|x-5|<= 2

Solution

∣ x - 5 ∣ \leq 2

3 \leq x \leq 7

Interval Notation

[3, 7]

Solution steps

∣ x - 5 ∣ \leq 2

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

- 2 \leq x - 5 \leq 2

x - 5 \geq - 2 and x - 5 \leq 2

x \geq 3 and x \leq 7

Merge Overlapping Intervals

3 \leq x \leq 7

g (2) = 4

\frac{( x ^{2} - 9 x + 11 )}{( x - 3 )}

6 x + 2 = 14

\frac{6.4 ( π ( \frac{0.000254}{2} ) ^{2} )}{0.3}