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###### Exercises

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Exercises 1 To calculate the difference between the expected attendance, 65 people, and actual attendance, we need to subtract the expected attendance from the actual attendance. ∣actual−expected∣=deviation If we call the actual attendance x and substitute the given expected attendance, our expression becomes: ∣x−65∣=deviation. We are told that the actual attendance can vary by up to 30. Thus, we can write our equation as the following. ∣x−65∣=30 The equation corresponds to A. | |

Exercises 2 aTo determine the value of b when a=5 and when the solution to this inequality is x≤-3, let's substitute a=5 in the inequality and solve for x. ax+4≤3x+ba=55x+4≤3x+b Solve for x LHS−3x=RHS−3x2x+4≤bLHS−4=RHS−42x≤b−4LHS/2=RHS/2 x≤2b−4 Since we want x to be less than or equal to -3, we have to find b, such that 2b−4=-3. Let's solve this equation for b. 2b−4=-3LHS⋅2=RHS⋅2b−4=-6LHS+4=RHS+4b=-2 When a=5 and b=-2, x≤-3.bSince we want the solution of the inequality to be all real numbers, we have to eliminate x from the inequality and obtain an statement that is always true. To eliminate x, we need the x-terms on both sides to be the same. Then they will cancel out. Since we have 3x on the right-hand side, we will substitute a=3. ax+4≤3x+ba=33x+4≤3x+bLHS−3x=RHS−3x4≤b If we want 4≤b to be always true, we can substitute any number for b that makes this inequality true. Let's choose b=5. Then we have 4≤5. This statement is always true and the solution to the original inequality will be all real numbers. In conclusion, when a=3 and b=5, the solution of the inequality is all real numbers.cSince we want the inequality to have no solution, we have to eliminate x from the inequality and obtain a contradiction. To eliminate x, we need the x-terms on both sides to be the same. Then they will cancel out. Again, since we have 3x on the right-hand side, we will substitute a=3. ax+4≤3x+ba=33x+4≤3x+bLHS−3x=RHS−3x4≤b If we want 4≤b to be a contradiction, we have to substitute a number for b that makes this inequality false. Let's choose b=3. Then we have 4≰3. This is always false and the original inequality will have no solution. Therfore, when a=3 and b=3, the inequality has no solution. | |

Exercises 3 To place each inequality in the correct box, we will solve each of them separately.5x−6+x≥2x−8 5x−6+x≥2x−8 Solve for x LHS+6≥RHS+65x+x≥2x−2Add terms6x≥2x−2LHS−2x≥RHS−2x4x≥-2LHS/4≥RHS/4x≥-42ba=b/2a/2 x≥-21 All values above -0.5 are solutions to the inequality. Therefore, any integer greater than this number is a solution, such as x=1 or x=2.2(3x+8)>3(2x+6) 2(3x+8)>3(2x+6) Solve for x Distribute 26x+16>3(2x+6)Distribute 36x+16>6x+18LHS−6x>RHS−6x 16≯18 Since 16 is never greater than 18, there is no solution to this inequality. This includes any possible integer solutions.17<4x+5<21 To solve this inequality, we have to separate it into two inequalities: 17<4x+5 and 4x+5<21. Now we solve each of them separately and then combine the solutions. 17<4x+5 Solve for x LHS−5<RHS−512<4xLHS/4<RHS/43<xRearrange inequality x>3 Let's solve the second inequality. 4x+5<21 Solve for x LHS−5<RHS−54x<16LHS/4<RHS/4 x<4 The final solution is 3<x<4. This means that x cannot be an integer because the solution set is all values between two consecutive integers.x−8+4x≤3(x−3)+2x x−8+4x≤3(x−3)+2x Solve for x Distribute 3x−8+4x≤3x−9+2xSimplify terms5x−8≤5x−9LHS−5x=RHS−5x-8≤-9Flip inequality and change signs 8≱9 Since 8 is not greater than or equal to 9, there are no solutions to this inequality. This includes any integer solutions.9x−3<12 or 6x+2>-10 We will solve each of the inequalities separately and then combine the solutions. 9x−3<12 Solve for x LHS+3<RHS+39x<15LHS/9<RHS/9x<915Use a calculatorx<1.66666… x<1.67 So far we have that x<1.67. Since the original inequality is 9x−3<12 or 6x+2>-10, we can already tell that it has at least one integer solution. For example, x=1 satisfies the first inequality and this is enough.5(x−1)≤5x−3 5(x−1)≤5x−3 Solve for x Distribute 55x−5≤5x−3LHS−5x≤RHS−5x-5≤-3Flip inequality and change signs 5≥3 Because 5 is always greater than 3, we have reached an identity. The solution to this inequality is all real numbers. Therefore there is at least one integer solution.Conclusion Let's fill in the table with the information we have found.At least one integer solutionNo integer solutions 5x−6+x≥2x−89x−3<12 or 6x+2>-105(x−1)≤5x−32(3x+8)>3(2x+6) 17<4x+5<21 x−8+4x≤3(x−3)+2x | |

Exercises 4 a In this part, we will write an inequality that represents the numbers x of plays we must attend for the season pass to be better deal. The cost for x plays can be written as the following. Cost of one play $25 × number of plays× x The cost 25×x can be also written as 25x. This must be greater than $180 if it will be cheaper to buy the season pass. Greater than is represented by > and the inequality can be written as the following. 25x>180bIn order to select the numbers, we need to solve the inequality that we found in Part A. Let's solve it by using inverse operations. 25x>180LHS/25>RHS/25x>25180Calculate quotientx>7.2 The solution tells us that the season pass is a better deal for more than 7.2 plays. This means that a season pass is not a better deal for the following numbers of plays. {0,1,2,3,4,5,6,7} | |

