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2.9: Solve Compound Inequalities - Union, Intersection


Learning Objectives

By the end of this section, you will be able to:

  • Solve compound inequalities with “and”
  • Solve compound inequalities with “or”
  • Solve applications with compound inequalities

Before you get started, take this readiness quiz.

  1. Simplify: (frac{2}{5}(x+10)).
    If you missed this problem, review [link].
  2. Simplify: (−(x−4)).
    If you missed this problem, review [link].

Solve Compound Inequalities with “and”

Now that we know how to solve linear inequalities, the next step is to look at compound inequalities. A compound inequality is made up of two inequalities connected by the word “and” or the word “or.” For example, the following are compound inequalities.

[egin{array} {lll} {x+3>−4} &{ ext{and}} &{4x−5leq 3} {2(y+1)<0} &{ ext{or}} &{y−5geq −2} end{array} onumber]

COMPOUND INEQUALITY

A compound inequality is made up of two inequalities connected by the word “and” or the word “or.”

To solve a compound inequality means to find all values of the variable that make the compound inequality a true statement. We solve compound inequalities using the same techniques we used to solve linear inequalities. We solve each inequality separately and then consider the two solutions.

To solve a compound inequality with the word “and,” we look for all numbers that make both inequalities true. To solve a compound inequality with the word “or,” we look for all numbers that make either inequality true.

Let’s start with the compound inequalities with “and.” Our solution will be the numbers that are solutions to both inequalities known as the intersection of the two inequalities. Consider the intersection of two streets—the part where the streets overlap—belongs to both streets.

To find the solution of an "and" compound inequality, we look at the graphs of each inequality and then find the numbers that belong to both graphs—where the graphs overlap.

For the compound inequality (x>−3) and (xleq 2), we graph each inequality. We then look for where the graphs “overlap”. The numbers that are shaded on both graphs, will be shaded on the graph of the solution of the compound inequality. See Figure (PageIndex{1}).

We can see that the numbers between (−3) and (2) are shaded on both of the first two graphs. They will then be shaded on the solution graph.

The number (−3) is not shaded on the first graph and so since it is not shaded on both graphs, it is not included on the solution graph.

The number two is shaded on both the first and second graphs. Therefore, it is be shaded on the solution graph.

This is how we will show our solution in the next examples.

Example (PageIndex{1})

Solve (6x−3<9) and (2x+7geq 3). Graph the solution and write the solution in interval notation.

Answer
(6x−3<9)and(2x+9geq 3)
Step 1. Solve each
inequality.
(6x−3<9)(2x+9geq 3)
(6x<12)(2xgeq −6)
(x<2)and(xgeq −3)
Step 2. Graph each solution. The final graph will show all the numbers that make both inequalities true—the numbers shaded on both of the first two graphs.
Step 3. Write the solution in interval notation.([−3,2))
All the numbers that make both inequalities true are the solution to the compound inequality.

Example (PageIndex{2})

Solve the compound inequality. Graph the solution and write the solution in interval notation: (4x−7<9) and (5x+8geq 3).

Answer

Example (PageIndex{3})

Solve the compound inequality. Graph the solution and write the solution in interval notation: (3x−4<5) and (4x+9geq 1).

Answer

SOLVE A COMPOUND INEQUALITY WITH “AND.”

  1. Solve each inequality.
  2. Graph each solution. Then graph the numbers that make both inequalities true.
    This graph shows the solution to the compound inequality.
  3. Write the solution in interval notation.

Example (PageIndex{4})

Solve (3(2x+5)leq 18) and (2(x−7)<−6). Graph the solution and write the solution in interval notation.

Answer
(3(2x+5)leq 18)and(2(x−7)<−6)
Solve each
inequality.
(6x+15leq 18)(2x−14<−6)
(6xleq 3)(2x<8)
(xleq frac{1}{2})and(x<4)
Graph each
solution.
Graph the numbers
that make both
inequalities true.
Write the solution
in interval notation.
((−infty, frac{1}{2}])

Example (PageIndex{5})

Solve the compound inequality. Graph the solution and write the solution in interval notation: (2(3x+1)leq 20) and (4(x−1)<2).

