abs(2*x-1)+3=abs(x-5) — step-by-step solution

1) Let's solve in

ℝ

abs (2 x - 1) + 3 = abs (x - 5)

The absolute-value arguments are zero at

{\frac{1}{2}, 5}

. Split ℝ at these points and solve on each interval

(-∞, \frac{1}{2}]

2 x - 1 \leq 0

, so

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

;

x - 5 \leq 0

, so

abs (x - 5) = - (x - 5)

(-∞, \frac{1}{2}]

, the equation becomes

- (2 x - 1) + 3 = - (x - 5)

2) Let's solve in

ℝ

- (2 x - 1) + 3 = - (x - 5)

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

- (2 x - 1)

- 2 x + 1

- (x - 5)

- x + 5

- 2 x + 1 + 3 = - x + 5

ⓘ Start from

1

and move forward

3

on the number line:

1 + 3

4

- 2 x + 4 = - x + 5

ⓘ We gather the terms in

x

on the right and the other terms on the left

4 - 5 = - x + 2 x

ⓘ Start from

4

and move back

5

on the number line, going below zero:

4 - 5

-1

■Like terms are grouped by adding their coefficients:

- x + 2 x

x

-1 = x

Solution obtained:

x = -1

-1 \leq \frac{1}{2}

, we can keep the solution

-1

[\frac{1}{2}, 5]

2 x - 1 \geq 0

, so

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

;

x - 5 \leq 0

, so

abs (x - 5) = - (x - 5)

[\frac{1}{2}, 5]

, the equation becomes

(2 x - 1) + 3 = - (x - 5)

3) Let's solve in

ℝ

(2 x - 1) + 3 = - (x - 5)

ⓘ Start from

-1

and move forward

3

on the number line, going past zero:

- 1 + 3

2

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

- (x - 5)

- x + 5

2 x + 2 = - x + 5

ⓘ We subtract

2

from both sides■Start from

5

and move back

2

on the number line:

5 - 2

3

2 x = - x + 3

ⓘ We add

x

to both sides

2 x + x = 3

ⓘ Like terms are grouped by adding their coefficients:

2 x + x

3 x

3 x = 3

ⓘ Both sides are divided by

3

x = \frac{3}{3}

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

3

, their greatest common divisor:

\frac{3}{3}

1

Solution found:

x = 1

1 \in [\frac{1}{2}, 5]

, we can keep the solution

1

[5, +∞)

2 x - 1 \geq 0

, so

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

;

x - 5 \geq 0

, so

abs (x - 5) = x - 5

[5, +∞)

, the equation becomes

(2 x - 1) + 3 = x - 5

4) Let's solve in

ℝ

(2 x - 1) + 3 = x - 5

ⓘ Start from

-1

and move forward

3

on the number line, going past zero:

- 1 + 3

2

2 x + 2 = x - 5

ⓘ We subtract

2

from both sides■Start from

-5

and move back

2

more on the number line, away from zero:

- 5 - 2

-7

2 x = x - 7

ⓘ We subtract

x

from both sides

2 x - x = -7

ⓘ Like terms are grouped by adding their coefficients:

2 x - x

x

Solution obtained:

x = -7

-7 \notin [5, +∞)

, we cannot keep the solution

-7

▶Solutions found:

x = {-1, 1}