Teach Me Maths

Linear Inequalities and Set Notation

Question 11 mark

Solve x+5<9.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Subtract the same number from both sides.

Worked solution

Solve x+5<9.

  1. Subtract 5 from both sides of the inequality.
  2. The solution is x<4.

Question 21 mark

Solve 3x\ge15.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Divide both sides by the positive coefficient of the unknown.

Worked solution

Solve 3x\ge15.

  1. Divide both sides by 3.
  2. Because 3 is positive, the inequality keeps its direction, giving x\ge5.

Question 32 marks

Solve 12-2x>6.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

When dividing by a negative number, reverse the inequality sign.

Worked solution

Solve 12-2x>6.

  1. Subtract 12 from both sides: -2x>-6.
  2. Divide by -2 and reverse the sign, giving x<3.

Question 42 marks

Solve 4x-7\le9.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Undo the subtraction before undoing the multiplication.

Worked solution

Solve 4x-7\le9.

  1. Add 7 to each side to obtain 4x\le16.
  2. Divide by 4: x\le4. The endpoint is included.

Question 52 marks

Solve 5x+2>2x+17.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Collect the unknown terms on one side and the constants on the other.

Worked solution

Solve 5x+2>2x+17.

  1. Subtract 2x and then 2 from each side to obtain 3x>15.
  2. Divide by 3, giving x>5.

Question 62 marks

Solve 2(x-3)\le10.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

You can divide by the number outside the bracket before removing the bracket.

Worked solution

Solve 2(x-3)\le10.

  1. Divide by 2: x-3\le5.
  2. Add 3, giving x\le8.

Question 72 marks

Write down the inequality shown on the number line.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

A filled circle means that number is included; an empty circle means it is not.

Worked solution

Write down the inequality shown on the number line.

  1. The line runs from -2 to 3, so x is between -2 and 3.
  2. The filled circle at -2 means -2 is included: -2\le x.
  3. The empty circle at 3 means 3 is not included: x<3.
  4. Together: -2\le x<3.

Question 82 marks

Solve \frac{x+2}{3}>4.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Multiply both sides by the positive denominator.

Worked solution

Solve \frac{x+2}{3}>4.

  1. Multiply by 3 to get x+2>12.
  2. Subtract 2, giving x>10.

Question 93 marks

Solve -2<3x+1\le10. Give your answer as one combined inequality.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Apply each operation to all three parts of the inequality.

Worked solution

Solve -2<3x+1\le10. Give your answer as one combined inequality.

  1. Subtract 1 throughout: -3<3x\le9.
  2. Divide all three parts by 3.
  3. The solution is -1<x\le3; the lower endpoint is excluded and the upper one is included.
    −1 < x ≤ 3A blue number-line segment between two boundary values. Open circles exclude a boundary and filled circles include it.−1 < x ≤ 3−13Open: excluded · Filled: included
    The open circle excludes the lower bound; the filled circle includes the upper bound.

Question 102 marks

A=\{x:x\text{ is an integer and }-3<x\le2\}. List all the members of A.

Give every value, separated by commas

Hint

Check which endpoints are included before listing the integers.

Worked solution

A=\{x:x\text{ is an integer and }-3<x\le2\}. List all the members of A.

  1. The strict lower inequality excludes -3, while the upper inequality includes 2.
  2. The integers between those bounds are -2,-1,0,1,2.
  3. So A=\{-2,-1,0,1,2\}.

Question 113 marks

Solve 4(2-x)>3x-6.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Expand the bracket carefully, keeping the negative coefficient of the unknown.

Worked solution

Solve 4(2-x)>3x-6.

  1. Expand to get 8-4x>3x-6.
  2. Add 4x and 6 to both sides, giving 14>7x.
  3. Divide by 7: x<2.

Question 123 marks

Solve \frac{2x-1}{3}\le\frac{x+4}{2}.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Multiply both sides by the lowest common multiple of the denominators.

Worked solution

Solve \frac{2x-1}{3}\le\frac{x+4}{2}.

  1. Multiply by 6, a positive number, to obtain 2(2x-1)\le3(x+4).
  2. Expand: 4x-2\le3x+12.
  3. Subtract 3x and add 2, giving x\le14.

Question 132 marks

A=\{x:x\ge-2\} and B=\{x:x<5\}. Which set is A\cap B?

Hint

An intersection contains only values that satisfy both conditions.

Worked solution

A=\{x:x\ge-2\} and B=\{x:x<5\}. Which set is A\cap B?

  1. \{x:x<-2\}
  2. \{x:-2\le x<5\}
  3. \{x:-2<x\le5\}
  4. \{x:x\ge5\}
  1. To belong to A, a value must be at least -2.
  2. To belong to B, it must also be less than 5.
  3. Combining these conditions gives A\cap B=\{x:-2\le x<5\}.
    Take the overlapThree number lines share the same scale: the first two conditions and their intersection.A: x ≥ −2−2B: x < 55A ∩ B: −2 ≤ x < 5−25Open: excluded · Filled: included
    The number lines share a scale. The final row keeps only the values allowed by both conditions.

Question 143 marks

The universal set is U=\{-4,-3,-2,-1,0,1,2,3,4,5,6\}. Let A=\{x\in U:2x+1>6\} and B=\{x\in U:x\le4\}. List the members of A\cap B.

