What is the answer to |α+(α-1)+(α+1)|=2 ?

The answer to |α+(α-1)+(α+1)|=2 is α=-2/3 \lor α= 2/3

Solve for α: |α+(α-1)+(α+1)|=2

The solution is α=-2/3 \lor α= 2/3

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∣ α + (α - 1) + (α + 1) ∣ = 2

α = - \frac{2}{3} or α = \frac{2}{3}

Decimal Notation

α = - 0.66666 \dots, α = 0.66666 \dots

Solution steps

∣ α + α - 1 + α + 1 ∣ = 2

Apply absolute rule: If

∣ u ∣ = a, a > 0

then

u = a

u = - a

α + α - 1 + α + 1 = - 2 or α + α - 1 + α + 1 = 2

α + α - 1 + α + 1 = - 2 or α + α - 1 + α + 1 = 2

α = - \frac{2}{3} or α = \frac{2}{3}

4 ∣ 3 x - 6 ∣ + 3 = 15

∣ 14 x + 8 ∣ + 7 = 2

∣ 2 x + 2 ∣ = 13

∣ a ∣ - 3 = 8

∣ 2 x - 3 ∣ - 3 = 4 x^{4} x

What is the answer to |α+(α-1)+(α+1)|=2 ?
The answer to |α+(α-1)+(α+1)|=2 is α=-2/3 \lor α= 2/3
Solve for α: |α+(α-1)+(α+1)|=2
The solution is α=-2/3 \lor α= 2/3