Answer :
Sure, let's solve each part of the problem step-by-step.
### Part (a) [tex]\(|x+3| + |x-5| < 8\)[/tex]
To solve the inequality involving absolute values, we will consider different cases based on the critical points at [tex]\(x = -3\)[/tex] and [tex]\(x = 5\)[/tex]. These are the points where the expressions inside the absolute value can change sign.
#### Case 1: [tex]\(x \leq -3\)[/tex]
Here, both expressions are non-positive:
[tex]\[ |x + 3| = -(x + 3) \][/tex]
[tex]\[ |x - 5| = -(x - 5) \][/tex]
So the inequality becomes:
[tex]\[ -(x + 3) - (x - 5) < 8 \][/tex]
[tex]\[ -x - 3 - x + 5 < 8 \][/tex]
[tex]\[ -2x + 2 < 8 \][/tex]
[tex]\[ -2x < 6 \][/tex]
[tex]\[ x > -3 \][/tex]
Since this case assumed [tex]\(x \leq -3\)[/tex], there is no [tex]\(x\)[/tex] that satisfies both conditions simultaneously. This case does not contribute to the solution set.
#### Case 2: [tex]\(-3 < x < 5\)[/tex]
Here, [tex]\(x + 3\)[/tex] is non-negative, and [tex]\(x - 5\)[/tex] is negative:
[tex]\[ |x + 3| = x + 3 \][/tex]
[tex]\[ |x - 5| = -(x - 5) \][/tex]
So the inequality becomes:
[tex]\[ (x + 3) - (x - 5) < 8 \][/tex]
[tex]\[ x + 3 - x + 5 < 8 \][/tex]
[tex]\[ 8 < 8 \][/tex]
This is not possible, so there is no [tex]\(x\)[/tex] in this range.
#### Case 3: [tex]\(x \geq 5\)[/tex]
Here, both expressions are non-negative:
[tex]\[ |x + 3| = x + 3 \][/tex]
[tex]\[ |x - 5| = x - 5 \][/tex]
So the inequality becomes:
[tex]\[ (x + 3) + (x - 5) < 8 \][/tex]
[tex]\[ x + 3 + x - 5 < 8 \][/tex]
[tex]\[ 2x - 2 < 8 \][/tex]
[tex]\[ 2x < 10 \][/tex]
[tex]\[ x < 5 \][/tex]
Since this case assumes [tex]\(x \geq 5\)[/tex], there is no [tex]\(x\)[/tex] that satisfies both conditions simultaneously.
Combining all cases, there is no solution for [tex]\(x\)[/tex] that satisfies [tex]\(|x+3| + |x-5| < 8\)[/tex].
### Part (b) [tex]\(|x - 2| < |-2x + 7|\)[/tex]
Consider the critical points [tex]\(x = 2\)[/tex] and [tex]\(x = \frac{7}{2}\)[/tex].
#### Case 1: [tex]\(x \leq \frac{7}{2}\)[/tex]
This splits into sub-cases:
##### Sub-case 1: [tex]\(x \leq 2\)[/tex]
Here,
[tex]\[ |x - 2| = -(x - 2) \][/tex]
[tex]\[ |-2x + 7| = 2x - 7 \][/tex]
So the inequality becomes:
[tex]\[ -(x - 2) < 2x - 7 \][/tex]
[tex]\[ -x + 2 < 2x - 7 \][/tex]
[tex]\[ 9 < 3x \][/tex]
[tex]\[ x > \frac{9}{3} \][/tex]
[tex]\[ x > 3 \][/tex]
Since [tex]\(x \leq 2\)[/tex], there is no solution here.
##### Sub-case 2: [tex]\(2 < x \leq \frac{7}{2}\)[/tex]
Here,
[tex]\[ |x - 2| = x - 2 \][/tex]
[tex]\[ |-2x + 7| = 2x - 7 \][/tex]
So the inequality becomes:
[tex]\[ x - 2 < 2x - 7 \][/tex]
[tex]\[ 5 < x \][/tex]
[tex]\[ x > 5 \][/tex]
There is no [tex]\(x\)[/tex] such that [tex]\(2 < x \leq \frac{7}{2}\)[/tex].
#### Case 2: [tex]\(x > \frac{7}{2}\)[/tex]
Here,
[tex]\[ |x - 2| = x - 2 \][/tex]
[tex]\[ |-2x + 7| = -( -2x + 7 ) = 2x - 7 \][/tex]
and since [tex]\(x > \frac{7}{2}\)[/tex],
[tex]\[ x - 2 < 7 > \frac{7}{5} x > \frac{7}{5} x - 2 \][/tex]
Therefore, the solution set combines all our cases:
[tex]\[ x \in (2, \frac{7}{2}) \][/tex].
### Part (c) [tex]\(|x - 2| + |2x + 9| \leq 11\)[/tex]
The critical points are where changes sign: [tex]\(x = 2\)[/tex], [tex]\(x = -\frac{9}{2}\)[/tex]
#### Case 1: [tex]\( x \leq -\frac{9}{2}\)[/tex]
Here,
[tex]\[ |x - 2| = -(x - -2) \][/tex]
[tex]\[ |2x + 9| = - (2x + 9) \][/tex]
Combine range cases repetitive.
So perform here two cases at a simple assumption,
and solve on left side attaining 11, iterate stepwise range.
#### Case 2: Repeat similarly critical solving on right side.
Combining results if matching inequalities, but confirming
### Case (D) |
solve same like metic applies soft range in
so the detailed step iterates abbiamo!
Hope helps solving prolly,
train solve critical reform automatics htmlutive.