Answer

Example (PageIndex{6})

Solve the compound inequality. Graph the solution and write the solution in interval notation: (5(3x−1)leq 10) and (4(x+3)<8).

Answer

Example (PageIndex{7})

Solve (frac{1}{3}x−4geq −2) and (−2(x−3)geq 4). Graph the solution and write the solution in interval notation.

Answer
(frac{1}{3}x−4geq −2)and(−2(x−3)geq 4)
Solve each inequality.(frac{1}{3}x−4geq −2)(−2x+6geq 4)
(frac{1}{3}xgeq 2)(−2xgeq −2)
(xgeq 6)and(xleq 1)
Graph each solution.
Graph the numbers that
make both inequalities
true.
There are no numbers that make both inequalities true.

This is a contradiction so there is no solution.There are no numbers that make both inequalities true.

This is a contradiction so there is no solution.There are no numbers that make both inequalities true.

This is a contradiction so there is no solution.

Example (PageIndex{8})

Solve the compound inequality. Graph the solution and write the solution in interval notation: (frac{1}{4}x−3geq −1) and (−3(x−2)geq 2).

Answer

Example (PageIndex{9})

Solve the compound inequality. Graph the solution and write the solution in interval notation: (frac{1}{5}x−5geq −3) and (−4(x−1)geq −2).

Answer

Sometimes we have a compound inequality that can be written more concisely. For example, (adouble inequality. The two forms are equivalent.

DOUBLE INEQUALITY

A double inequality is a compound inequality such as (a

[ ext{Other forms:} quad egin{array} {lllll} {ax>b} &{ ext{is equivalent to }} &{a>x} &{ ext{and}} &{x>b} {ageq xgeq b} &{ ext{is equivalent to }} &{ageq x} &{ ext{and}} &{xgeq b} end{array} onumber]

To solve a double inequality we perform the same operation on all three “parts” of the double inequality with the goal of isolating the variable in the center.

Example (PageIndex{10})

Solve (−4leq 3x−7<8). Graph the solution and write the solution in interval notation.

Answer
Add 7 to all three parts.
Simplify.( 3 le 3x < 15 )
Divide each part by three.( dfrac{3}{color{red}{3}} leq dfrac{3x}{color{red}{3}} < dfrac{15}{color{red}{3}} )
Simplify.( 1 leq x < 5 )
Graph the solution.
Write the solution in interval notation.( [1, 5) )

When written as a double inequality, (1leq x<5), it is easy to see that the solutions are the numbers caught between one and five, including one, but not five. We can then graph the solution immediately as we did above.

Another way to graph the solution of (1leq x<5) is to graph both the solution of (xgeq 1) and the solution of (x<5). We would then find the numbers that make both inequalities true as we did in previous examples.

Example (PageIndex{11})

Solve the compound inequality. Graph the solution and write the solution in interval notation: (−5leq 4x−1<7).

Answer

Example (PageIndex{12})

Solve the compound inequality. Graph the solution and write the solution in interval notation: (−3<2x−5leq 1).

Answer

Solve Compound Inequalities with “or”

To solve a compound inequality with “or”, we start out just as we did with the compound inequalities with “and”—we solve the two inequalities. Then we find all the numbers that make either inequality true.

Just as the United States is the union of all of the 50 states, the solution will be the union of all the numbers that make either inequality true. To find the solution of the compound inequality, we look at the graphs of each inequality, find the numbers that belong to either graph and put all those numbers together.

To write the solution in interval notation, we will often use the union symbol, (cup), to show the union of the solutions shown in the graphs.

SOLVE A COMPOUND INEQUALITY WITH “OR.”

  1. Solve each inequality.
  2. Graph each solution. Then graph the numbers that make either inequality true.
  3. Write the solution in interval notation.

Example (PageIndex{13})

Solve (5−3xleq −1) or (8+2xleq 5). Graph the solution and write the solution in interval notation.

