| Exercises 1 Comparing the given inequalities, we can see that the second inequality is more or less identical to the first one but with a 5 subtracted from both sides:
xx − 5≤6≤6 − 5
If we preform the same operation on both sides of an inequality, the solution set remains unchanged. These actions are allowed by the Addition, Subtraction, Multiplication, and Division Properties of Inequality. Therefore, the inequalities are equivalent. | |
| Exercises 2 Solving an equation with addition and solving an inequality with addition looks very similar. In fact, the only difference will be the solution. Meaning, we apply the inverse operation of addition in both equations and inequalities in the same ways. Let's consider the following examples.Solving an equation with addition
Solve x−2=8. Solving this equation requires adding 2 to both sides to isolate x.
x−2=8LHS+2=RHS+2x−2+2=8+2Add termsx=10
The solution to the equation is
x=10.Solve an inequality with addition
Solve x−2>8. Solving this inequality requires adding 2 to both sides to isolate x.
x−2>8LHS+2>RHS+2x−2+2>8+2Add termsx>10
The solution to the inequality is
x>10.Conclusion
From the work above, we can see that solving equations and inequalities with addition is very similar. The only difference is the solution, there is one value for an equation and an interval for an inequality. | |
| Exercises 3 To solve an inequality, we use the same strategy as if it was an equation, applying inverse operations to isolate the variable. For the inequality
k+11<-3,
we would subtract 11 from both sides to isolate the variable. | |
| Exercises 4 To solve an inequality, we use the same strategy as if they were equations. In other words, we apply inverse operations to isolate the variable. For the inequality
v−2>14,
we would add 2 to both sides to isolate the variable. | |
| Exercises 5 To solve an inequality, we use the same strategy as if it was an equation, applying inverse operations to isolate the variable. For the inequality
-1≥b−9,
we would add 9 to both sides to isolate the variable. | |
| Exercises 6 To solve an inequality, we use the same strategy as if it was an equation, applying inverse operations to isolate the variable. For the inequality
-6≤17+p,
we would subtract 17 from both sides to isolate the variable. | |
| Exercises 7 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
x−4<-5LHS+4<RHS+4x<-1
This inequality tells us that all values less than -1 will satisfy the inequality. Notice that x cannot equal -1, which we show with an open dot on the number line. | |
| Exercises 8 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
1≤s−8LHS+8≤RHS+89≤sRearrange inequalitys≥9
This inequality tells us that all values greater than or equal to 9 will satisfy the inequality. Notice that s can equal 9, which we show with a closed dot on the number line. | |
| Exercises 9 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
6≥m−1LHS+1≥RHS+17≥mRearrange inequalitym≤7
This inequality tells us that all values less than or equal to 7 will satisfy the inequality. Notice that m can equal 7, which we show with a closed dot on the number line. | |
| Exercises 10 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
c−12>-4LHS+12>RHS+12c>8
This inequality tells us that all values greater than 8 will satisfy the inequality. Notice that c can equal 8, which we show with an open dot on the number line. | |
| Exercises 11 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
r+4<5LHS−4<RHS−4r<1
This inequality tells us that all values les than 1 will satisfy the inequality. Notice that r cannot equal 1, which we show with an open dot on the number line. | |
| Exercises 12 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
-8≤8+yLHS−8≤RHS−8-16≤yRearrange inequalityy≥-16
This inequality tells us that all values greater than or equal to -16 will satisfy the inequality. Notice that y can equal -16, which we show with a closed dot on the number line. | |
| Exercises 13 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
9+w>7LHS−9>RHS−9w>-2
This inequality tells us that all values greater than -2 will satisfy the inequality. Notice that w cannot equal -2, which we show with an open dot on the number line. | |
| Exercises 14 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
15≥q+3LHS−3≥RHS−312≥qRearrange inequalityq≤12
