot2x \geq \dfrac{1}{2}\cdot 6

x is a solution of the equation if and only if:

x \geq 3

\lfloor 3x-\dfrac{1}{2} \rfloor \geq 4

x \geq \dfrac{5}{2}

x \geq \dfrac{7}{3}

x \geq \dfrac{7}{5}

x \geq \dfrac{3}{2}

\lfloor 3x-\dfrac{1}{2} \rfloor \geq 4 can be rewritten as 3x-\dfrac{1}{2} \geq 4 since the greatest integer of all numbers greater than or equal to 4 will be greater than or equal to 4.

Solve 3x-\dfrac{1}{2} \geq 4 by adding \dfrac{1}{2} to both sides:

3x-\dfrac{1}{2} +\dfrac{1}{2} \geq 4 +\dfrac{1}{2}

Therefore:

3x \geq \dfrac{9}{2}

Multiply both sides by \dfrac{1}{3} :

\dfrac{1}{3}\cdot3x \geq \dfrac{1}{3}\cdot \dfrac{9}{2}

x is a solution of the equation if and only if:

x \geq \dfrac{3}{2}

\lfloor 5x-\dfrac{1}{6} \rfloor \leq 10

x \lt \dfrac{67}{30}

x \lt \dfrac{47}{21}

x \lt \dfrac{43}{32}

x \lt \dfrac{53}{23}

\lfloor 5x-\dfrac{1}{6} \rfloor \leq 10 can be rewritten as 5x-\dfrac{1}{6} \lt 11 since the greatest integer of all numbers less than 11 will be less than or equal to 10.

Solve 5x-\dfrac{1}{6} \lt 11 by adding \dfrac{1}{6} to both sides:

5x-\dfrac{1}{6}+\dfrac{1}{6} \lt 11+\dfrac{1}{6}

Therefore:

5x \lt \dfrac{67}{6}

Multiply both sides by \dfrac{1}{5} :

\dfrac{1}{5}\cdot5x \lt \dfrac{1}{5}\cdot\dfrac{67}{6}

x is a solution of the equation if and only if:

x \lt \dfrac{67}{30}