Answer :
Sure! Let's solve the given equation step by step.
The equation we have is:
[tex]\[ |3x + 5| + |2x - 3| = 25 \][/tex]
To solve this absolute value equation, we must consider different cases based on the points where the expressions inside the absolute values change sign. The critical points are where [tex]\(3x + 5 = 0\)[/tex] and [tex]\(2x - 3 = 0\)[/tex].
1. [tex]\(3x + 5 = 0\)[/tex]:
[tex]\[ x = -\frac{5}{3} \][/tex]
2. [tex]\(2x - 3 = 0\)[/tex]:
[tex]\[ x = \frac{3}{2} \][/tex]
We need to consider three intervals determined by these points:
- [tex]\(x < -\frac{5}{3}\)[/tex]
- [tex]\(-\frac{5}{3} \leq x \leq \frac{3}{2}\)[/tex]
- [tex]\(x > \frac{3}{2}\)[/tex]
### Case 1: [tex]\( x < -\frac{5}{3} \)[/tex]
Here, both [tex]\(3x + 5 < 0\)[/tex] and [tex]\(2x - 3 < 0\)[/tex]. Therefore, the absolute value expressions become:
[tex]\[ |3x + 5| = -(3x + 5) = -3x - 5 \][/tex]
[tex]\[ |2x - 3| = -(2x - 3) = -2x + 3 \][/tex]
So the equation becomes:
[tex]\[ -3x - 5 - 2x + 3 = 25 \][/tex]
[tex]\[ -5x - 2 = 25 \][/tex]
[tex]\[ -5x = 27 \][/tex]
[tex]\[ x = -\frac{27}{5} \][/tex]
We need to check if [tex]\( x = -\frac{27}{5} \)[/tex] falls in the interval [tex]\( x < -\frac{5}{3} \)[/tex]. Since [tex]\(-\frac{27}{5} = -5.4\)[/tex] and it is less than [tex]\(-\frac{5}{3} = -1.67\)[/tex], this value is valid.
### Case 2: [tex]\( -\frac{5}{3} \leq x \leq \frac{3}{2} \)[/tex]
Here, [tex]\(3x + 5 \geq 0\)[/tex] and [tex]\(2x - 3 < 0\)[/tex] when [tex]\(x < \frac{3}{2}\)[/tex]. Therefore, the absolute value expressions become:
[tex]\[ |3x + 5| = 3x + 5 \][/tex]
[tex]\[ |2x - 3| = -(2x - 3) = -2x + 3 \][/tex]
So the equation becomes:
[tex]\[ 3x + 5 - 2x + 3 = 25 \][/tex]
[tex]\[ x + 8 = 25 \][/tex]
[tex]\[ x = 17 \][/tex]
We need to check if [tex]\( x = 17 \)[/tex] falls in the interval [tex]\( -\frac{5}{3} \leq x \leq \frac{3}{2} \)[/tex]. Since [tex]\(17\)[/tex] is not in this interval, this value is not valid.
### Case 3: [tex]\( x > \frac{3}{2} \)[/tex]
Here, both [tex]\(3x + 5 \geq 0\)[/tex] and [tex]\(2x - 3 \geq 0\)[/tex]. Therefore, the absolute value expressions are:
[tex]\[ |3x + 5| = 3x + 5 \][/tex]
[tex]\[ |2x - 3| = 2x - 3 \][/tex]
So the equation becomes:
[tex]\[ 3x + 5 + 2x - 3 = 25 \][/tex]
[tex]\[ 5x + 2 = 25 \][/tex]
[tex]\[ 5x = 23 \][/tex]
[tex]\[ x = \frac{23}{5} \][/tex]
We need to check if [tex]\( x = \frac{23}{5} \)[/tex] falls in the interval [tex]\( x > \frac{3}{2} \)[/tex]. Since [tex]\(\frac{23}{5} = 4.6\)[/tex] and it is greater than [tex]\( \frac{3}{2} = 1.5 \)[/tex], this value is valid.
### Summary of Solutions
The valid solutions are:
- [tex]\(x = -\frac{27}{5}\)[/tex]
- [tex]\(x = \frac{23}{5}\)[/tex]
### Sum of Highest and Lowest Values
The highest value is [tex]\( \frac{23}{5} \)[/tex].
The lowest value is [tex]\(-\frac{27}{5}\)[/tex].
