Solving Inequalities

 Intro to Inequalities Basic Rule of Inequalities Quadratic Inequalities Rational Inequalities Absolute Value Inequalities Practice Solutions

In previous lessons, we learned to solve equations -- algebraic sentences that describe an equality relation ( = ) between variables and constants (ex: 2x – 5 = 11). Now, we learn to solve inequality relations.
Now we talk about a binary relation in which one variable is greater than (>),
less than (<), greater than or equal to (
) and less than or equal to () another variable or constant.

When I say "That dictionary weighs at least 5 kilograms", I'm defining an inequality relation between the weight of the dictionary and the constant, 5 kilograms.

The inequality for this statement reads: Let d = the weight of the dictionary;

d 5

Had I said "at most" instead of "at least" the statement would have been d 5.

The equality is included since, both "at most" and "at least" include the extreme value or "cut-off point." In other words, the book could weigh exactly 5 kilos. It could also weigh more.

When I say "That book weighs more than 5 kilos, I eliminate the extreme value.The book can no longer weigh exactly 5 kilos -- it can only weigh more than 5 kilos. How much more is not defined.

In this case, the inequality would read d > 5.

memory memo:
How do we remember which sign is which? Does > mean greater than or less than?

Look at the signs: each one has a bigger end and a smaller end.

It's the left end of the sign that says it.

> is bigger on the left than on the right, so it means bigger than.

< starts small -- so it means smaller than.

memory memo:
x > 0 means x is positive -- because only positive numbers live right of 0.

x < 0 means x is negative -- because only negative numbers live left of 0.

In general, inequalities are solved in the same manner as equations
except for one very important rule.

 The basic rule of inequalities states: when we divide or multiply an inequality by a negative, we reverse the direction of the inequality.

Why is this? Because when you divide or multiply through by a negative value, you are switching from one side of the number line to the other.

Look at this number line. Its obvious that: left of 0, the larger the number, the less its value.
The opposite is true right of 0.

In other words, – 6 is less than – 3, but 6 is greater than 3.

– 6 < – 3 but multiply both sides by –1 to get 6 > 3

Since the value relation is reversed when we switch sides of 0, we must reverse the direction of the inequality when we multiply or divide by a negative, for it is this which makes us switch sides.

Here's an example: – 3x > 18 so x < – 6.

Note: we divided through by – 3, so we changed the greater than (>) sign to the less than (<) sign.

This rule applies to all inequalities!!

When solving linear, absolute value and fractional inequalities with constant denominators, we proceed exactly as we do for equations except for this one rule. (see examples)

However, quadratic and fractional inequalities with variables in the denominators are a different story. We now discuss how to solve these kinds of inequalities.

helpful hint: always keep the variables on the left and the constants on the right of the inequality sign so that you don't have to read the statements in reverse to know what " x " is.

Let's get the sillyness out of the way right off the bat.

Since we are working in the set of Real Numbers, perfect squares are all positive (> 0),
so don't waste time trying to find a solution for:

5x 2 + 125 < 0

A positive constant (5), multiplying a perfect square ( x 2 ), plus a positive constant (125) cannot be negative !!! in the set of Real Numbers.

There isn't a single term in the expression that has a hope of being negative.

Simply write "No Solutions" -- hilite it with 17 different colors and move on!

The same is true for the negation of such a statement:

– (5x 2 + 125) > 0 has no solutions,

since the left side is always negative for all values of x in the known universe, but the statement says that it is positive.In philosophy this statement could merit some thought, but not in real-value math.There are no solutions in · , the set of real numbers. There are solutions in the set of imaginary numbers which we will not cover here. Reality is enough for now.

Usually, quadratic inequalities are solved like quadratic equations with one exception. Once we find the zeros (if there are any), either by factoring or formula, it is most efficient to use a number line rather than an algebraic approach to find the solution intervals.

The first step in solving quadratic inequalities is to relate to zero just as we did with quadratic equations.

If the expression can be factored, we'll get a statement that says the product of the factors is either greater than zero or less than zero. In other words, we will have either a positive or negative product. Since we know that a positive product results from multiplying all positives or an even number of negatives, and a negative product is the result of multiplying an odd number of negatives, it is simple to find the correct solutions using a number line.

.

example

Solve the inequality x2 + 3x > 10

Relate to zero x2 + 3x – 10 > 0

Factor the trinomial (x + 5)(x – 2) > 0

We want a positive product from (x + 5) and (x – 2).