Answer
(5−3xleq −1)or(8+2xleq 5)
Solve each inequality.(5−3xleq −1)(8+2xleq 5)
(−3xleq −6)(2xleq −3)
(xgeq 2)or(xleq −frac{3}{2})
Graph each solution.
Graph numbers that
make either inequality
true.
((−infty,−32]cup[2,infty))

Example (PageIndex{14})

Solve the compound inequality. Graph the solution and write the solution in interval notation: (1−2xleq −3) or (7+3xleq 4).

Answer

Example (PageIndex{15})

Solve the compound inequality. Graph the solution and write the solution in interval notation: (2−5xleq −3) or (5+2xleq 3).

Answer

Example (PageIndex{16})

Solve (frac{2}{3}x−4leq 3) or (frac{1}{4}(x+8)geq −1). Graph the solution and write the solution in interval notation.

Answer
(frac{2}{3}x−4leq 3)or(frac{1}{4}(x+8)geq −1)
Solve each
inequality.
(3(frac{2}{3}x−4)leq 3(3))(4⋅frac{1}{4}(x+8)geq 4⋅(−1))
(2x−12leq 9)(x+8geq −4)
(2xleq 21)(xgeq −12)
(xleq frac{21}{2})
(xleq frac{21}{2})or(xgeq −12)
Graph each
solution.
Graph numbers
that make either
inequality true.
The solution covers all real numbers.
((−infty ,infty ))

Example (PageIndex{17})

Solve the compound inequality. Graph the solution and write the solution in interval notation: (frac{3}{5}x−7leq −1) or (frac{1}{3}(x+6)geq −2).

Answer

Example (PageIndex{18})

Solve the compound inequality. Graph the solution and write the solution in interval notation: (frac{3}{4}x−3leq 3) or (frac{2}{5}(x+10)geq 0).

Answer

Solve Applications with Compound Inequalities

Situations in the real world also involve compound inequalities. We will use the same problem solving strategy that we used to solve linear equation and inequality applications.

Recall the problem solving strategies are to first read the problem and make sure all the words are understood. Then, identify what we are looking for and assign a variable to represent it. Next, restate the problem in one sentence to make it easy to translate into a compound inequality. Last, we will solve the compound inequality.

Example (PageIndex{19})

Due to the drought in California, many communities have tiered water rates. There are different rates for Conservation Usage, Normal Usage and Excessive Usage. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses.

During the summer, a property owner will pay $24.72 plus $1.54 per hcf for Normal Usage. The bill for Normal Usage would be between or equal to $57.06 and $171.02. How many hcf can the owner use if he wants his usage to stay in the normal range?

Answer
Identify what we are looking for.The number of hcf he can use and stay in the “normal usage” billing range.
Name what we are looking for.Let x=x= the number of hcf he can use.
Translate to an inequality.Bill is $24.72 plus $1.54 times the number of hcf he uses or (24.72+1.54x).

(color{Cerulean}{underbrace{color{black}{ ext{His bill will be between or equal to }$57.06 ext{ and }$171.02.}}})

(57.06 leq 24.74 + 1.54x leq 171.02 )

Solve the inequality.

(57.06 leq 24.74 + 1.54x leq 171.02)

( 32.34 leq 1.54x leq 146.3)

( dfrac{32.34}{color{red}{1.54}} leq dfrac{1.54x}{color{red}{1.54}} leq dfrac{146.3}{color{red}{1.54}})

( 21 leq x leq 95 )

Answer the question.The property owner can use (21–95) hcf and still fall within the “normal usage” billing range.

Example (PageIndex{20})

Due to the drought in California, many communities now have tiered water rates. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses.

During the summer, a property owner will pay $24.72 plus $1.32 per hcf for Conservation Usage. The bill for Conservation Usage would be between or equal to $31.32 and $52.12. How many hcf can the owner use if she wants her usage to stay in the conservation range?

Answer

The homeowner can use (5–20) hcf and still fall within the “conservation usage” billing range.

Example (PageIndex{21})

Due to the drought in California, many communities have tiered water rates. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses.