Their sum is:
[tex]\[ \frac{23}{5} + \left(-\frac{27}{5}\right) = \frac{23 - 27}{5} = \frac{-4}{5} = -0.8 \][/tex]
Therefore, the sum of the highest and lowest values is:
[tex]\[ \boxed{-0.8} \][/tex]
The equation we have is:
[tex]\[ |3x + 5| + |2x - 3| = 25 \][/tex]
To solve this absolute value equation, we must consider different cases based on the points where the expressions inside the absolute values change sign. The critical points are where [tex]\(3x + 5 = 0\)[/tex] and [tex]\(2x - 3 = 0\)[/tex].
1. [tex]\(3x + 5 = 0\)[/tex]:
[tex]\[ x = -\frac{5}{3} \][/tex]
2. [tex]\(2x - 3 = 0\)[/tex]:
[tex]\[ x = \frac{3}{2} \][/tex]
We need to consider three intervals determined by these points:
- [tex]\(x < -\frac{5}{3}\)[/tex]
- [tex]\(-\frac{5}{3} \leq x \leq \frac{3}{2}\)[/tex]
- [tex]\(x > \frac{3}{2}\)[/tex]
### Case 1: [tex]\( x < -\frac{5}{3} \)[/tex]
Here, both [tex]\(3x + 5 < 0\)[/tex] and [tex]\(2x - 3 < 0\)[/tex]. Therefore, the absolute value expressions become:
[tex]\[ |3x + 5| = -(3x + 5) = -3x - 5 \][/tex]
[tex]\[ |2x - 3| = -(2x - 3) = -2x + 3 \][/tex]
So the equation becomes:
[tex]\[ -3x - 5 - 2x + 3 = 25 \][/tex]
[tex]\[ -5x - 2 = 25 \][/tex]
[tex]\[ -5x = 27 \][/tex]
[tex]\[ x = -\frac{27}{5} \][/tex]
We need to check if [tex]\( x = -\frac{27}{5} \)[/tex] falls in the interval [tex]\( x < -\frac{5}{3} \)[/tex]. Since [tex]\(-\frac{27}{5} = -5.4\)[/tex] and it is less than [tex]\(-\frac{5}{3} = -1.67\)[/tex], this value is valid.
### Case 2: [tex]\( -\frac{5}{3} \leq x \leq \frac{3}{2} \)[/tex]
Here, [tex]\(3x + 5 \geq 0\)[/tex] and [tex]\(2x - 3 < 0\)[/tex] when [tex]\(x < \frac{3}{2}\)[/tex]. Therefore, the absolute value expressions become:
[tex]\[ |3x + 5| = 3x + 5 \][/tex]
[tex]\[ |2x - 3| = -(2x - 3) = -2x + 3 \][/tex]
So the equation becomes:
[tex]\[ 3x + 5 - 2x + 3 = 25 \][/tex]
[tex]\[ x + 8 = 25 \][/tex]
[tex]\[ x = 17 \][/tex]
We need to check if [tex]\( x = 17 \)[/tex] falls in the interval [tex]\( -\frac{5}{3} \leq x \leq \frac{3}{2} \)[/tex]. Since [tex]\(17\)[/tex] is not in this interval, this value is not valid.
### Case 3: [tex]\( x > \frac{3}{2} \)[/tex]
Here, both [tex]\(3x + 5 \geq 0\)[/tex] and [tex]\(2x - 3 \geq 0\)[/tex]. Therefore, the absolute value expressions are:
[tex]\[ |3x + 5| = 3x + 5 \][/tex]
[tex]\[ |2x - 3| = 2x - 3 \][/tex]
So the equation becomes:
[tex]\[ 3x + 5 + 2x - 3 = 25 \][/tex]
[tex]\[ 5x + 2 = 25 \][/tex]
[tex]\[ 5x = 23 \][/tex]
[tex]\[ x = \frac{23}{5} \][/tex]
We need to check if [tex]\( x = \frac{23}{5} \)[/tex] falls in the interval [tex]\( x > \frac{3}{2} \)[/tex]. Since [tex]\(\frac{23}{5} = 4.6\)[/tex] and it is greater than [tex]\( \frac{3}{2} = 1.5 \)[/tex], this value is valid.
### Summary of Solutions
The valid solutions are:
- [tex]\(x = -\frac{27}{5}\)[/tex]
- [tex]\(x = \frac{23}{5}\)[/tex]
### Sum of Highest and Lowest Values
The highest value is [tex]\( \frac{23}{5} \)[/tex].
The lowest value is [tex]\(-\frac{27}{5}\)[/tex].
Their sum is:
[tex]\[ \frac{23}{5} + \left(-\frac{27}{5}\right) = \frac{23 - 27}{5} = \frac{-4}{5} = -0.8 \][/tex]
Therefore, the sum of the highest and lowest values is:
[tex]\[ \boxed{-0.8} \][/tex]