Exercises 5 First, let's solve the given equation for x. 3(2x−4)=4(ax−2) Solve for x Distribute 36x−12=4(ax−2)Distribute 46x−12=4ax−8LHS+12=RHS+126x=4ax+4LHS−4ax=RHS−4ax6x−4ax=4Factor out xx(6−4a)=4LHS/4=RHS/4 x=6−4a4 We can notice that the numerator is a positive number, 4. Because the fraction needs to simplify to be a positive number, we need the denominator to be positive as well. In other words, we must have that 6−4a>0. Let's solve this inequality for a. 6−4a>0 Solve for a LHS−6>RHS−6-4a>-6Flip inequality and change signs4a<6LHS/4<RHS/4a<46ba=b/2a/2a<23Write as a decimal a<1.5 When a<1.5, the solution of the equation is positive. From the given values, we identify the ones which are less than 1.5 as: -2,-1,0,1. | |

Exercises 6 Let's begin by determining the compound inequality that the solution set represents.Identifying the compound inequality Starting from the left, we see a closed circle at -2 which means x has to be greater than or equal to -2. We can write this as -2≤x. On the right, we have an open circle at 3 which means x has to be less than 3. This is written as x<3. Combining these inequalities, we have the following compound inequality x≥-2andx<3.Identifying the correct sign Next, we have to determine which signs we should fill the two gaps with. Since x is not isolated, we cannot fill them in right away. Let's begin by isolating x in both inequalities. 4x−18 22 -x−3LHS+18 22 RHS+184x 22 -x+15LHS+x 22 RHS+x5x 22 15LHS/5 22 RHS/5x 22 3 Notice that we never divided or multiplied by a negative number. This means that the inequality sign stayed the same direction throughout solving it. Therefore, when comparing this to the compound inequality, we have: x 22 3⇒x < 3. Let's solve the next inequality. -3x−9 22 -3LHS+3x 22 RHS+3x-9 22 -3+3xLHS+3 22 RHS+3-6 22 3xLHS/3 22 RHS/3-2 22 x Again, notice that we did not divide or multiply by a negative number and the inequality sign stayed the same direction throughout. So when comparing this to the compound inequality, we get: -2 22 x⇒-2 ≤ x. | |

Exercises 7 aThe price of one pair of sneakers is $80, so if we want 2 pairs we have to pay 80⋅2=$160. If we also buy x pairs of socks, at $12 per pair, all-in-all we pay 160+12x. Having a $250 gift card, this expression has to be less than or equal to 250, which gives us the following inequality: 160+12x≤250. Let's isolate x to find the possible numbers of pairs of socks we can buy in this situation. 160+12x≤250LHS−160≤RHS−16012x≤90LHS/12≤RHS/12x≤7.5 Since we cannot buy 0.5 pairs of socks, and we don't have enough money to buy 8, the number of pairs we can buy can be written as: x≤7. Hence, we can't buy 8 pairs of socks.bThe unknown number x is multiplied by 80. Since $80 is the price for a pairs of sneakers, we know that x pairs of sneakers costs 80x. Also, 60 is added to the above expression. This has to be the amount we pay for socks. We can determine how many pairs of socks we can buy for $60 by dividing it by 12: 1260=5. The whole expression is less than or equal to 250 because this is the amount of money we have on our gift card. Therefore, this inequality represents the number of pairs of sneakers we can buy when we buy 5 pairs of socks. | |

Exercises 8 aStudent A claims that the equation will always have one solution. To support this claim, we have to select values for a, b, c and d such that x doesn't cancel out of the equation when simplifying. This necessitates that a and c are different numbers. For example, let's suppose that a=-1 and c=2. Now we can choose any values for b and d. We will choose: b=1 and d=0. Let's substitute those values in our equation and make sure we have solved the equation correctly. ax+b=cx+dSubstitute values-1x+1=2x+0 Solve for x LHS−1=RHS−1-1x=2x−1LHS−2x=RHS−2x-3x=-1Change signs3x=1LHS/3=RHS/3 x=31 There is one solution to this equation, so our selection of the values was correct.bStudent B claims that the equation will always have no solution. This time, we have to select values for a, b, c and d such that x cancels out of the equation and leaves us with a contradiction. This means a and c have to be the same numbers. We can choose a=c=3. After x cancels out, we are left with the equation b=d. Since we want a contradiction, those numbers have to be different, so let's take: b=4 and d=5. Let's substitute those values in the equation and check if we made appropriate selections. ax+b=cx+dSubstitute values3x+4=3x+5LHS−3x=RHS−3x4≠5 Yes, we are correct.cFinally, we want both students to be incorrect. This means that we want the equation to have infinitely many solutions, we want it to be an identity. To create an identity, a and c should be equal and b and d should be equal. We can choose a=c=6 and b=d=1. Again, let's substitute those values in the equation and check if we made appropriate selections. ax+b=cx+dSubstitute values6x+1=6x+1LHS−6x=RHS−6x1=1 We got an identity, our equation has infinitely many solutions. Now both students are incorrect. |

##### Other subchapters in Solving Linear Inequalities

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Writing and Graphing Inequalities
- Solving Inequalities Using Addition or Subtraction
- Solving Inequalities Using Multiplication or Division
- Solving Multi-Step Inequalities
- Quiz
- Solving Compound Inequalities
- Solving Absolute Value Inequalities
- Chapter Review
- Chapter Test