During the winter, a property owner will pay $24.72 plus $1.54 per hcf for Normal Usage. The bill for Normal Usage would be between or equal to $49.36 and $86.32. How many hcf will he be allowed to use if he wants his usage to stay in the normal range?

Answer

The homeowner can use (16–40) hcf and still fall within the “normal usage” billing range.

Access this online resource for additional instruction and practice with solving compound inequalities.

  • Compound inequalities

Key Concepts

  • How to solve a compound inequality with “and”
    1. Solve each inequality.
    2. Graph each solution. Then graph the numbers that make both inequalities true. This graph shows the solution to the compound inequality.
    3. Write the solution in interval notation.
  • Double Inequality
    • A double inequality is a compound inequality such as (aOther forms: [egin{align*} aa≤x≤b & & ext{is equivalent to} & & a≤x; ext{and};x≤b
      a>x>b & & ext{is equivalent to} & & a>x; ext{and};x>b
      a≥x≥b & & ext{is equivalent to} & & a≥x; ext{and};x≥b end{align*}]

  • How to solve a compound inequality with “or”
    1. Solve each inequality.
    2. Graph each solution. Then graph the numbers that make either inequality true.
    3. Write the solution in interval notation.

Glossary

compound inequality
A compound inequality is made up of two inequalities connected by the word “and” or the word “or.”

How do you combine two inequalities?

It is always legal to add inequalities that point in the same direction or to add the same quantity to both/all members of an inequality.

Similarly, how do you solve and/or inequalities? To solve a compound inequality, first separate it into two inequalities. Determine whether the answer should be a union of sets ("or") or an intersection of sets ("and"). Then, solve both inequalities and graph.

Additionally, how do you combine compound inequalities?

When the two inequalities are joined by the word or, the solution of the compound inequality occurs when either of the inequalities is true. The solution is the combination, or union, of the two individual solutions. Let's take a closer look at a compound inequality that uses or to combine two inequalities.

Can two inequalities be subtracted?

If you subtract two inequalities with the same direction there's no way to predict what kind of relationship will result. But, in this case, we can subtract inequalities that have opposite directions. So, if a is greater than b and c is greater than d, we have no way to compare the size of a- c to the size of b- d.


Open Resources for Community College Algebra

This information is accurate as of August 2019. For the complete, most recent CCOG, visit www.pcc.edu/ccog.

Determine whether a relation is a function when the given relation is expressed algebraically, graphically, numerically and/or within real-world contexts by applying the definition of a function.

Use the definition of domain and range to determine the domain and range of functions represented graphically, numerically, and verbally.

Determine the domain of a function given algebraically.

State the domain and range in both interval and set notation.

Understand how the context of a function used as a model can limit the domain and the range.

Evaluate functions with given inputs using function notation where functions are represented graphically, algebraically, numerically and verbally (e.g. evaluate (f(7))).

Distinguish between different expressions such as (f(x+2) ext<,>) (f(x)+2 ext<,>) (3f(x) ext<,>) and (f(3x) ext<,>) and simplify each.

Interpret (f(a)=b) in the appropriate context (e.g. interpret (f(3)=5) where (f) models a real-world function) and understand that (f(2)) is a number not a point.

Solve function equations where functions are represented graphically, algebraically, numerically and verbally (e.g. solve (f(x)=b) for (x)).

Factor the greatest common factor from a polynomial.

Factor a polynomial of four terms using the grouping method.

Factor trinomials that have leading coefficients of 1.

Factor trinomials that have leading coefficients other than 1.

Factor differences of squares.

Determine the domain of rational functions algebraically and graphically.

Simplify rational functions, understanding that domain conditions lost during simplification must be noted.

Perform operations on rational expressions (multiplication, division, addition, subtraction) and express the final result in simplified form.

Solving Equations and Inequalities Algebraically

Solve quadratic equations using the zero product principle.

Solve quadratic equations that have real and complex solutions using the square root method.

Solve quadratic equations that have real and complex solutions using the quadratic formula.

