|2*x-1|+3=|5-x|+4 — step-by-step solution

1) We solve in

ℝ

| 2 x - 1 | + 3 = | 5 - x | + 4

| 2 x - 1 |

+ 3

- | 5 - x |

- 4 = 0

ⓘ Start from

3

and move back

4

on the number line, going below zero:

3 - 4

-1

| 2 x - 1 | - 1 - | 5 - x | = 0

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

| 2 x - 1 | = - (2 x - 1)

;

5 - x \geq 0

, so

| 5 - x | = 5 - x

(-∞, \frac{1}{2}]

, the equation becomes

- (2 x - 1) - 1 - (5 - x) = 0

2) We solve in

ℝ

- (2 x - 1) - 1 - (5 - x) = 0

- 2 x

+ 1

- 1

- 5

+ x = 0

ⓘ 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:

- 2 x + x

- x

- x + 0 - 5 = 0

ⓘ Adding 0 preserves the original value:

0 - 5

-5

- x - 5 = 0

ⓘ Add

x

to both sides

-5 = x

Solution obtained:

x = -5

Since

-5 \leq \frac{1}{2}

, we can keep the solution

-5

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

2 x - 1 \geq 0

, so

| 2 x - 1 | = 2 x - 1

;

5 - x \geq 0

, so

| 5 - x | = 5 - x

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

, the equation becomes

(2 x - 1) - 1 - (5 - x) = 0

3) We solve in

ℝ

(2 x - 1) - 1 - (5 - x) = 0

ⓘ Start from

-1

and move back

1

more on the number line, away from zero:

- 1 - 1

-2

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

- (5 - x)

- 5 + x

2 x

- 2

- 5

+ x = 0

ⓘ Start from

-2

and move back

5

more on the number line, away from zero:

- 2 - 5

-7

■Like terms are grouped by adding their coefficients:

2 x + x

3 x

3 x - 7 = 0

ⓘ Add

7

to both sides

3 x = 7

ⓘ Both sides are divided by

3

Solution obtained:

x = \frac{7}{3}

\frac{7}{3} \in [\frac{1}{2}, 5]

, we can keep the solution

\frac{7}{3}

[5, +∞)

2 x - 1 \geq 0

, so

| 2 x - 1 | = 2 x - 1

;

5 - x \leq 0

, so

| 5 - x | = - (5 - x)

[5, +∞)

, the equation becomes

(2 x - 1) - 1 + (5 - x) = 0

4) We solve in

ℝ

(2 x - 1) - 1 + (5 - x) = 0

ⓘ Start from

-1

and move back

1

more on the number line, away from zero:

- 1 - 1

-2

■Group like terms by adding their coefficients:

2 x - x

x

x - 2 + 5 = 0

ⓘ Start from

-2

and move forward

5

on the number line, going past zero:

- 2 + 5

3

x + 3 = 0

ⓘ Subtract

3

from both sides

Solution obtained:

x = -3

-3 \notin [5, +∞)

, we cannot keep the solution

-3

▶Solutions obtained:

x = {-5, \frac{7}{3}}

5) The numeric value of:

x = {-5, \frac{7}{3}}

is:

▷

x = {-5, \approx 2.3333}