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2x + 5 > 7

2x > 2 subtract 5 from both sides

x > 1. divide both sides by 2

The only - although incredibly significant - difference is that you must flip the sign of the inequality whenever you multiply or divide by a negative! Inequalities can be manipulated like equations and follow very similar rules, but there is one important exception.

If you add/subtract/multiply the same number to both sides of an inequality, the inequality remains true. **But if you multiply or divide both sides of an inequality by a negative number, the inequality becomes reversed.**

This is quite easy to see because we can write that 4 > 2. But if we multiply both sides of this inequality by −1, we get −4 > −2, which is not true. We have to reverse the inequality, writing as −4 < −2 in order for it to be true.

This leads to difficulties when dealing with variables because a variable can be either positive or negative.

Many SAT inequality problems involve more than one inequality. To solve such problems, you may need to convert several inequalities to a compound inequality, which is a series of inequalities strung together, such as 2 < 3 < 4.

To convert multiple inequalities to a compound inequality, first line up the variables, then combine.

EXAMPLE: If x > 8 , x < 17, and x + 5 < 19, what is the range of possible values for x.

First, solve any inequalities that need to be solved. In this example, only the last inequality needs to be solved:

x + 5 < 19

x < 14

Second, simplify the inequalities so that all the inequality symbols point in the same direction:

8 < x

x < 17

x < 14

Third, line up the common variables in the inequalities.

8 < x

x < 17

x < 14

Notice that x < 14 is more limiting than x < 17 (in other words, whenever x < l4, x will always be less than 17, but not vice versa.). That is why you choose 8 < x < 14 rather than 8 < x < 17 as the compound inequality that solves the problem. **The correct answer is 8 < x < 14**

****Sometimes a problem with compound inequalities will require you to manipulate the inequalities in order to solve the problem. You can perform operations on a compound inequality as long as you remember to perform those operations on every term in the inequality, not just the outside terms.

If 1 > 1 - ab > 0 which of the following must be true?

I. \(\frac{a}{b}>0\)

II. \(\frac{a}{b}

III. ab < 1

** A.** I only

You can manipulate the original compound inequality as follows, making sure to perform each manipulation on every term:

1 > 1 - ab > 0

0 > -ab > -1. Multiply all three terms by -1

0 < ab < 1. and flip the inequality signs.

Therefore you know that 0 < ab < 1. This tells you that ab is positive, so must \(\frac{a}{b}\) be positive (a and b have the same sign). Therefore, it must be true. However, you do not know whether \(\frac{a}{b}**THEREFORE, the correct answer is D.**

Related to extreme values are problems involving optimization, specifically, minimization or maximization problems. In these problems, you need to focus on the largest and smallest possible values for each of the variables, as some combination of them will usually lead to the largest or smallest possible result.

If 2y + 3 < 11 and 1 < x < 5, what is the maximum possible value for xy?

You need to test the extreme values for x and for y to determine which combinations of extreme values will maximize xy:

2y + 3 ≤ 11 2y ≤ 8 y ≤ 4

The lowest value for x is 1 There is no lower limit to y. The highest value for x is 5 The highest value for y is 4.

Now consider the extreme value scenarios for X, Y, and XY:

- Since Y has no lower limit and x is positive, the product XY has no lower limit.
- Using Y’s highest value (4), test the extreme values of x(1 and 5).
- The first extreme value generates a product XY = (1)(4) = 4. The second extreme value generates XY = (5)(4) = 20.

Clearly, XY is maximized when X = 5 and Y = 4, with a result that XY = 20.

If -7 < a < 6 and -7 < b < 8, what is the maximum possible value for ab?

Once again, you are looking for a maximum possible value, this time for ab. You need to test the extreme values for a and for b to determine which combinations of extreme values will maximize ab.

The lowest value for a is -7. The lowest value for b is -7 The highest value for a is 6. The highest value for b is 8.

Now consider the extreme value scenarios for a > b, and ab:

This time, ab is maximized when you take the extreme negative values for both a and b, resulting in ab = 49. Notice that you could have focused right away on the first and fourth scenarios because they are the only scenarios that produce positive products.