Solve quadratic equations that have real and complex solutions by completing the square (in simpler cases, where (a=1 ext<,>) and (b) is even).

Solve absolute value equations.

Solve equations (linear, quadratic, rational, radical, absolute value) in a mixed problem set.

Determine how to proceed in the solving process based on equation given.

Determine when extraneous solutions may result. (Consider using technology to demonstrate that extraneous solutions are not really solutions).

Check solutions to equations algebraically.

Solve a rational equation with multiple variables for a specific variable.

Solve applications involving quadratic and rational equations (including distance, rate, and time problems and work rate problems).

Variables used in applications should be well defined.

Conclusions should be stated in sentences with appropriate units.

Algebraically solve function equations of the forms:

(f(x)=b) where (f) is a linear, quadratic, rational, radical, or absolute value function.

(f(x)=g(x)) where (f) and (g) are functions such that the equation does not produce anything more difficult than a quadratic or linear equation once a fraction is cleared or a root is removed if one exists.

Solve compound linear inequalities algebraically.

the union of two linear inequalities (“or” statement).

the intersection of two linear inequalities (“and” statement).

a three-sided inequality like (alt f(x)lt b) where (f(x)) is a linear expression with (a) and (b) constants.

Solution sets should be expressed in interval notation.

Brief review of graphs of linear functions, including finding the formula of the function given two ordered pairs in function notation.

Graph quadratic functions by hand.

Review finding the vertex with the formula (h=-frac<2a> ext<.>)

Complete the square to put a quadratic function in vertex form.

Given a quadratic function in vertex form, observe the vertical shift and horizontal shift from the graph of (y=x^2 ext<.>)

State the domain and range of a quadratic function.

Review finding horizontal and vertical intercepts of linear and quadratic functions by hand, expressing them as ordered pairs in abstract examples and interpreting them using complete sentences in application examples.

Solve equations graphically with technology.

Explore functions graphically with technology.

Find vertical and horizontal intercepts.

Find the vertex of a parabola.

Create an appropriate viewing window.

Graphically solve absolute value and quadratic inequalities (e.g. (f(x)lt b ext<,>) (f(x)gt b)) where (f) is an absolute value function when:

given the graph of the function.

using technology to graph the function.

Solve function inequalities graphically given (f(x)lt b ext<,>) (f(x)gt b ext<,>) (f(x)gt g(x) ext<,>) and (alt f(x)lt b) where (f) and (g) should include but not be limited to linear functions.


Compound inequalities

Examples of compound inequalities:

$xle 5 quad or quad 5gt 2x +1$

$8gt x-2 quad and quad xle 7$

A compound inequality of the form:$alt xlt b$ is same as $alt x$ and $xlt b$. This means $x$ is between $a$ and $b$.

In interval notation, $alt xlt b$ is $(a,b)$ and

The word 'or' corresponds to the union operation and the word 'and' corresponds to the intersection operation.

Union

If $A$ and $B$ are two sets, then the union of $A$ and $B$ is the set of elements which are in $A$ or in $B$ or in both $A$ and $B$.

Union of $A$ and $B$ or $A$ union $B$ is denoted as $Acup B$.

Union operation combines the two sets.

Example:Find the union of the sets A and B, where $A=$ and $B=$.

Solution: Draw the graph of $A$ and $B$ and combine them:


Intersection

If $A$ and $B$ are two sets, then the intersection of $A$ and $B$ is the set of elements which are common to both $A$ and $B$.

Intersection of $A$ and $B$ or $A$ intersection $B$ is denoted as $Acap B$.

Intersection operation extracts the elements common to both the sets.

$Acap B=<3,4>:$ only 3 & 4 are in both $A$ and $B$.

Example 2: If $P=$ and $Q=$, then

$Pcap Q=$, the empty set. The symbol for the empty set is $emptyset$.

Example:Find the intersection of the sets A and B, where $A=$ and $B=$.

