Exercise practice: Linear Inequalities

3.1 How the exercises are laid out

Exercise 5.1 has four blocks. Questions 1 to 4 are one step inequalities solved separately over natural numbers and integers. Questions 5 to 16 are linear inequalities for real x. Questions 17 to 20 ask for the graph on a number line. Questions 21 to 26 are word problems about averages, consecutive numbers, a triangle and a board cut into pieces. The miscellaneous exercise adds double inequalities, systems of two inequalities and mixture problems.

This page works representative questions from each block. The method is the same each time: simplify, apply the two rules, read the universe, and write the answer in the requested form.

3.2 Exercise 5.1: one and two step inequalities

Question 1 asks for 24x less than 100 when x is natural. Dividing by the positive 24 gives x less than 25 by 6, so the natural solutions are 1, 2, 3 and 4. When x is an integer, the solutions are all integers less than 4.166, including the negative ones, so the set is {..., -2, -1, 0, 1, 2, 3, 4}.

Question 2 asks for minus 12x greater than 30. Dividing by minus 12 flips the sign and gives x less than minus 5 by 2, that is x less than minus 2.5. Over the natural numbers there is no solution, so the solution set is empty. Over the integers the solutions are -3, -4, -5 and so on, written as all integers at most minus 3.

Question 7 asks for 3(x minus 1) at most 2(x minus 3). Expanding gives 3x minus 3 at most 2x minus 6, so x at most minus 3. The solution set is the interval from minus infinity to minus 3, closed at the right.

Question 10 asks for x over 2 greater than x plus 1 over 3. Multiplying by the positive 6 gives 3x greater than 2x plus 2, so x greater than 2. The result is the open interval from 2 to infinity. Checking a value such as x equal to 3 in the original confirms the answer.

3.3 Exercise 5.1: graphical questions

Question 17 asks for 3x minus 2 less than 2x plus 1, which reduces to x less than 3. Draw a number line, put an open circle at 3, and shade to the left with an arrow. The open circle records the strict sign and the arrow shows the solution continues without bound.

Question 19 asks for 3(1 minus x) less than 2(x plus 4). Expanding gives 3 minus 3x less than 2x plus 8, so minus 5x less than 5, which after dividing by minus 5 gives x greater than minus 1. The graph is a line with an open circle at minus 1 and shading to the right.

Question 20 combines a difference of fractions with a comparison. Multiply by the positive common denominator to clear both fractions, collect the terms, and then draw the result. In each graphical question, the marks are for the correct boundary, the correct circle type and the correct direction.

3.4 Exercise 5.1: word problems

Question 21 gives marks 70 and 75 in two tests and asks for the least third score so that the average is at least 60. Let the third score be x. Then (70 plus 75 plus x) by 3 at least 60, which gives 145 plus x at least 180, so x at least 35. The least score is 35 marks.

Question 22 asks for grade A, which needs an average of at least 90 over five examinations. Sunita's first four marks are 87, 92, 94 and 95, whose total is 368. For an average of 90 over five papers the total must be at least 450, so the fifth mark must be at least 82.

Question 23 asks for consecutive odd positive integers below 10 whose sum is more than 11. Let the smaller be x and the other x plus 2, with x positive and x plus 2 less than 10. The sum condition gives 2x plus 2 greater than 11, so x greater than 4.5. The odd values satisfying both bounds are 5 and 7, giving the pairs (5, 7) and (7, 9).

Question 25 describes a triangle in which the longest side is triple the shortest, while the remaining side falls 2 cm short of the longest. Let the shortest be x, the longest 3x and the third 3x minus 2. The perimeter is at least 61, so 7x minus 2 is at least 61, giving x at least 9. The minimum shortest side is 9 cm. Check that the triangle inequality holds for the three sides before closing the answer.

3.5 Miscellaneous exercise

Question 1 asks for 2 at most 3x minus 4 at most 5. Add 4 to all three parts to get 6 at most 3x at most 9, then divide by 3 to get 2 at most x at most 3. The solution set is the closed interval [2, 3].

Question 2 asks for 6 at most minus 3(2x minus 4) less than 12. Divide all parts by minus 3 and flip both signs: minus 2 at least 2x minus 4 greater than minus 4. Add 4 and divide by 2 to get 1 less than x at most 2, which is the half open interval (1, 2].

Question 7 asks for the common part of 5x plus 1 greater than minus 24 and 5x minus 1 less than 24. The first gives x greater than minus 5 and the second gives x less than 5, so the solution is the open interval from minus 5 to 5.

Question 11 converts a Fahrenheit range to Celsius. From F equal to 9C by 5 plus 32, solve for C to get C equal to 5(F minus 32) by 9. With F between 68 and 77, C lies between 20 and 25. The range is 20 degrees Celsius to 25 degrees Celsius, with both endpoints included or not according to the wording of the question.

Question 12 is a mixture problem. An 8 percent boric acid solution of 640 litres is diluted with a 2 percent solution of x litres so that the mixture is more than 4 percent and less than 6 percent acid. The acid amount is 0.08 times 640 plus 0.02x, and the total volume is 640 plus x. The two inequalities give x between 320 and 1280 litres after solving.

3.6 Common slips in this exercise

Three errors repeat. Dividing by a negative without flipping the sign. Forgetting the universe word such as natural or integer, so an infinite real interval is written where a short list was required. Writing a strict inequality with a closed endpoint on the number line.

A fourth slip appears in word problems: reversing at least and at most. Read the sentence twice and underline the bound word before writing the inequality. When the answer is a range, state both ends with their circles or brackets, because the boundary treatment is part of the answer.

A final check on every solution is to substitute one value from the claimed set and one value outside it into the original inequality. If the first works and the second fails, the answer is almost certainly right.