Graphing linear inequalities is similar to graphing linear equations, but the process involves two major differences. The first difference is with regard to the line itself, depending on whether it is an inclusive or non-inclusive inequality.

- For inequalities that are inclusive (≤ or ≥), a solid line is used.
- For inequalities that are non-inclusive (< or >), a dashed line is used.
- The next aspect of graphing an inequality involves shading a region.
- If an inequality reads y is less than or y is less than or equal to, the area below the line will be shaded.
- If an inequality reads y is greater than or y is greater than or equal to, the area above the line will be shaded.

If you’re ever unsure about which side of inequality to shade, you can always choose a point to test. Consider the inequality y > x − 4. Let’s say that we’re unsure about which side of the inequality to shade. We can choose the value of (0,0) and plug it into the equation:

0 > 0 − 4

0 > − 4

With these values inserted, the inequality holds true, so we want to shade the side of the line on which the origin (0,0) falls. Any point can function as your test point; the origin is just an example. The following are graphs of linear inequalities.

Systems of inequalities are similar to systems of equations with regard to the algebraic operations needed to solve them, with the exception of the multiplication and division of negative numbers. The main difference occurs when one arrives at the solution – specifically how it is seen in the final equation and the corresponding graph. Instead of having a specific value for each variable in the system, a solution to a system of inequalities has a range of values for each variable.

“Jane is selling cookies and brownies for a fundraiser. Each brownie is $2.50 and each cookie is $1.85. Jane wants to make at least $80 and sell at minimum 10 brownies.”

If c refers to the number of cookies Jane sells and b to the number of brownies she sells, which of the following systems of inequalities models the word problem?

** A.** 2.50b + 1.85c ≤ 80

b ≥ 10

b ≥ 10

b ≥ 10

b ≥ 10

** SOLUTION**: First, let’s make sure that the correct coefficients are used for the variables c and b, or in other words, that the correct prices for the cookies and brownies are represented. According to the word problem, each brownie (b) costs $2.50, and each cookie (c) costs $1.85, so the first inequality should include the terms 2.50b and 1.85c. Only answer choices A and B do this.

The only detail that distinguishes A from B is in the direction of the inequality sign used in the first inequality. So, which is true - do we want the sum of the variables to be **greater than or equal to or less than or equal to 80**? Keep in mind what the 80 is representing in this context: the amount of money that Jane wants to make at the fundraiser. What type of wording is used to convey this amount? We’re told that Jane wants to make “at least” eighty dollars. **This tells us that we want the variables to equal or be greater than 80,** making **B the correct answer.**

Bob rides his bike to and from school. It takes him more than four minutes but less than six minutes to bike the first 3 blocks to school, and he is able to get to school in less than 20 minutes.

If x represents the number of minutes it takes Bob to ride the first three blocks to school and y represents the number of minutes it takes him to ride the rest of the way to his school, which of the following systems of inequalities models the word problem.

** A.** 4 < x < 6

x + y < 20

x + y < 20

x - y < 20

x - y < 20

** Solution: **Let’s focus on the x-variable first. We’re told that it takes Bob “more than 4 minutes but less than 6 minutes” to ride the first 3 blocks to school. This language indicates a non-inclusive inequality; “more than 4 minutes” does not include exactly 4 minutes as a possibility. So, we can ignore answer choices C and D, because they use inclusive inequalities.

How do answer choices A and B differ? Their first inequalities are identical, but A uses the sum of x and y in its second inequality, and B uses the difference between x and y in its. Which models the problem? Consider what x and y represent in relation to 20.

Both variables are measuring the amount of time it takes Bob to ride his bike a certain distance to school, and the 20 represents how Bob “is able to get to school in under 20 minutes.” We need to add the variables together to find the total time it takes Bob to ride his bike to school and then state the amount of time is less than 20. Finding the difference between the two variables does not relate to the phrasing of the word problem. **A IS THE CORRECT ANSWER.**