Inequality & Modulus: Level 1
1.
Which of the following is equivalent to the statement \( |x – 3| < 2 \)?
Solution (C):
The inequality \( |x – 3| < 2 \) unwraps to \( -2 < x - 3 < 2 \).
To solve for \(x\), add 3 to all parts:
\( -2 + 3 < x < 2 + 3 \)
\( 1 < x < 5 \).
2.
If \(3x – 5 > 10\), which of the following could be a value of \(x\)?
Solution (D):
\( 3x – 5 > 10 \)
Add 5 to both sides: \( 3x > 15 \)
Divide by 3: \( x > 5 \).
The only value in the options that is greater than 5 is 5.1.
3.
If \(|a| = 5\) and \(|b| = 2\), what is the smallest possible value of \(a + b\)?
Solution (D):
\(|a| = 5 \implies a = 5\) or \(a = -5\).
\(|b| = 2 \implies b = 2\) or \(b = -2\).
To find the smallest sum, we add the two smallest values:
\( -5 + (-2) = -7 \).
4.
If \(|x + 1| > -3\), which of the following represents all possible values of \(x\)?
Solution (D): The absolute value of any expression, \(|x+1|\), must be greater than or equal to 0.
Since 0 is always greater than -3, the inequality \(|x+1| \ge 0 > -3\) is true for any value of \(x\).
Therefore, the solution is all real numbers.
5.
If \(x – y > x + y\), which of the following must be true?
Solution (D):
\( x – y > x + y \)
Subtract \(x\) from both sides:
\( -y > y \)
Add \(y\) to both sides:
\( 0 > 2y \)
Divide by 2: \( 0 > y \), or \( y < 0 \).
6.
If \(x > 5\), which of the following must be true?
Solution (E): We start with the inequality \(x > 5\).
(A) \(x > 5 \implies x+1 > 6\). This statement is false.
(B) \(x > 5 \implies -x < -5\). (Multiplying by -1 flips the sign). This statement is false.
(C) \(x > 5 \implies 2x > 10\). This statement is false.
(D) \(x > 5 \implies x-5 > 0\). This statement is false.
(E) \(x > 5 \implies x+1 > 6\). This statement is true.
7.
If \(|a| = |b|\) and \(ab < 0\), which of the following must be true?
Solution (B):
1. \(|a| = |b|\) means that \(a = b\) or \(a = -b\).
2. \(ab < 0\) means that \(a\) and \(b\) have opposite signs.
3. If \(a = b\), they would have the same sign, so \(ab\) would be positive. This case is not possible.
4. If \(a = -b\), they have opposite signs, so \(ab\) would be negative. This case must be true.
8.
If \(|x + 1| = 3\), what is the sum of all possible values of \(x\)?
Solution (C): An absolute value equation has two solutions:
1. \(x + 1 = 3 \implies x = 2\).
2. \(x + 1 = -3 \implies x = -4\).
The sum of the values is \(2 + (-4) = -2\).
9.
If \(5 – 2x > 11\), which of the following could be a value of \(x\)?
Solution (D):
\( 5 – 2x > 11 \)
Subtract 5: \( -2x > 6 \)
Divide by -2 and flip the inequality sign:
\( x < -3 \).
The only value in the options that is less than -3 is -4.
10.
If \(|x| = 4\) and \(|y| = 3\), what is the largest possible value of \(x – y\)?
Solution (C):
\(|x| = 4 \implies x = 4\) or \(x = -4\).
\(|y| = 3 \implies y = 3\) or \(y = -3\).
To get the largest possible value of \(x – y\), we must use the largest possible \(x\) (which is 4) and subtract the smallest possible \(y\) (which is -3).
\( 4 – (-3) = 4 + 3 = 7 \).
Score: 0 / 10