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Find the set of values of x- write in set notation

We are tasked with solving the system of inequalities:

\[
x^2 - x - 12 > 0 \quad \text{and} \quad 3x + 17 > 2
\]

### Step 1: Solve \( x^2 - x - 12 > 0 \)
First, factor the quadratic expression:

\[
x^2 - x - 12 = (x - 4)(x + 3)
\]

The inequality becomes:

\[
(x - 4)(x + 3) > 0
\]

To solve this, we find the critical points where the expression is equal to zero, which are \( x = 4 \) and \( x = -3 \).

Now, we examine the sign of the expression in the intervals divided by these critical points: \( (-\infty, -3) \), \( (-3, 4) \), and \( (4, \infty) \).

- For \( x \in (-\infty, -3) \), both factors \( (x - 4) \) and \( (x + 3) \) are negative, so their product is positive.
- For \( x \in (-3, 4) \), one factor is negative and the other is positive, so their product is negative.
- For \( x \in (4, \infty) \), both factors are positive, so their product is positive.

Thus, the solution to \( x^2 - x - 12 > 0 \) is:

\[
x \in (-\infty, -3) \cup (4, \infty)
\]

### Step 2: Solve \( 3x + 17 > 2 \)
Simplify the inequality:

\[
3x + 17 > 2
\]

Subtract 17 from both sides:

\[
3x > -15
\]

Divide by 3:

\[
x > -5
\]

### Step 3: Find the intersection of the two solutions
We need to find the intersection of \( (-\infty, -3) \cup (4, \infty) \) and \( x > -5 \).

- \( x > -5 \) limits the solution to \( (-5, \infty) \).
- Therefore, the intersection is \( (-3, -3) \cup (4, \infty) \), since \( x > -5 \) removes values less than \( -5 \).

### Final Solution in Set Notation:
\[
x \in (-3, -3)  or (4 , -tables\in

Find the set of values of x for the inequality: x^2 - x - 12 > 0 and 3x + 17 > 2.

Learn how to solve a system of inequalities involving a quadratic equation and a linear inequality. This problem involves factoring the quadratic expression x^2 - x - 12 > 0 and solving the inequality 3x + 17 > 2. Suitable for students in Grades 10-12.

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