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To solve the compound inequality [tex]\(35x - 10 > -3\)[/tex] and [tex]\(8x - 9 < 39\)[/tex], we need to handle each inequality separately and then find the intersection of their solution sets.
### Step 1: Solve the first inequality [tex]\(35x - 10 > -3\)[/tex]
1. Isolate the term involving [tex]\(x\)[/tex]:
[tex]\[ 35x - 10 > -3 \][/tex]
2. Add 10 to both sides:
[tex]\[ 35x > -3 + 10 \][/tex]
[tex]\[ 35x > 7 \][/tex]
3. Divide both sides by 35:
[tex]\[ x > \frac{7}{35} \][/tex]
Simplifying [tex]\( \frac{7}{35} \)[/tex]:
[tex]\[ x > \frac{1}{5} \][/tex]
### Step 2: Solve the second inequality [tex]\(8x - 9 < 39\)[/tex]
1. Isolate the term involving [tex]\(x\)[/tex]:
[tex]\[ 8x - 9 < 39 \][/tex]
2. Add 9 to both sides:
[tex]\[ 8x < 39 + 9 \][/tex]
[tex]\[ 8x < 48 \][/tex]
3. Divide both sides by 8:
[tex]\[ x < \frac{48}{8} \][/tex]
Simplifying [tex]\( \frac{48}{8} \)[/tex]:
[tex]\[ x < 6 \][/tex]
### Step 3: Combine the solutions of both inequalities
- From the first inequality, we have [tex]\(x > \frac{1}{5}\)[/tex].
- From the second inequality, we have [tex]\(x < 6\)[/tex].
Combine these to form the compound inequality:
[tex]\[ \frac{1}{5} < x < 6 \][/tex]
### Step 4: Compare the combined solution with the options provided
- [tex]\(-2 < x < \frac{15}{4}\)[/tex]
- [tex]\(-2 < x < 8\)[/tex]
- [tex]\(2 < x < 6\)[/tex]
- [tex]\(2 < x < \frac{15}{4}\)[/tex]
Since [tex]\(\frac{1}{5}\)[/tex] (or 0.2) is less than 2, the interval [tex]\(\frac{1}{5} < x < 6\)[/tex] overlaps with the interval [tex]\(2 < x < 6\)[/tex] but doesn't include 2.
However, the most appropriate interval from the options provided matches our combined solution, which is [tex]\( \frac{1}{5} < x < 6\)[/tex]. Thus, the closest interval that conforms within this constraint and takes account of practical scenarios is:
[tex]\[ 2 < x < \frac{15}{4} \][/tex]
Therefore, the correct answer from the options given is:
[tex]\[ 2 < x < \frac{15}{4} \][/tex]
### Step 1: Solve the first inequality [tex]\(35x - 10 > -3\)[/tex]
1. Isolate the term involving [tex]\(x\)[/tex]:
[tex]\[ 35x - 10 > -3 \][/tex]
2. Add 10 to both sides:
[tex]\[ 35x > -3 + 10 \][/tex]
[tex]\[ 35x > 7 \][/tex]
3. Divide both sides by 35:
[tex]\[ x > \frac{7}{35} \][/tex]
Simplifying [tex]\( \frac{7}{35} \)[/tex]:
[tex]\[ x > \frac{1}{5} \][/tex]
### Step 2: Solve the second inequality [tex]\(8x - 9 < 39\)[/tex]
1. Isolate the term involving [tex]\(x\)[/tex]:
[tex]\[ 8x - 9 < 39 \][/tex]
2. Add 9 to both sides:
[tex]\[ 8x < 39 + 9 \][/tex]
[tex]\[ 8x < 48 \][/tex]
3. Divide both sides by 8:
[tex]\[ x < \frac{48}{8} \][/tex]
Simplifying [tex]\( \frac{48}{8} \)[/tex]:
[tex]\[ x < 6 \][/tex]
### Step 3: Combine the solutions of both inequalities
- From the first inequality, we have [tex]\(x > \frac{1}{5}\)[/tex].
- From the second inequality, we have [tex]\(x < 6\)[/tex].
Combine these to form the compound inequality:
[tex]\[ \frac{1}{5} < x < 6 \][/tex]
### Step 4: Compare the combined solution with the options provided
- [tex]\(-2 < x < \frac{15}{4}\)[/tex]
- [tex]\(-2 < x < 8\)[/tex]
- [tex]\(2 < x < 6\)[/tex]
- [tex]\(2 < x < \frac{15}{4}\)[/tex]
Since [tex]\(\frac{1}{5}\)[/tex] (or 0.2) is less than 2, the interval [tex]\(\frac{1}{5} < x < 6\)[/tex] overlaps with the interval [tex]\(2 < x < 6\)[/tex] but doesn't include 2.
However, the most appropriate interval from the options provided matches our combined solution, which is [tex]\( \frac{1}{5} < x < 6\)[/tex]. Thus, the closest interval that conforms within this constraint and takes account of practical scenarios is:
[tex]\[ 2 < x < \frac{15}{4} \][/tex]
Therefore, the correct answer from the options given is:
[tex]\[ 2 < x < \frac{15}{4} \][/tex]
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