This inequality tells us that all values less than or equal to 12 will satisfy the inequality. Notice that q can equal 12, which we show with a closed dot on the number line. | |
| Exercises 15 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
h−(-2)≥10a−(-b)=a+bh+2≥10LHS−2≥RHS−2h≥8
This inequality tells us that all values greater than or equal to 8 will satisfy the inequality. Notice that h can equal 8, which we show with a closed dot on the number line. | |
| Exercises 16 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
-6>t−(-13)a−(-b)=a+b-6>t+13LHS−13>RHS−13-19>tRearrange inequalityt<-19
This inequality tells us that all values less than -19 will satisfy the inequality. Notice that t cannot equal -19, which we show with an open dot on the number line. | |
| Exercises 17 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
j+9−3<8LHS+3<RHS+3j+9<11LHS−9<RHS−9j<2
This inequality tells us that all values less than 2 will satisfy the inequality. Notice that j cannot equal 2, which we show with an open dot on the number line. | |
| Exercises 18 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
1−12+y≥-5Subtract term−11+y≥-5LHS+11≥RHS+11y≥6
This inequality tells us that all values greater than or equal to 6 will satisfy the inequality. Notice that y can equal 6, which we show with a closed dot on the number line. | |
| Exercises 19 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
10≥3p−2p−7Subtract term10≥p−7LHS+7≥RHS+717≥pRearrange inequalityp≤17
This inequality tells us that all values less than or equal to 17 will satisfy the inequality. Notice that p can equal 17, which we show with a closed dot on the number line. | |
| Exercises 20 Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign.
18−5z+6z>3+6Add and subtract terms18+z>9LHS−18>RHS−18z>-9
This inequality tells us that all values greater than -9 will satisfy the inequality. Notice that z cannot equal -9, which we show with an open dot on the number line. | |
| Exercises 21 To write a verbal inequality in algebraic terms, we will start by identifying the inequality symbol we need to use and placing it in the center of our statement. In this case, we have "greater than," which is represented by the symbol >.
…>…
On the left-hand side of the inequality we have "a number plus 8," which we will express as a sum.
n+8>…
On the right-hand side of the inequality, we have "11." We can now form our final answer.
A number plus 8 is greater than 11n + 8>11
Now we can solve for n.
n+8>11LHS−8>RHS−8n+8−8>11−8Subtract termsn>3
All real numbers greater than 3 will satisfy this inequality. | |
| Exercises 22 To write a verbal inequality in algebraic terms, we will start by identifying the inequality symbol we need to use and placing it in the center of our statement. In this case, we have "at least," which is represented by the symbol ≥.
…≥…
On the left-hand side of the inequality we have "a number minus 3," which we will express as a difference.
n−3≥…
On the right-hand side of the inequality, we have "-5." We can now form our final answer.
A number minus 3 is at least -5n−3≥-5
Now we can solve for n.
n−3≥-5LHS+3≥RHS+3n−3+3≥-5+3Add termsn≥-2
All real numbers greater than or equal to -2 will satisfy this inequality. | |
| Exercises 23 To write a verbal inequality in algebraic terms, we will start by identifying the inequality symbol we need to use and placing it in the center of our statement. In this case, we have "fewer than," which is represented by the symbol <.
…<…
On the left-hand side of the inequality we have "the difference of a number and 9".
n−9<…
On the right-hand side of the inequality, we have "4." We can now form our final answer.
The difference of a number and 9 is fewer than 4n−9 < 4
Now we can solve for n.
n−9<4LHS+9<RHS+9n−9+9<4+9Add termsn<13
All real numbers less than 13 will satisfy this inequality. | |
| Exercises 24 To write a verbal inequality in algebraic terms, we will start by identifying the inequality symbol we need to use and placing it in the center of our statement. In this case, we have "less than or equal to," which is represented by the symbol ≤.
…≤…
On the left-hand side of the inequality we have "six."
6≤…
On the right-hand side of the inequality, we have "the sum of a number and 15." We can now form our final answer.
Six is less than or equal to the sum of a number and 156≤n+15
Now we can solve for n.