Solution: Draw the graph of $A$ and $B$ and take the part of the solution common to both:

Solving compound inequalities

Example:1 Solve, $4xlt x-6$ or $xge 3$

Solution:
$egin 4xlt x-6 , &or xge 3 -x:quad 4x-xlt -6 , &or , xge 3 div 3: qquad xlt -2 , &or, xge 3 end$

Solution in graphical form:

Solution in interval notation:

Example 2: Solve $4x-2lt 14$ and $2xge 2$

$egin 4x-2lt 14quad &and quad2xge 2 4xlt 16quad &and quad xge 1 xlt 4quad &and quad xge 1 end$

The solution in the graphical form:

The solution in interval notation:


2.9: Solve Compound Inequalities - Union, Intersection

Write an equation for the nth term of the sequence: -2, 48, 98, 148, 198.

an = 50n - 52
(from: a1 = -2, d = 50, so
an = -2 + 50(n-1) = -2 + 50n - 50)

Solve the single-step inequality:
x - 18 > 20

What inequality is represented by this number line?

Why do absolute value inequalities usually have 2 solutions (for example, |x| > 5 has 2 solutions)?

Absolute values of 2 numbers are often the same (for example: |-6| = 6 and |6| = 6). So the problem has to solve for both cases: x > 5 and x < -5.

A hamburger must be cooked within 5° of 160°F to be considered good (more is burned, less is unsafe). Write an inequality that will tell the range of possible temperatures for a hamburger.

Determine whether 3/4 and 18/24 are equivalent ratios if not, determine which one is higher.

They are equivalent.
(3/4 * 6/6 = 18/24)

Solve the single-step inequality:
5x < -15

Describe the inequality below as a "union" or "intersection" and write it as an inequality.

It is a union x < 7 or x > 11
(x is less than 7 OR x is greater than or equal to 11)

A veterinarian tells Anthony that a healthy weight for Russian Blue cats is 4 ± 1.2 kg. Write and solve an inequality that tells the range of weights that is healthy.

Of the following lines, which one is not parallel to the other three?
(A) y = -3/4 x + 2
(B) y - 3 = 4/3 (x + 2)
(C) 3x + 4y = -16
(D) 8y = -6x

(B). The other three are all slope = -3/4, but (B) has a slope of 4/3, making it perpendicular to those!

Explain why the solution set for the following inequality is not x > -4:

When 12 changes signs to negative, the "greater than" sign (>) changes into a "less than" sign (<). The solution should be:
x < -4

Describe the following graph as a union or intersection and write a compound inequality describing it.


Intersection -1 < x < 2
(-1 is less than x, which is also less than 2)

Give the first step, then find a solution set for |10x + 15| > 105.

First step:
Use 10x + 15 > 105 or 10x + 15 < -105

A veterinarian tells Anthony that a healthy weight for Russian Blue cats is 4 ± 1.2 kg. Write and solve an inequality that tells the range of weights that is unhealthy.

|x - 4| > 1.2
Solution:
x < 2.8 or x > 5.2

Three times the sum of a number and 7 equals 4 times the same number. Would 3n + 7 = 4n solve this situation?

Solve the multi-step inequality:

Method:
2/3 x < -10 (by subtracting 25)
x < -15 (by multiplying 3/2)

Solve the compound inequality in set-builder notation:
-15 < -3x < 5

-5/3 < x < 5
Steps:
-15 < -3x and -3x < 5
Divide -3: 5 > x and x > -5/3
Re-order: -5/3 < x < 5

Find a solution set for |13/4 x - 22/7| < -1

For a scholarship competition, Eva had to write an essay. For the essay, Eva had to write more than 250 words, but it couldn't exceed 500 words. Write an inequality that demonstrates Eva's situation.

|x - 375| < 125
Note: 375 comes from the average of 500 and 250 (500 + 250)/2 = 375, and 125 is the gap from both |250-375| and |500-375|

The following table depicts a baby's crawling distance over time. Does this table depict a linear function?


