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Sagot :
To solve the inequality [tex]\(2|x-1| < 2x + 5\)[/tex], we need to consider the definition of the absolute value function. Absolute value cases will split our problem into two inequalities which stem from the nature of the absolute value function:
1. For [tex]\(x - 1 \geq 0\)[/tex], which simplifies to [tex]\(x \geq 1\)[/tex], the inequality becomes:
[tex]\[2(x - 1) < 2x + 5\][/tex]
2. For [tex]\(x - 1 < 0\)[/tex], which simplifies to [tex]\(x < 1\)[/tex], the inequality becomes:
[tex]\[2(1 - x) < 2x + 5\][/tex]
Let's solve each case separately:
### Case 1: [tex]\(x \geq 1\)[/tex]
The inequality is:
[tex]\[2(x - 1) < 2x + 5\][/tex]
First, simplify the left-hand side:
[tex]\[2x - 2 < 2x + 5\][/tex]
Next, subtract [tex]\(2x\)[/tex] from both sides:
[tex]\[-2 < 5\][/tex]
This inequality is always true, meaning that for all [tex]\(x \geq 1\)[/tex], the inequality holds.
### Case 2: [tex]\(x < 1\)[/tex]
The inequality is:
[tex]\[2(1 - x) < 2x + 5\][/tex]
First, expand the left-hand side:
[tex]\[2 - 2x < 2x + 5\][/tex]
Next, add [tex]\(2x\)[/tex] to both sides:
[tex]\[2 < 4x + 5\][/tex]
Now, subtract 5 from both sides:
[tex]\[-3 < 4x\][/tex]
Finally, divide both sides by 4:
[tex]\[-\frac{3}{4} < x\][/tex]
or
[tex]\[x > -\frac{3}{4}\][/tex]
### Combining the results
From case 1, we have [tex]\(x \geq 1\)[/tex].
From case 2, we have [tex]\(-\frac{3}{4} < x < 1\)[/tex].
Combining these two cases, we get the complete solution:
[tex]\[-\frac{3}{4} < x < 1 \text{ or } x \geq 1\][/tex]
Thus, the solution to the inequality [tex]\(2|x-1| < 2x + 5\)[/tex] is:
[tex]\[ x > -\frac{3}{4} \][/tex]
1. For [tex]\(x - 1 \geq 0\)[/tex], which simplifies to [tex]\(x \geq 1\)[/tex], the inequality becomes:
[tex]\[2(x - 1) < 2x + 5\][/tex]
2. For [tex]\(x - 1 < 0\)[/tex], which simplifies to [tex]\(x < 1\)[/tex], the inequality becomes:
[tex]\[2(1 - x) < 2x + 5\][/tex]
Let's solve each case separately:
### Case 1: [tex]\(x \geq 1\)[/tex]
The inequality is:
[tex]\[2(x - 1) < 2x + 5\][/tex]
First, simplify the left-hand side:
[tex]\[2x - 2 < 2x + 5\][/tex]
Next, subtract [tex]\(2x\)[/tex] from both sides:
[tex]\[-2 < 5\][/tex]
This inequality is always true, meaning that for all [tex]\(x \geq 1\)[/tex], the inequality holds.
### Case 2: [tex]\(x < 1\)[/tex]
The inequality is:
[tex]\[2(1 - x) < 2x + 5\][/tex]
First, expand the left-hand side:
[tex]\[2 - 2x < 2x + 5\][/tex]
Next, add [tex]\(2x\)[/tex] to both sides:
[tex]\[2 < 4x + 5\][/tex]
Now, subtract 5 from both sides:
[tex]\[-3 < 4x\][/tex]
Finally, divide both sides by 4:
[tex]\[-\frac{3}{4} < x\][/tex]
or
[tex]\[x > -\frac{3}{4}\][/tex]
### Combining the results
From case 1, we have [tex]\(x \geq 1\)[/tex].
From case 2, we have [tex]\(-\frac{3}{4} < x < 1\)[/tex].
Combining these two cases, we get the complete solution:
[tex]\[-\frac{3}{4} < x < 1 \text{ or } x \geq 1\][/tex]
Thus, the solution to the inequality [tex]\(2|x-1| < 2x + 5\)[/tex] is:
[tex]\[ x > -\frac{3}{4} \][/tex]
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