What is (|3+x|)/7 <= 5 ?

The solution to (|3+x|)/7 <= 5 is -38<= x<= 32

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(|3+x|)/7 <= 5

\frac{∣ 3 + x ∣}{7} \leq 5

- 38 \leq x \leq 32

Interval Notation

[- 38, 32]

Solution steps

\frac{∣ 3 + x ∣}{7} \leq 5

Multiply both sides by

7

∣ 3 + x ∣ \leq 35

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

- 35 \leq 3 + x \leq 35

3 + x \geq - 35 and 3 + x \leq 35

x \geq - 38 and x \leq 32

Merge Overlapping Intervals

- 38 \leq x \leq 32

x^{3} + 8 x^{2} + 5 x - 14

\frac{1}{2} 2

x^{2} - 4 x - 3 = 0

- 3 - 1

8 x^{3} + 125