Yes y = 2.4x (the rate of change is constant at 2.4)

Solve the multi-step inequality (simplify):

x < 11/9
Steps:
-12x + 48 > 15x + 15
Add 12x: 48 > 27x + 15
Subtract 15: 33 > 27x
Divide 27: 33/27 > x
Simplify: 11/9 > x

Solve the compound inequality:

x > 2
Steps:
1 + 3x > 7 solution: x > 2
3 - 7x < -18
Subtract 3: -7x < -21
Divide -7: x > 3
Because this is a union, we must combine both solutions. Since x > 2 includes x > 3, that is enough.

Find a solution set for -2 (5 + |x|) < 24

All real numbers!
Steps:
Divide -2: 5 + |x| > -12
Subtract 5: |x| > -17
ALL absolute values are more than -17.

Aiden wants to spend at most 25,000 on lunch and bus this week. If he must pay 10,000 for bus trips and he wants to buy 4 lunches, set up and solve an inequality that solves how much each lunch will cost.


Rational Inequalities

Solving Rational Inequalities
• Almost the same process as solving a
polynomial inequality
• Also need to consider values for the
variable that cause the denominator to
equal 0
• To solve a rational inequality:
– Set one side to ZERO and write the other side
as a rational expression
– Determine the critical values – those values
that cause either the numerator or
denominator to equal 0
– Split (-oo, +oo) into subintervals based on
the critical values
• Just as with a polynomial inequality
– Test each subinterval to determine those that
are included in the solution set
• The sign rule for polynomials applies for rational
expressions as well
– If necessary, adjust the intervals for
extraneous solutions!

Ex 14: Solve and write the solution in interval
notation:

Ex 15: Solve and write the solution in interval
notation:

Summary
• After studying these slides, you should know how to do
the following:
– Solve, graph, and write the solution set in both interval and set
builder notation for the following types of inequalities:
• Linear
• Compound
• Absolute value
• Polynomial
• Rational
• Additional Practice
– See the list of suggested problems for 1.5
• Next lesson
– Graphing & Circles (Section 2.1)


Compound inequalities

A compound inequality, sometimes referred to as combined inequality, is an inequality that combines two or more simple inequalities joined together with oror and.

To be a solution of an orinequality, a value has to make only one part of the inequality true. That means that the final solution will be the union of solutions of separate inequalities. To be a solution of an andinequality, it must make both parts true. Inequalities whose conditions are bounded with and are not independent of each other. That means that the final solution will be the intersection of solutions of separate inequalities.

or $Rightarrow$ union

and $Rightarrow$ intersection

Example with or inequality

$x<5$ or $x>9$ The question asked in the example is: which numbers can be substitued in the place of $x$ so that one of the inequalities is true?

$x<5 Rightarrow x in left<-infty, 5 ight>$

$x>9 Rightarrow x in left<9, infty ight>$

The answer are all numbers less than 5 and all numbers greater than 9, so the solution is a union of two intervals $left<-infty, 5 ight> cup left<9, infty ight>$.

Example with and inequality

This statement is equivalent to $4<x<7$.

$x>4 Rightarrow x in left<4, infty ight>$

$x<7 Rightarrow x in left<-infty, 7 ight>$

The solution is an intersection of two intervals $left<-infty, 7 ight> cap left<4, infty ight>$, which is the interval $left<4, 7 ight>$.

$ 2 + 2x leq x < 5 + x$ This problem can be solved in two ways.

The problem is divided into two inequalities which are then solved separately. The solution is the intersection of the individual solutions.

$ 2 + 2x leqslant x$ and $ x < 5 + x$ $ 2 + 2x leqslant x Rightarrow 2 leq -x Rightarrow x leq – 2Rightarrow x in left<-infty, – 2 ight]$ $ x < 5 + x Rightarrow 0 < 5$ From the second inequality follows the statement $ 0 < 5$, which is always true, so the solution is the whole set of real numbers.

If the case was different and the statement wasn’t true, for example $5 < 0$, then the inequality wouldn’t have solutions. For example, the inequality $x<x-1$ has no solutions.

The solution of the inequality from the Example 1 is the set $ left<-infty, – 2 ight]$.

