Solve the following equations using operations.
\lfloor x-\dfrac{2}{7} \rfloor =3
The equation:
\lfloor x-\dfrac{2}{7} \rfloor =3
is equivalent to:
3\leq x-\dfrac{2}{7} \lt 4
Adding \dfrac{2}{7} to all sides gives:
3+ \dfrac {2} {7} \leq x- \dfrac {2} {7}+\dfrac {2} {7} \lt 4+ \dfrac {2} {7}
\dfrac{23}{7} \leq x \lt \dfrac{30}{7}
x is a solution of the equation if and only if:
\dfrac{23}{7} \leq x \lt \dfrac{30}{7}
\lfloor x-\dfrac{2}{3} \rfloor =7
The equation:
\lfloor x-\dfrac{2}{3} \rfloor =7
is equivalent to:
6\leq x-\dfrac{2}{3} \lt 7
Adding \dfrac{2}{3} to all sides gives:
6+\dfrac{2}{3} \leq x-\dfrac{2}{3} +\dfrac{2}{3} \lt 7+\dfrac{2}{3}
\dfrac{20}{3} \leq x \lt \dfrac{23}{3}
x is a solution of the equation if and only if:
\dfrac{18}{3} \leq x \lt \dfrac{23}{3}
\lfloor x-\dfrac{1}{5} \rfloor =12
The equation:
\lfloor x-\dfrac{1}{5} \rfloor =12
is equivalent to:
11\leq x-\dfrac{1}{5} \lt 12
Adding \dfrac{1}{5} to all sides gives:
11+ \dfrac{1}{5} \leq x-\dfrac{1}{5}+ \dfrac{1}{5} \lt 12+ \dfrac{1}{5}
\dfrac{56}{5} \leq x \lt \dfrac{61}{5}
x is a solution of the equation if and only if:
\dfrac{56}{5} \leq x \lt \dfrac{61}{5}
\lfloor 3x-\dfrac{2}{9} \rfloor =2
The equation:
\lfloor 3x-\dfrac{2}{9} \rfloor =2
is equivalent to:
1\leq 3x-\dfrac{2}{9} \lt 2
Adding \dfrac{2}{9} to all sides gives:
1 +\dfrac{2}{9} \leq 3x-\dfrac{2}{9}+\dfrac{2}{9} \lt 2+\dfrac{2}{9}
\dfrac{11}{9} \leq 3x \lt \dfrac{20}{9}
Multiply all sides by \dfrac{1}{3} :
\dfrac{1}{3}\cdot \dfrac{11}{9} \leq \dfrac{1}{3}\cdot 3x \lt \dfrac{1}{3}\cdot \dfrac{20}{9}
\dfrac{11}{27} \leq x \lt \dfrac{20}{27}
x is a solution of the equation if and only if:
\dfrac{11}{27} \leq x \lt \dfrac{20}{27}
\lfloor 5x-\dfrac{2}{7} \rfloor =4
The equation:
\lfloor 5x-\dfrac{2}{7} \rfloor =4
is equivalent to:
3\leq 5x-\dfrac{2}{7} \lt 4
Adding \dfrac{2}{7} to all sides gives:
3+ \dfrac{2}{7} \leq 5x-\dfrac{2}{7}+ \dfrac{2}{7} \lt 4 + \dfrac{2}{7}
\dfrac{23}{7} \leq 5x \lt \dfrac{30}{7}
Multiply all sides by \dfrac{1}{5} :
\dfrac{1}{5}\cdot \dfrac{23}{7} \leq \dfrac{1}{5}\cdot 5x \lt \dfrac{1}{5}\cdot \dfrac{30}{7}
\dfrac{23}{35} \leq x \lt \dfrac{30}{35}
x is a solution of the equation if and only if:
\dfrac{23}{35} \leq x \lt \dfrac{30}{35}
\lfloor 7x+\dfrac{1}{2} \rfloor =-1
The equation:
\lfloor 7x+\dfrac{1}{2} \rfloor =-1
is equivalent to:
-2\leq 7x+\dfrac{1}{2} \lt -1
Subtracting \dfrac{1}{2} from all sides gives:
-2 -\dfrac{1}{2} \leq 7x+\dfrac{1}{2}-\dfrac{1}{2} \lt -1-\dfrac{1}{2}
\dfrac{-5}{2} \leq 7x \lt \dfrac{-3}{2}
Multiply all sides by \dfrac{1}{7} :
\dfrac{1}{7}\cdot \dfrac{-5}{2} \leq \dfrac{1}{7}\cdot 7x \lt \dfrac{1}{7}\cdot \dfrac{-3}{2}
\dfrac{-5}{14} \leq x \lt \dfrac{-3}{14}
x is a solution of the equation if and only if:
\dfrac{-5}{14} \leq x \lt \dfrac{-3}{14}
\lfloor x+\dfrac{2}{7} \rfloor =-5
The equation:
\lfloor x+\dfrac{2}{7} \rfloor =-5
is equivalent to:
-6\leq x+\dfrac{2}{7} \lt -5
Subtracting \dfrac{2}{7} from all sides gives:
-6 -\dfrac{2}{7} \leq x+\dfrac{2}{7} -\dfrac{2}{7}\lt -5-\dfrac{2}{7}
\dfrac{-44}{7} \leq x \lt \dfrac{-37}{7}
x is a solution of the equation if and only if:
\dfrac{-44}{7} \leq x \lt \dfrac{-37}{7}