abs(x-1)+abs(x+1)=2 — step-by-step solution

1) Let's solve in

ℝ

abs (x - 1) + abs (x + 1) = 2

The absolute-value arguments are zero at

{-1, 1}

. Split ℝ at these points and solve on each interval

(-∞, -1]

x - 1 \leq 0

, so

abs (x - 1) = - (x - 1)

;

x + 1 \leq 0

, so

abs (x + 1) = - (x + 1)

(-∞, -1]

, the equation becomes

- (x - 1) - (x + 1) = 2

2) Let's solve in

ℝ

- (x - 1) - (x + 1) = 2

- x

+ 1

- x

- 1 = 2

ⓘ Start from

1

and move back

1

on the number line, down to zero:

1 - 1

0

■Like terms are grouped by adding their coefficients:

- x - x

-2 x

- 2 x + 0 = 2

ⓘ Adding 0 changes nothing:

- 2 x + 0

-2 x

- 2 x = 2

ⓘ Both sides are divided by

-2

x = \frac{2}{-2}

ⓘ The sign of a fraction goes in front of the bar, whether it comes from the numerator or the denominator:

\frac{2}{-2}

- \frac{2}{2}

x = - \frac{2}{2}

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

2

, their greatest common divisor:

- \frac{2}{2}

-1

Solution found:

x = -1

-1 \leq -1

, we can keep the solution

-1

[-1, 1]

x - 1 \leq 0

, so

abs (x - 1) = - (x - 1)

;

x + 1 \geq 0

, so

abs (x + 1) = x + 1

[-1, 1]

, the equation becomes

- (x - 1) + (x + 1) = 2

3) Let's solve in

ℝ

- (x - 1) + (x + 1) = 2

ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (x - 1)

- x + 1

- x

+ 1

+ x

+ 1 = 2

ⓘ Start from

1

and move forward

1

on the number line:

1 + 1

2

■Like terms are grouped by adding their coefficients:

- x + x

0

0 + 2 = 2

ⓘ Adding 0 changes nothing:

0 + 2

2

2 = 2

ⓘ We subtract

2

from both sides

2 - 2 = 0

ⓘ Start from

2

and move back

2

on the number line, down to zero:

2 - 2

0

0 = 0

Solution found:

x \in ℝ

The equation holds for every

x

[-1, 1]

: keep the whole interval

[1, +∞)

x - 1 \geq 0

, so

abs (x - 1) = x - 1

;

x + 1 \geq 0

, so

abs (x + 1) = x + 1

[1, +∞)

, the equation becomes

(x - 1) + (x + 1) = 2

4) Let's solve in

ℝ

(x - 1) + (x + 1) = 2

ⓘ Start from

-1

and move forward

1

on the number line, up to zero:

- 1 + 1

0

■Like terms are grouped by adding their coefficients:

x + x

2 x

2 x + 0 = 2

ⓘ Adding 0 changes nothing:

2 x + 0

2 x

2 x = 2

ⓘ Both sides are divided by

2

x = \frac{2}{2}

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

2

, their greatest common divisor:

\frac{2}{2}

1

Solution found:

x = 1

1 \geq 1

, we can keep the solution

1

▶Solution found:

x \in [-1, 1]