We know that the product = 0 when x = – 5 and x = 2.

1. We mark those values on a number line and substitute values for x from the 3 intervals created by the zeros.
2. Check to see if the product is positive or negative, choose values for x that make the product positive.

As you can see, the three intervals to check are numbers:

left of – 5, between – 5 and + 2, and right of + 2.

We plug in x = – 6:

(x + 5)(x – 2) = (– 6 + 5)(– 6 – 2) = (– 1)(– 8) > 0;

which is what we want.

The interval x < – 5 makes the inequality true.

We continue plugging in values for x in the 2 remaining intervals.

We find the values on the interval – 5 < x < 2 give us a negative product:

Values for x on the interval x > 2 produce a positive product.

The solution therefore is all real numbers on the intervals x < – 5 and x > 2.

When we study the Analytic Geometry of the Quadratic Function, the graph will make it obvious that our solution is correct, since all the y-values on the 2 intervals will be positive.

Note: A common error is to write the solution for an inequality such as the previous example as a single interval. As you can see from the number line above, the solution consists of two independent intervals: that is, the numbers that are less than –5 and those that are greater than +2. Though you may be tempted to express your answer as the interval –5< x > 2, it is incorrect.

The correct ways to express this solution are:

1. x < –5 or x > 2
2. ] , – 5 [ 4 ] 2, [
3. (, – 5) 4 (2 , )

Note: when using brackets and parentheses to indicate intervals instead of points:

1. with square brackets: " [ " means the endpoint is included, " ] " means it is not.
2. with parentheses: " [ " means the endpoint is included, " ( " means it is not.

Here's one where we use the quadratic formula to locate the zeros:

Solve x 2 – 5x – 3 0

The trinomial doesn't factor.

With the quadratic formula find: x 1 = – 0.54 and x 2 = 5.54 , the zeros.

Mark these points on the number line and solve.

The solution is the interval – 0. 54 [ x [ 5. 54

.

Rational or fractional inequalities with constants in the denominators are solved exactly the same way that fractional equations with constants in the denominators are solved.

We multiply through by the lcm, eliminate the denominators, then we solve the inequality in the usual way.

Fractional inequalities with variables in the denominator are a different story however.

We cannot solve them the same way we do equations of the same sort. Because the basic rule for inequalities says that multiplying through by a negative changes the direction of the inequality, we cannot multiply through by the common denominator since we don't know if it is positive or negative. Instead, we relate to zero, combine all fractions into a single fraction, and then solve a statement that says our single fraction is either positive ( > 0 )
or negative ( < 0 )
.

The conditions which determine the sign of a fraction are exactly the same as those that determine the sign of a product since a fraction is just division and division is the inverse of multiplication.

In other words, a positive fraction is the result of either all positives or an even number of negatives, and a negative fraction is the result of an odd number of negatives.

Once again, the most efficient way of solving such a problem is a number line like we do for quadratic inequalities.

.

Example:

Solve

Relate to zero

Find lcd and combine

Collect like terms

Mark the zeros, –3, 1/5, and 5 on a number line

creating four intervals: x < –3, –3 < x < 1/5, 1/5 < x < 5, and x > 5.

We test the four intervals to see where the fraction is positive.

We set x = –4, 0, 2, and 6.

We find that the fraction is positive ( > 0) when –3 < x < 1/5 and when x > 5.

Had the question included the equality relationship, in other words had the greater than sign been a greater than or equal to sign (), the solution would have included x = 1/5, since that is the value of x that makes the numerator = 0.

Recall that for a fraction to equal zero, the numerator must be equal to zero.

A denominator of zero makes the fraction undefined or equal to infinity.

Let's do one more example:

Solve

Relate to zero

Find lcd and combine

Collect like terms

Mark the zeros –41, –7 and 3/2 on a number line and test the four intervals.

We're looking for values of x that make the fraction negative or zero.

The solution intervals are x – 41 and – 7 < x < 3/2.

.

Since there are two types of inequalities: greater than and less than, there are two situations to look at in absolute value inequalities. We'll first investigate how to deal with a less than absolute value inequality and then we'll look at the greater than relation.

What does | 2x – 3 | < 5 really mean?

If we look at a number line, it is easy to see that this says that 2x – 3 is sandwiched
between –5 and + 5, since that interval includes all the numbers with an absolute value < 5.

Therefore, we solve an absolute value, "less than" relation between an expression and a constant by sandwiching the expression between the negative and positive values of the constant.