6≤n+15LHS−15≤RHS−156−15≤n+15−15Subtract terms-9≤nRearrange inequalityn≥-9
All real numbers greater than or equal to -9 will satisfy this inequality. | |
| Exercises 25 aOur carry-on bag cannot exceed 50 pounds, the total weight of the stuff in our bag must be less than that. We can express this mathematically as an inequality.
…≤50
Let x be the additional weight we can add to our bag. Then we can write the inequality with the left-hand side is the current weight of our carry-on plus any additional weight we can fit.
38+x≤50
Let's solve this inequality for x.
38+x≤50LHS−38≤RHS−3838+x−38≤50−38Subtract termx≤12
We can add up to 12 pounds to our bag.bThe total weight of the laptop and the pair of boots is is the sum of their weights.
9+5=14 pounds
Their total weight is more than 12 pounds and, therefore, we cannot add both of them to our bag. | |
| Exercises 26 The price of the book you ordered is $19.76. To be eligible for free shipping, you need to spend at least $25. We can express this more mathematically as:
…≥25.
Let x be the additional money you need to spend. Now we can write the inequality as:
19.76+x≥25.
Let's solve this inequality for x.
19.76+x≥25LHS−19.76≥RHS−19.7619.76+x−19.76≥25−19.76Subtract termsx≥5.24
You need to purchase at least $5.24 more worth of books to be eligible for the free shipping. | |
| Exercises 27 The inequality is solved in the correct way. However, the result -3<x does not mean that x is less than -3. Let's take a look at the inequality rearranged:
-3<x⇔x>-3
Instead this means x must be greater than -3. But when we look at the graph, it shows values that are less than. The solution set should instead look like the graph shown below. | |
| Exercises 28 The inequality is not solved correctly. Anything added to one side of an inequality must also be added to the other side, just like in equations.
-10+x≥-9LHS+10≥RHS+10-10+x + 10≥-9 + 10Add termsx≥1
The correct answer is all numbers greater than or equal to 1. | |
| Exercises 29 We are told that the NHL record for goals in a season is 92. For the given NHL hockey player to match or break that record, he needs at least 92 goals. We can express this mathematically as:
…≥92.
On the left-hand side of this inequality should be the sum of the player's current goal count and the number of goals he needs to match or break the record. The player has scored 59 goals so far. Let's use the variable x to represent the number of additional goals the player needs to score.
59+x≥92.
Let's solve the inequality for x.
59+x≥92
LHS−59≥RHS−59
LHS−59≥RHS−5959+x−59≥92−59Subtract terms
x≥33
The player needs to score at least 33 goals during the rest of the season to match or break the NHL record. | |
| Exercises 30 aTo begin, we can assume that to beat our competitor, our score must be greater than theirs. Adding our opponents two scores, we can find their overall score:
117.1+119.8=236.9.
To win, our score must thus be greater than 236.9. Our first score is 119.5 and we can let s be our second score. Then, our total score can be determined by the expression:
119.5+s.
This sum must be greater than 236.9 in order for us to win. We can thus write the following inequality:
119.5+s>236.9.
Solving this inequality will tell us the score necessary for us to beat our opponent.
119.5+s>236.9LHS−119.5>RHS−119.5119.5+s−119.5>236.9−119.5Subtract termss>117.4
Our second score must be greater than 117.4 for us to win.bWe found that our second score must be greater than 117.4 for us to win. This means any value greater than 117.4 will allow us to win. We could score 117.5, and we'd win. We could score 118.4 and win. Both our coach and our teammate are thus correct. | |
| Exercises 31 Let's check each option one by one, trying to rewrite them as the given inequality.A
We can start with the first option.
x−b−3<0LHS+3<RHS+3x−b<3
This is exactly the same inequality.B
Now let's check the second option.
0>b−x+3LHS−b>RHS−b-b>-x+3LHS+x>RHS+xx−b>3
This is not equivalent to the given inequality.C
Next, let's check the third option.
x<3−bLHS+b<RHS+bx+b<3
This is not equivalent to the given inequality.D
Finally, let's check the fourth option.