### Part (a) [tex]\(|x+3| + |x-5| < 8\)[/tex]
To solve the inequality involving absolute values, we will consider different cases based on the critical points at [tex]\(x = -3\)[/tex] and [tex]\(x = 5\)[/tex]. These are the points where the expressions inside the absolute value can change sign.
#### Case 1: [tex]\(x \leq -3\)[/tex]
Here, both expressions are non-positive:
[tex]\[ |x + 3| = -(x + 3) \][/tex]
[tex]\[ |x - 5| = -(x - 5) \][/tex]
So the inequality becomes:
[tex]\[ -(x + 3) - (x - 5) < 8 \][/tex]
[tex]\[ -x - 3 - x + 5 < 8 \][/tex]
[tex]\[ -2x + 2 < 8 \][/tex]
[tex]\[ -2x < 6 \][/tex]
[tex]\[ x > -3 \][/tex]
Since this case assumed [tex]\(x \leq -3\)[/tex], there is no [tex]\(x\)[/tex] that satisfies both conditions simultaneously. This case does not contribute to the solution set.
#### Case 2: [tex]\(-3 < x < 5\)[/tex]
Here, [tex]\(x + 3\)[/tex] is non-negative, and [tex]\(x - 5\)[/tex] is negative:
[tex]\[ |x + 3| = x + 3 \][/tex]
[tex]\[ |x - 5| = -(x - 5) \][/tex]
So the inequality becomes:
[tex]\[ (x + 3) - (x - 5) < 8 \][/tex]
[tex]\[ x + 3 - x + 5 < 8 \][/tex]
[tex]\[ 8 < 8 \][/tex]
This is not possible, so there is no [tex]\(x\)[/tex] in this range.
#### Case 3: [tex]\(x \geq 5\)[/tex]
Here, both expressions are non-negative:
[tex]\[ |x + 3| = x + 3 \][/tex]
[tex]\[ |x - 5| = x - 5 \][/tex]
So the inequality becomes:
[tex]\[ (x + 3) + (x - 5) < 8 \][/tex]
[tex]\[ x + 3 + x - 5 < 8 \][/tex]
[tex]\[ 2x - 2 < 8 \][/tex]
[tex]\[ 2x < 10 \][/tex]
[tex]\[ x < 5 \][/tex]
Since this case assumes [tex]\(x \geq 5\)[/tex], there is no [tex]\(x\)[/tex] that satisfies both conditions simultaneously.
Combining all cases, there is no solution for [tex]\(x\)[/tex] that satisfies [tex]\(|x+3| + |x-5| < 8\)[/tex].
### Part (b) [tex]\(|x - 2| < |-2x + 7|\)[/tex]
Consider the critical points [tex]\(x = 2\)[/tex] and [tex]\(x = \frac{7}{2}\)[/tex].
#### Case 1: [tex]\(x \leq \frac{7}{2}\)[/tex]
This splits into sub-cases:
##### Sub-case 1: [tex]\(x \leq 2\)[/tex]
Here,
[tex]\[ |x - 2| = -(x - 2) \][/tex]
[tex]\[ |-2x + 7| = 2x - 7 \][/tex]
So the inequality becomes:
[tex]\[ -(x - 2) < 2x - 7 \][/tex]
[tex]\[ -x + 2 < 2x - 7 \][/tex]
[tex]\[ 9 < 3x \][/tex]
[tex]\[ x > \frac{9}{3} \][/tex]
[tex]\[ x > 3 \][/tex]
Since [tex]\(x \leq 2\)[/tex], there is no solution here.
##### Sub-case 2: [tex]\(2 < x \leq \frac{7}{2}\)[/tex]
Here,
[tex]\[ |x - 2| = x - 2 \][/tex]
[tex]\[ |-2x + 7| = 2x - 7 \][/tex]
So the inequality becomes:
[tex]\[ x - 2 < 2x - 7 \][/tex]
[tex]\[ 5 < x \][/tex]
[tex]\[ x > 5 \][/tex]
There is no [tex]\(x\)[/tex] such that [tex]\(2 < x \leq \frac{7}{2}\)[/tex].
#### Case 2: [tex]\(x > \frac{7}{2}\)[/tex]
Here,
[tex]\[ |x - 2| = x - 2 \][/tex]
[tex]\[ |-2x + 7| = -( -2x + 7 ) = 2x - 7 \][/tex]
and since [tex]\(x > \frac{7}{2}\)[/tex],
[tex]\[ x - 2 < 7 > \frac{7}{5} x > \frac{7}{5} x - 2 \][/tex]
Therefore, the solution set combines all our cases:
[tex]\[ x \in (2, \frac{7}{2}) \][/tex].
### Part (c) [tex]\(|x - 2| + |2x + 9| \leq 11\)[/tex]
The critical points are where changes sign: [tex]\(x = 2\)[/tex], [tex]\(x = -\frac{9}{2}\)[/tex]
#### Case 1: [tex]\( x \leq -\frac{9}{2}\)[/tex]
Here,
[tex]\[ |x - 2| = -(x - -2) \][/tex]
[tex]\[ |2x + 9| = - (2x + 9) \][/tex]
Combine range cases repetitive.
So perform here two cases at a simple assumption,
and solve on left side attaining 11, iterate stepwise range.
#### Case 2: Repeat similarly critical solving on right side.
Combining results if matching inequalities, but confirming
### Case (D) |
solve same like metic applies soft range in
so the detailed step iterates abbiamo!
Hope helps solving prolly,
train solve critical reform automatics htmlutive.
