Chapter 7 - Inequalities Flashcards
Inequalties are much like equation with One big exception
1) add or subtract a constant on both sides
2) add or subtract a variable expression on both sides
3) Multiply or divide by a positive # on both sides.
Difference: When you multiply or divide an inequality by a negative #, the inequality sign flips.
Combining Inequalities: Line them up!
If x > 8, x < 17, and x + 5 < 19, what is the range of possible values of x?
**Step 1: **x + 5 < 19 ⇒ x < 14 ⇒ Solve inequality
**Step 2: **8 < x, x < 17, x < 14 ⇒ Simplify all inequality symbols so all symbols point in the same direction ( Less than)
**Step 3: **8 < x ⇒ Line up the common variables in the
x \< 17 inequality. Since x \< 14 is more
x \< 14 limiting than x \< 17, chose **8 \< x \< 14**
** 8 < x < 14 ** as the compound inequality.
Given the u < t, b > r, f < t, and r > t, is b > u?
Q: u < t, b > r, f < t, and r > t, is b > u?
Simplify: u < t, r < b, f < t, t < r ⇒ Inequality sign rearranged
**Line up the variables Combine **
u < t u < t < r < b
r \< b f \< t
f < t ** ANS: ** Yes, b > u
t \< r
You cannot fit the inequality f < t into the long combo. You do know that both u and f are < t but you dont know the relationship bw u and f.
Manipulating Compound Inequalities:
You can perform operations on a compound inequality as long as you remember to perform those operations on EVERY TERM in the inequality and not just the outside terms
If 1 > 1 - ab > 0, which of the following must be true?
I: a/b > 0 II: a/b < 1 III: ab < 1
1 > 1 - ab > 0
0 > - ab > -1 ⇒ Subtract 1 from ALL 3 terms
0 < ab < 1 ⇒Multiply ALL 3 terms by (-1) & flip inequality sign
So 0 < ab < 1. ab is pos, so a/b must be positive (a & b have the same sign). I must be true. But you dont know whether a/b < 1, so II is not true necessarily. But ab must be less than 1, so III must be true. So I and III are true.
Combining Inequalities: Add Them Up
complete card once understood
Is mn < 10?
1) m < 2
2) n < 5
Variation: If m and n are both positive, is mn < 10?
It is tempting to multiply these two stmts together & conclude that mn < 10. That would be a mistake, however b/c both m and n could be neg #’s that yield a # larger than 10 when multiplied together.
Ex: if m= -2 and n = -6, then mn = 12 which is greater than 10. Since you can find cases with mn < 10 and with mn > 10, the two stmts are insufficient to ans the ques. Choice E.
Variation ANS: Since the variables are both positive, you CAN multiply these inequalities together & conclude that mn < 10. **Choice C. **
** NOTE: Only multiply inequalities together if both sides of both inequalities are positive.
Inequalities and Abs Value:
GR: When | x **+ b | = c, the center point of our graph is *-b so x must be EXACTLY c units away from *-b.
x + b | < c, the center point of the graph is -b and the “less than” symbol tells us that x must be LESS THAN c units away from -b.
What is the graph of | x - 4 | < 3?
Based on formula, the center point of the graph is -(-4) = **4 **and x must be less than 3 units away from the center point.
Alegracially: Given that | x - 2 | < 5, what is the range of possible values of x?
First Scenario _ Second Scenario _(reverse sign)
x-2 < 5 - (x - 2 ) < 5 x > -3
**x < 7 ** -x + 2 < 5
x - 2 | < 5 | x - 2 | < 5 -x < 3
Square Rooting Inequalities
If 10 + x2 ≥ 19, what is the range of possible values of x?
10 + x2 ≥ 19
x2 ≥ 9
|x| ≥ 3
If 0 < ab < ac, is a negative?
1) c < 0
2) b > c
By the transitive property of inequalities, if 0 < ab < ac, then 0 < ac. Therefore a and c must have the same sign. Stmt 1 tells us that c is neg. Therefore, a is neg = SUFF
Stmt 2 indicates that b > c but the question stem also told you that ab < ac. When you multiply both sides of b > c by a, the sign gets flipped. For inequalities, if you multiply by a negative #, the sign of the inequality flips. Therefore, a must be negative bc multiplying the two sides of the equation by a results in a flipped inequality sign = **SUFF **