-3<b−xLHS+3<RHS+30<b−x+3LHS+x<RHS+xx<b+3LHS−b<RHS−bx−b<3
This is exactly the same inequality. | |
| Exercises 32 Let's first find the perimeter by adding the three sides of the given triangle.
Perimeter=14.2+15.5+x
We are also given that the perimeter is less than 51.3 inches.
Perimeter<51.3
Substituting the expression for the perimeter we formed above, we can form an inequality.
14.2+15.5+x<51.3
We can solve this inequality by isolating x.
14.2+15.5+x<51.3Add terms29.7+x<51.3LHS−29.7<RHS−29.7x<21.6
The length of the unknown side x must be less than 21.6 inches. | |
| Exercises 33 Let's first find the perimeter by adding the four sides of the given quadrilateral.
Perimeter=6.4+4.9+4.1+x
We are told that the perimeter is less than or equal to 18.7 feet.
Perimeter≤18.7
Substituting the expression for the perimeter, we can form an inequality.
6.4+4.9+4.1+x≤18.7
We can solve this inequality by isolating x.
4.9+4.1+6.4+x≤18.7Add terms15.4+x≤18.7LHS−15.4≤RHS−15.4x≤3.3
The length of the unknown side x must be less than or equal to 3.3 feet. | |
| Exercises 34 The inequality graph shows all values less than or equal to 16. If we call the values that are part of the solution's set x, we have the following inequality:
x≤16.
Now let's describe a real-life situation. A group of 20 children is going to the movies. The oldest child is 16 years old. Write an inequality to describe the age of all the children in the group. | |
| Exercises 35 The solution set of an inequality is usually an interval. Since there are infinitely many real numbers in any interval, it isn't possible to check all the numbers of the solution set.
To answer the second question, let's solve the inequality first.
x−11≥-3LHS+11≥RHS+11x≥8
To verify our solution, we can check 8 as it's the first number of the solution set. We also check a few numbers greater than and less than 8 to be sure that the solution set was calculated correctly. For example, let's check 6, 7, 8, and 9.xx−11≥-3Simplify
66−11≥-3-5≱-3
77−11≥-3-4≱-3
88−11≥-3-3≥-3
99−11≥-3-2≥-3
Substituting the numbers that are less than 8, we produced false statements. For the numbers greater than or equal to 8, we produced true statements. Therefore, x≥8 is a correct solution set. | |
| Exercises 36 aThis inequality is not necessarily true as we do not know the size of sets H or E. We have no idea how many students have brown hair or how many have brown eyes. As such, we cannot compare their sizes.bJust as with Part A, this inequality is not necessarily true as we do not know the size of sets H or E. We still have no idea how many students have brown hair or how many have brown eyes. As such, we cannot compare their sizes.cThis inequality is true because it contains the "or equal to." It is possible for all of the students with brown hair to have brown eyes. It is also possible for any number of the students with brown hair to not have brown eyes.dBecause H contains X, this inequality is true. If you add 10 to a set that is already larger than another set, the larger set will continue being larger.eThe given inequality implies that the number of students with brown hair will always be greater than the number of students with brown hair and brown eyes. However, it is possible for all of the students with brown hair to have brown eyes, H=X. The given inequality states that this is not possible.fJust as in Part D, this inequality is true because H contains X. The largest possible value of X is when H=X. Adding 10 to H would just make it even larger than X. | |
| Exercises 37 a
Let's begin by solving the inequality.
x+8<14LHS−8<RHS−8x+8−8<14−8Subtract termsx<6
All numbers that are less than 6 are in the solution set. This means that all numbers that are greater than or equal to 6 are not in the solution set. We write this as:
x≥6.
We can graph this inequality by placing a closed dot on 6 on a number line and shading to the right.bLet's begin by solving the inequality.
x−12≥5.7LHS+12≥RHS+12x−12+12≥5.7+12Add termsx≥17.7
All numbers that are greater than or equal to 17.7 are in the solution set. This means that all numbers that are less than 17.7 are not in the solution set. We can write this as:
x<17.7.
