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To solve the inequalities [tex]\( x \cdot \frac{x}{2} - 3 \geq -1 \)[/tex] and [tex]\( 2x + 3 < 7 \)[/tex], let's tackle each inequality step-by-step.
### Solving the First Inequality
1. Rewrite the inequality:
[tex]\[ x \cdot \frac{x}{2} - 3 \geq -1 \][/tex]
2. Simplify the expression:
[tex]\[ x \cdot \frac{x}{2} - 3 + 1 \geq 0 \][/tex]
[tex]\[ \frac{x^2}{2} - 2 \geq 0 \][/tex]
3. Multiply through by 2 to clear the fraction:
[tex]\[ x^2 - 4 \geq 0 \][/tex]
4. Factorize the quadratic expression:
[tex]\[ (x - 2)(x + 2) \geq 0 \][/tex]
5. Determine the critical points and test intervals:
The quadratic expression [tex]\( (x - 2)(x + 2) \)[/tex] changes sign at [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex].
Test intervals are:
[tex]\[ (-\infty, -2), (-2, 2), (2, \infty) \][/tex]
6. Test a point in each interval:
- For [tex]\( x \in (-\infty, -2) \)[/tex], choose [tex]\( x = -3 \)[/tex]:
[tex]\[ (-3 - 2)(-3 + 2) = (-5)(-1) = 5 \geq 0 \quad (\text{True}) \][/tex]
- For [tex]\( x \in (-2, 2) \)[/tex], choose [tex]\( x = 0 \)[/tex]:
[tex]\[ (0 - 2)(0 + 2) = (-2)(2) = -4 < 0 \quad (\text{False}) \][/tex]
- For [tex]\( x \in (2, \infty) \)[/tex], choose [tex]\( x = 3 \)[/tex]:
[tex]\[ (3 - 2)(3 + 2) = (1)(5) = 5 \geq 0 \quad (\text{True}) \][/tex]
The solution for this inequality is:
[tex]\[ x \in (-\infty, -2] \cup [2, \infty) \][/tex]
### Solving the Second Inequality
1. Rewrite and simplify the inequality:
[tex]\[ 2x + 3 < 7 \][/tex]
[tex]\[ 2x < 4 \][/tex]
[tex]\[ x < 2 \][/tex]
The solution for this inequality is:
[tex]\[ x < 2 \][/tex]
### Combining the Solutions
We need to find the intersection of the solutions to both inequalities:
- First inequality: [tex]\( x \in (-\infty, -2] \cup [2, \infty) \)[/tex]
- Second inequality: [tex]\( x < 2 \)[/tex]
The intersection is:
[tex]\[ x \in (-\infty, -2] \][/tex]
Thus, the solution to the system of inequalities [tex]\( x \cdot \frac{x}{2} - 3 \geq -1 \)[/tex] and [tex]\( 2x + 3 < 7 \)[/tex] is:
[tex]\[ x \in (-\infty, -2] \][/tex]
Among the provided options, the solution matches:
[tex]\[ \boxed{-4 \leq x < 2} \][/tex]
But this was simplified to the wrong range — it should be more precise:
[tex]\[ \boxed{\text{{None of the provided options directly}} \][/tex]
### Solving the First Inequality
1. Rewrite the inequality:
[tex]\[ x \cdot \frac{x}{2} - 3 \geq -1 \][/tex]
2. Simplify the expression:
[tex]\[ x \cdot \frac{x}{2} - 3 + 1 \geq 0 \][/tex]
[tex]\[ \frac{x^2}{2} - 2 \geq 0 \][/tex]
3. Multiply through by 2 to clear the fraction:
[tex]\[ x^2 - 4 \geq 0 \][/tex]
4. Factorize the quadratic expression:
[tex]\[ (x - 2)(x + 2) \geq 0 \][/tex]
5. Determine the critical points and test intervals:
The quadratic expression [tex]\( (x - 2)(x + 2) \)[/tex] changes sign at [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex].
Test intervals are:
[tex]\[ (-\infty, -2), (-2, 2), (2, \infty) \][/tex]
6. Test a point in each interval:
- For [tex]\( x \in (-\infty, -2) \)[/tex], choose [tex]\( x = -3 \)[/tex]:
[tex]\[ (-3 - 2)(-3 + 2) = (-5)(-1) = 5 \geq 0 \quad (\text{True}) \][/tex]
- For [tex]\( x \in (-2, 2) \)[/tex], choose [tex]\( x = 0 \)[/tex]:
[tex]\[ (0 - 2)(0 + 2) = (-2)(2) = -4 < 0 \quad (\text{False}) \][/tex]
- For [tex]\( x \in (2, \infty) \)[/tex], choose [tex]\( x = 3 \)[/tex]:
[tex]\[ (3 - 2)(3 + 2) = (1)(5) = 5 \geq 0 \quad (\text{True}) \][/tex]
The solution for this inequality is:
[tex]\[ x \in (-\infty, -2] \cup [2, \infty) \][/tex]
### Solving the Second Inequality
1. Rewrite and simplify the inequality:
[tex]\[ 2x + 3 < 7 \][/tex]
[tex]\[ 2x < 4 \][/tex]
[tex]\[ x < 2 \][/tex]
The solution for this inequality is:
[tex]\[ x < 2 \][/tex]
### Combining the Solutions
We need to find the intersection of the solutions to both inequalities:
- First inequality: [tex]\( x \in (-\infty, -2] \cup [2, \infty) \)[/tex]
- Second inequality: [tex]\( x < 2 \)[/tex]
The intersection is:
[tex]\[ x \in (-\infty, -2] \][/tex]
Thus, the solution to the system of inequalities [tex]\( x \cdot \frac{x}{2} - 3 \geq -1 \)[/tex] and [tex]\( 2x + 3 < 7 \)[/tex] is:
[tex]\[ x \in (-\infty, -2] \][/tex]
Among the provided options, the solution matches:
[tex]\[ \boxed{-4 \leq x < 2} \][/tex]
But this was simplified to the wrong range — it should be more precise:
[tex]\[ \boxed{\text{{None of the provided options directly}} \][/tex]
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