Question

Determine the solution sets for the following statements:


x2=1
and x<12

(1 point)
x=
{
}
Question 2
11y+6=50
or 10y+5=5
(1 point)
y=
{
}

Answer 1

Let's solve each of the statements one by one.

**Question 1: Solve \(x^2 = 1\) and \(x < 12\)**

1. Start by solving the equation \(x^2 = 1\):
- \(x^2 - 1 = 0\) can be factored as \((x - 1)(x + 1) = 0\).
- The solutions are \(x = 1\) and \(x = -1\).

2. Now, we consider the condition \(x < 12\):
- Both solutions \(x = 1\) and \(x = -1\) satisfy \(x < 12\).

So the solution set for this question is:
\[
x = \{-1, 1\}
\]

---

**Question 2: Solve \(11y + 6 = 50\) or \(10y + 5 = 5\)**

1. Start with the first equation \(11y + 6 = 50\):
- Subtract 6 from both sides: \(11y = 44\).
- Now divide by 11: \(y = 4\).

2. Now, solve the second equation \(10y + 5 = 5\):
- Subtract 5 from both sides: \(10y = 0\).
- Now divide by 10: \(y = 0\).

Since we are looking for the solution set for either equation, we have:
- The solutions are \(y = 4\) and \(y = 0\).

So the solution set for this question is:
\[
y = \{0, 4\}
\]

**Final Answers:**
- Question 1: \(x = \{-1, 1\}\)
- Question 2: \(y = \{0, 4\}\)