We can graph this inequality by placing an open dot at 17.7 on a number line and shading to the left. | |
| Exercises 38 We need to order the four variables from least to greatest. Let's start by isolating one of the variables in each of the given inequalities and draw a conclusion from what we find. In
c−3≥d,
d has already been isolated. It says that d is less than or equal to a reduction in c. This means we know that c>d. Now, let's look at the second inequality.
b+4<a+1LHS−1<RHS−1b+3<a
This inequality says that, even if we increase b, a is still larger. Hence, we know that a>b. Finally, let's look at the third inequality.
a−2≤d−7LHS+7≤RHS+7a+5≤d
This inequality says that, even if we increase a, d is still larger. Thus, we know that d>a. We have the following individual comparisons.
b<aanda<dandd<c
We can use these three pieces of information to compare the values of the variables from least to greatest.
{b,a,d,c} | |
| Exercises 39 When you multiply a positive number by a negative number, the product is always negative.
7⋅(-9)a(-b)=-a⋅b-7⋅9Multiply-63
The product is -63. | |
| Exercises 40 When calculating the product, remember that multiplying two negative numbers results in a positive number.
-11(-12)-a(-b)=a⋅b11⋅12Multiply132 | |
| Exercises 41 For this exercise, both the divisor and dividend are negative. When calculating, remember that the quotient of two negative numbers is always positive.
-27÷-3a÷b=ba-3-27-b-a=ba327Calculate quotient9 | |
| Exercises 42 The expression a÷b is the same thing as ba. Notice that, in this case, the numerator and denominator have opposite signs. The quotient of a positive number and a negative number is always negative.
20÷(-5)Write as a fraction-520Put minus sign in front of fraction-520Calculate quotient-4 | |
| Exercises 43 To solve the equation, we have to isolate x on one side of the equation. By dividing both sides of the equation by 6, we can remove the coefficient from x.
6x=24LHS/4=RHS/466x=624Calculate quotientx=4
To check that we have the correct solution, we can substitute x=4 into the original equation.
6x=24x=46(4)=?24Multiply24=24
Since both sides are equal, we can conclude that x=4 is correct. | |
| Exercises 44 To solve the equation, we have to isolate y on one side of the equation. By dividing both sides of the equation by -3, we can remove the coefficient from y.
-3y=-18LHS/(-3)=RHS/(-3)-3-3y=-3-18-b-a=bay=318Calculate quotienty=6
To check that we have the correct solution, we can substitute y=6 into the original equation.
-3y=-18y=6-3(6)=?-18(-a)b=-ab-3⋅6=?-18Multiply18=18
Since both sides are equal, we can conclude that y=6 is correct. | |
| Exercises 45 To solve the equation, we will isolate the variable. To undo the division in the given equation, we will use the Multiplication Property of Equality to multiply both sides of the equation by -8.
-8s=13LHS⋅(-8)=RHS⋅(-8)-8s(-8)=13(-8)(-8)a⋅(-8)=as=13(-8)Multiplyx=-104
We will check our solution by substituting this result back into the original equation and simplifying.
-8s=13x=-104-8-104=?13-b-a=ba8104=?13Calculate quotient13=13
Substituting s=-104 resulted in a true statement, so our solution is correct. | |
| Exercises 46 To solve the equation, we will isolate the variable. To undo the division in the given equation, we will use the Multiplication Property of Equality to multiply both sides of the equation by 4.
4n=-7.3LHS⋅(4)=RHS⋅(4)4n⋅4=-7.3⋅4(4)a⋅(4)=an=-7.3⋅4Multiplyn=-29.2
We will check our solution by substituting this result back into the original equation and simplifying.
4n=-7.3n=-29.24-29.2=?-7.3Calculate quotient-7.3=-7.3
Since substituting n=-29.2 resulted in a true statement, our solution is correct. | |