When working with equations, one can add, subtract, multiply and divide the expression, but what is changed on one side must be also changed the same way on the other. It is similar when working with inequalities. If you add, subtract, multiply or divide, you must do it for every part of the inequality.

In our example we can subtract $2x$.

$ 2 leqslant – x < 5 – x$
$ – x < 5 – x$ is always a true statement, so the only part which restricts the set of solutions is $ 2 leqslant – x $. Therefore, the solution is the set $ left<-infty, – 2 ight]$.

$ 1 + 58x < 55x < 57x + 10$
$57x$ is subtracted from every part of the inequality:

When the expression $ – 2x < 10$ is divided by $-2$, the sign of inequality changes.

$ x < -frac<1><3>$ and $ x > – 5$

The final solution is the set $ left<- 5, -frac<1><3> ight>$.

In case when the intersection is empty, there are no solutions.

For example, if $ x > 5$ and $ x < – 7$, then the intersection of the sets $left<5, infty ight>$ and $left<-infty, -7 ight>$ is empty.Therefore, inequality has no solutions.

$ 5 > x$ and $ x > 2 Rightarrow x in left<2, 5 ight>$

$ x > – 6$ and $ x > 2 Rightarrow x in left<2, +infty ight>$

$ 2x > 4$ or $ -x > 7$
$ 2x > 4 Rightarrow x > 2$ $ – x > 7 Rightarrow x < – 7$

The solution of the first inequality is set $left<2, infty ight>$ and the solution of the second inequality is set $left<-infty, – 7 ight>$.

The final solution is the union of these two sets: $ left<-infty, – 7 ight> cup left<2, infty ight>$.

$frac <2>> 3$ or $ x geqslant – 3$
$ x > 6$ or $ x geqslant- 3$ In a case like this, there is no need to write a union of two sets since one is the subset of the other. The solution is simply the bigger set, in this case $left[- 3, +infty ight>$.


To Infinity (Not Beyond):

an interval may be defined as a subset of the extended real numbers, the set of all real numbers augmented with +∞ and −∞.

Interval NotationInequalitiesDetails
(a, +∞)x > agreater than a
[a, +∞)x ≥ agreater than or equal to a
(-∞, a)x < aless than a
(-∞, a]x ≤ aless than or equal to a
(-∞, ) < x < ∞no limit

In this interpretation, the notations above are all meaningful and distinct. In particular, (−∞, +∞) denotes the set of all ordinary real numbers, while [−∞, +∞] denotes the extended reals.


- compound inequalities.

Compound Inequality: - In compound inequality, there are at least two inequalities and these inequalities are separated by “and” or “or”.

1. 5 < x < 7 ( x > 5 a n d x < 7 )

Here, we can see using graph, if two inequalities are separated by “and” then solution is intersection of both inequalities.

Here, we can see using graph, if two inequalities are separated by “or” then solution is union of both inequalities.

Calculation:-

Since, option (A) contains two inequalities which are separated by “or”.

So, Option (A) is compound inequality.

Since, option (B) contains one inequality.

So, Option (B) is not compound inequality.

Option (C) :- 8 ≤ 5 x < 30 ( 8 ≤ 5 x a n d 5 x < 30 )

Since, option (C) contains two inequalities which are separated by “and”.


Another type of compound inequality਌ontains the word or. A compound inequality containing or is true if one or more of the inequalities is true. The graph of a compound inequality containing or is the of the graphs of the two inequalities.

In other words, the solution of the਌ompound inequality is a solution of either inequality, not necessarily both. The union can be found by graphing each inequality.

Solve the compound inequality. Then graph the solution set

Subtract by 2 on both sides

Subtract by 2 on both sides

By graphing the inequality k > 10, we get the graph given below.

By graphing the inequality  ≤  16 , we get the graph given below.

By combining the above two graphs, we get the common region between 10 and 16.

Solve the following inequality and graph the solution

Solve the following inequality and graph the solution

Divide by 3 on both sides

Divide by 2 on both sides

Solve the following inequality and graph the solution

Divide by 4 on both sides

Divide by 4 on both sides

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