What is |(3y-3)/3 |+4<= 7 ?

The solution to |(3y-3)/3 |+4<= 7 is -2<= y<= 4

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|(3y-3)/3 |+4<= 7

\frac{3 y - 3}{3} + 4 \leq 7

- 2 \leq y \leq 4

Interval Notation

[- 2, 4]

Solution steps

\frac{3 y - 3}{3} + 4 \leq 7

Move

4

to the right side

\frac{3 y - 3}{3} \leq 3

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

- 3 \leq \frac{3 y - 3}{3} \leq 3

\frac{3 y - 3}{3} \geq - 3 and \frac{3 y - 3}{3} \leq 3

y \geq - 2 and y \leq 4

Merge Overlapping Intervals

- 2 \leq y \leq 4

7 v^{3} \cdot 10 u^{3} v^{5} \cdot 8 u v^{3}

- 3 \leq x < 7

\frac{3}{2} - 8

13 x - 5 = 4 (3 x + 4) - 11