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###### Exercises

Exercise name | Free? |
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Exercises 1 The phrase greater than or equal to is symbolically represented by the symbol ≥. We can write "a number z minus 6" as A number z minus 6 z − 6. Writing a complete inequality, we have z−6≥11. | |

Exercises 2 In this exercise, we are asked to translate a statement into symbolic form. To do this, we will break the given statement into pieces. The given statement is Twelve is no more than the sumof−1.5 times a number w and 4. No more than means less than or equal to. Thus, we so far have 12≤… -1.5 times a number w means we should multiply. Symbolically this gives -1.5w. Our inequality is now 12≤-1.5w… The sum of -1.5 times a number w and 4 means that we should add -1.5w and 4. Combining everything gives the following inequality. 12≤-1.5w+4 | |

Exercises 3 Since the circle at 0 is open, the graph describes all values less than but not equal to 0. In other words, we have a strict inequality and should exclude 0. If we let x represent these values, we can write the following inequality. x<0 | |

Exercises 4 Since the circle at 8 is closed, the graph describes all values greater than or equal to 8. In other words, we have an inequality which is not strict and should include 8. If we let x represent these values, we can write the following inequality. x≥8 | |

Exercises 5 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+q≤15LHS−9≤RHS−9q≤6 This inequality tells us that all values less than or equal 6 will satisfy the inequality. Notice that q can equal 6, which we show with a closed dot on the number line. | |

Exercises 6 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. z−(-7)<5a−(-b)=a+bz+7<5LHS−7<RHS−7z<-2 This inequality tells us that all values less than -2 will satisfy the inequality. Notice that z cannot equal -2, which we show with an open dot on the number line. | |

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. -3<y−4 LHS+4<RHS+4 LHS+4<RHS+4-3+4<y−4+4Add terms 1<yRearrange inequalityy>1 This inequality tells us that all values greater than 1 will satisfy the inequality. Notice that y 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. 3p≥18LHS/3≥RHS/3p≥6 This inequality tells us that all values greater than or equal to 6 will satisfy the inequality. Notice that p can equal 6, which we show with a closed circle 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>-2wMultiply by -2 and flip inequality sign-12<wRearrange inequalityw>-12 This inequality tells us that all values greater than -12 will satisfy the inequality. Notice that w cannot equal -12, which we show with an open circle on the number line. | |

Exercises 10 Inequalities can be solved in the same way as equations. By dividing both sides by -20, we can isolate x. Because we are dividing the inequality by a negative number, we have to reverse the inequality sign. -20x>5Divide by -20 and flip inequality signx<-205ba=b/5a/5x<-41 This inequality tells us that all values less than -41 will satisfy the inequality. Notice that x cannot equal -41, which we show with an open circle on the number line. | |

Exercises 11 Remember that solving inequalities is done in the same way as solving equations, using inverse operations. Just be careful to flip the inequality symbol when multiplying or dividing the inequality by a negative number. We need to isolate y to solve the inequality. 3y−7≥17LHS+7≥RHS+73y≥24LHS/3≤RHS/3y≥8 | |

Exercises 12 Inequalities are solved in the same way as equations, by isolating the variable. To begin solving the given inequality, we should distribute on both sides of the inequality and then move the variable terms to one side. 8(3g−2)≤12(2g+1)Distribute 824g−16≤12(2g+1)Distribute 1224g−16≤24g+24LHS−24g≤RHS−24g-16≤24 The statement -16≤24 is always true. Thus, the solution to the inequality is all real numbers. | |

Exercises 13 Remember that solving inequalities is done in the same way as solving equations, we just need to be careful to flip the inequality symbol if we multiply or divide the inequality by a negative number. To begin solving the given inequality, we should distribute on both sides of the inequality and then isolate the variable terms on one side. 6(2x−1)≥3(4x+1)Distribute 612x−6≥3(4x+1)Distribute 312x−6≥12x+3LHS−12x≥RHS−12x-6≱3 The statement -6≥3 is always false. Thus, the inequality has no solution. | |

Exercises 14 | |

Exercises 15 If the maximum volume of a pelican's bill is 700 cubic inches, the total volume inside of pelican's bill must be less than or equal to that. We can express this algebraically as: …≤700. The pelican has already scooped 100 cubic inches of water into its bill, so the sum of that and any additional volume must be less than or equal to 700. If we let v represent the additional volume the pelican can scoop, our inequality becomes: 100+v≤700. Solving this inequality for v will tell us how much more water the pelican can scoop. 100+v≤700 LHS−100≤RHS−100 LHS−100≤RHS−100100+v−100≤700−100Subtract terms v≤600 The pelican's bill can contain up to 600 additional cubic inches of water. | |

Exercises 16 |

##### 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
- Solving Compound Inequalities
- Solving Absolute Value Inequalities
- Chapter Review
- Chapter Test
- Cumulative Assessment