3=1+|2-12y| — step-by-step solution

1) Let's solve in

ℝ

1 + | 2 - 12 y | = 3

ⓘ We subtract

1

from both sides

| 2 - 12 y | = 3 - 1

ⓘ Start from

3

and move back

1

on the number line:

3 - 1

2

| 2 - 12 y | = 2

Since

2 \geq 0

| 2 - 12 y | = 2

is equivalent to

2 - 12 y = 2

2 - 12 y = -2

2) Let's solve in

ℝ

2 - 12 y = 2

2 - 12 y - 2 = 0

ⓘ Start from

2

and move back

2

on the number line, down to zero:

2 - 2

0

0 - 12 y = 0

ⓘ Adding 0 changes nothing:

0 - 12 y

- 12 y

- 12 y = 0

We simplify the expression by

-12

⇥Let's solve in

ℝ

y = 0

⇥Solution obtained:

y = 0

3) Let's solve in

ℝ

2 - 12 y = -2

ⓘ We gather the terms in

y

on the right and the other terms on the left

2 + 2 = 12 y

ⓘ Start from

2

and move forward

2

on the number line:

2 + 2

4

4 = 12 y

ⓘ Both sides are divided by

12

y = \frac{4}{12}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

4

, their greatest common divisor:

\frac{4}{12}

\frac{1 \cdot 4}{3 \cdot 4}

\frac{1}{3}

Solution found:

y = \frac{1}{3}

▶Solution:

y = {0, \frac{1}{3}}

4) The numeric value of:

y = {0, \frac{1}{3}}

equals:

▷

y = {0, \approx 0.3333}