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Solve [tex]\(2x - 1 \ \textless \ 7\)[/tex] and [tex]\(5x + 3 \ \textless \ 3\)[/tex].

A. [tex]\(\{x \mid 0 \ \textless \ x \ \textless \ 4\}\)[/tex]
B. [tex]\(\{x \mid x \ \textless \ 0\}\)[/tex]
C. [tex]\(\{x \mid x \ \textless \ 4\}\)[/tex]


Sagot :

To solve the given inequalities [tex]\(2x - 1 < 7\)[/tex] and [tex]\(5x + 3 < 3\)[/tex], we will address each inequality step-by-step and then find the intersection of their solution sets.

### Solving the First Inequality: [tex]\(2x - 1 < 7\)[/tex]

1. Add 1 to both sides of the inequality to isolate the term with [tex]\(x\)[/tex]:
[tex]\[ 2x - 1 + 1 < 7 + 1 \][/tex]
Simplifying, we get:
[tex]\[ 2x < 8 \][/tex]

2. Divide both sides of the inequality by 2 to solve for [tex]\(x\)[/tex]:
[tex]\[ \frac{2x}{2} < \frac{8}{2} \][/tex]
Simplifying, we get:
[tex]\[ x < 4 \][/tex]

### Solving the Second Inequality: [tex]\(5x + 3 < 3\)[/tex]

1. Subtract 3 from both sides of the inequality to isolate the term with [tex]\(x\)[/tex]:
[tex]\[ 5x + 3 - 3 < 3 - 3 \][/tex]
Simplifying, we get:
[tex]\[ 5x < 0 \][/tex]

2. Divide both sides of the inequality by 5 to solve for [tex]\(x\)[/tex]:
[tex]\[ \frac{5x}{5} < \frac{0}{5} \][/tex]
Simplifying, we get:
[tex]\[ x < 0 \][/tex]

### Intersection of the Solution Sets

We have now obtained the solution sets for both inequalities:
1. From [tex]\(2x - 1 < 7\)[/tex], we have [tex]\(x < 4\)[/tex].
2. From [tex]\(5x + 3 < 3\)[/tex], we have [tex]\(x < 0\)[/tex].

The final solution is the intersection of these two sets, which means the values of [tex]\(x\)[/tex] that satisfy both inequalities simultaneously.

The solution to [tex]\(x < 4\)[/tex] and [tex]\(x < 0\)[/tex] is:
[tex]\[ x < 0 \][/tex]

Hence, the solution set is [tex]\(\{x \mid x < 0\}\)[/tex].
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