 If | thing | < a then – a < thing < a.

Let's finish the example above: | 2x – 3 | < 5 becomes –5 < 2x – 3 < 5 so –2 < 2x < 8

which means that –1 < x < 4,

so the answer is all values on the interval from –1 to 4.

the solution is an interval denoted (–1, 4) or ] –1, 4 [

The opposite is true of a "greater than" absolute value relation between an expression and a constant. In this case, the expression is either less than the negative value of the constant (left tail) or greater than the positive value of the constant (right tail).

In other words, the expression falls in the tails of the number line.

Say we look at | 2x – 3 | > 5. In this case, 2x – 3 would be < –5 or > 5 as shown.

So, when we solve | 2x – 3 | > 5 we break it into two separate inequalities like this:

| 2x – 3 | > 5 becomes 2x – 3 < –5 or 2x – 3 > 5

so, 2x < –2 becomes x < –1 or 2x > 8 becomes x > 4

Note: the solution is two separate intervals, x's less than –1, or x's greater than 4.

.

examples

linear inequalities:

 a) 3x – 7 < – 5x + 17 8x < 24 so x < 3 b) 2x – 172 < 8x – 100– 6x < 72x > – 12 (divide by –6 -- switch inequality)

fractions with constant denominators

c)

4(x + 5) + 12(2) > 3(x + 16) {lcm = 12, multiply everybody by 12, reduce as you go!}

4x + 44 > 3x + 48

x > 4

 absolute value less than absolute value greater than d) | 3x + 1 | < 7 –7 < 3x + 1 < 7–8/3 < x < 2 e) | 5x – 4 | > 215x – 4 < –21 or 5x – 4 > 215x < –17 or 5x > 25x < –17/5 or x > 5

.

1) Solve these Inequalities:

 a) 3x – 2 < x + 6 b) 4x + 3 > 5x – 17 c) x2 – 4x < 5 d) 2x2 > 5x + 3 e) f) g) h) i) j) | 2x – 3 | < 7 k) | 4 – x | > 5 l) | 17x – 256 | < –3

2) In 2003 I was the proud owner of some shares in the Noise-R-Us Corp.

The value in dollars (V) of my shares was V = 2 | x – 5 | + 30, where x represents the number of months since I bought the stock. I sold the stock after 12 months.

For how many of those 12 months was the value of my stock less than \$35?

.

 a) 3x – 2 < x + 6 2x < 8 x < 4 b) 4x + 3 > 5x – 17–x > –20 x < 20 c) x2 – 4x < 5x2 – 4x – 5 < 0(x – 5)(x + 1) < 0–1 < x < 5 d) 2x2 > 5x + 32x2 – 5x – 3 > 0(2x + 1)(x – 3) > 0x < –1/2, x > 3 e) 4( 2x + 6) < 3(x – 4) 8x + 24 < 3x – 12 5x < –36 so x < –36/5 f) the 0's are x = –2, –5 and 4–5 < x < –2 or x > 4. g) x > –3 since the denominator must be > 0 to make the fraction positive h) set up a number line with the zeros of x = 2, –5/2 and 4. The fraction is > 0 when x < –5/2 or 2 < x < 4. i) set up a number line with the zeros of x = –5, 4, 7 and 0.The fraction is negative (< 0) when x < –5, 0 < x < 4, x > 7 j) | 2x – 3 | < 7 –7 < 2x – 3 < 7 –2 < x < 5 k) | 4 – x | > 5 4 – x < –5 or 4 – x > 5 – x < –9 or – x > 1 so x > 9 or x < – 1 l) | 17x – 256 | < –3 ridiculous since abs. val > 0 always!

2) In 2003 I was the proud owner of some shares in the Noise-R-Us Corp.

The value in dollars (V) of my shares was V = 2 | x – 5 | + 30, where x represents the number of months since I bought the stock. I sold the stock after 12 months.

For how many of those 12 months was the value of my stock less than \$35?

We want to know the values of x (the # of months) that make V less than \$35,

so we have to solve 2 | x – 5 | + 30 < 35

we transpose the 30: 2 | x – 5 | < 5

divide both sides by 2: | x – 5 | < 2. 5

< = sandwich, so: – 2. 5 < x – 5 < 2. 5

transpose the –5 to get: 2. 5 < x < 7. 5

This says the value of the shares was less than \$35.00 from
half way through the 2nd month until half way through the 7th month.

So our answer is 5 months.

The value of the stock was less than \$35.00 for 5 months.