Give every value, separated by commas

Hint

Solve the inequality for the first set, then keep values that also belong to the second.

Worked solution

The universal set is U=\{-4,-3,-2,-1,0,1,2,3,4,5,6\}. Let A=\{x\in U:2x+1>6\} and B=\{x\in U:x\le4\}. List the members of A\cap B.

  1. Solve 2x+1>6: subtract 1 to get 2x>5, then divide by 2: x>\frac52.
  2. So A is the members of U greater than 2.5: A=\{3,4,5,6\}.
  3. A\cap B means the members in both A and B, so they must also satisfy x\le4.
  4. Therefore A\cap B=\{3,4\}.

Question 153 marks

A ticket order costs £3 for booking and £4.50 for each ticket. A customer can spend at most £30. What is the greatest number of tickets they can buy?

Hint

Write an inequality including the one booking charge before finding the whole-number maximum.

Worked solution

A ticket order costs £3 for booking and £4.50 for each ticket. A customer can spend at most £30. What is the greatest number of tickets they can buy?

  1. For n tickets, the total cost is 3+4.5n, so 3+4.5n\le30.
  2. Subtract 3 and divide by 4.5 to obtain n\le6.
  3. The greatest possible whole number is 6 tickets.

Question 16Challenge4 marks

Find the values of x that satisfy both 2(3x-1)>x+8 and 5-x\ge1. Give your answer as one combined inequality.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Solve the two inequalities separately, then take the overlap.

Worked solution

Find the values of x that satisfy both 2(3x-1)>x+8 and 5-x\ge1. Give your answer as one combined inequality.

  1. The first inequality expands to 6x-2>x+8, giving 5x>10 and hence x>2.
  2. The second inequality gives -x\ge-4, so reversing the sign on division by -1 gives x\le4.
  3. Both conditions hold when 2<x\le4.
    Take the overlapThree number lines share the same scale: the first two conditions and their intersection.First: x > 22Second: x ≤ 44Both: 2 < x ≤ 424Open: excluded · Filled: included
    The number lines share a scale. The final row keeps only the values allowed by both conditions.

Question 17Challenge4 marks

Solve -5\le\frac{7-2x}{3}<3. Give your answer as one combined inequality.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Dividing by a negative reverses both inequality signs, after which you can write the bounds in increasing order.

Worked solution

Solve -5\le\frac{7-2x}{3}<3. Give your answer as one combined inequality.

  1. Multiply all three parts by 3: -15\le7-2x<9.
  2. Subtract 7 throughout: -22\le-2x<2.
  3. Divide by -2 and reverse both signs to get 11\ge x>-1.
  4. In increasing order, this is -1<x\le11.
    −1 < x ≤ 11A blue number-line segment between two boundary values. Open circles exclude a boundary and filled circles include it.−1 < x ≤ 11−111Open: excluded · Filled: included
    The open circle excludes the lower bound; the filled circle includes the upper bound.

Question 18Challenge3 marks

List all integer values of x satisfying -2\le\frac{3x-5}{4}<\frac74.

Give every value, separated by commas

Hint

Solve the combined inequality first, then list only its integer solutions.

Worked solution

List all integer values of x satisfying -2\le\frac{3x-5}{4}<\frac74.

  1. Multiply all three parts by 4: -8\le3x-5<7.
  2. Add 5 to all three parts: -3\le3x<12.
  3. Divide all three parts by 3: -1\le x<4.
  4. -1 is included and 4 is not, so the integers are -1,0,1,2,3.
    Keep the integer pointsThe interval includes minus1 and excludes4. The integer solutions are minus1,0,1,2,3.−1 ≤ x < 4−14Integer solutions−101234Filled dots: −1, 0, 1, 2, 3
    The continuous interval includes -1 and excludes 4. Keep only the filled integer dots.

Question 19Challenge4 marks

The universal set is U=\{-3,-2,-1,0,1,2,3,4,5,6,7,8\}. A=\{x\in U:19-3x\ge-2\} and B=\{x\in U:x<6\}. List the members of A\cap B^{\prime}.

Give every value, separated by commas

Hint

B^{\prime} means the members of U that are not in B.

Worked solution

The universal set is U=\{-3,-2,-1,0,1,2,3,4,5,6,7,8\}. A=\{x\in U:19-3x\ge-2\} and B=\{x\in U:x<6\}. List the members of A\cap B^{\prime}.

  1. Solve 19-3x\ge-2: subtract 19 from both sides to get -3x\ge-21.
  2. Divide by -3 and reverse the sign: x\le7. So A is every member of U except 8.
  3. B^{\prime} is the members of U that are not less than 6: B^{\prime}=\{6,7,8\}.
  4. A\cap B^{\prime} is the members in both. 8 is not in A, so A\cap B^{\prime}=\{6,7\}.
    A finite universal setRows show members of U, A, B complement, and the intersection. Filled dots are included in that row. The intersection contains6 and7 only.Use only the members of UU−3−2−1012345678A: x ≤ 7−3−2−1012345678B′: x ≥ 6−3−2−1012345678A ∩ B′−3−2−1012345678Filled dots are members of each set
    Work only within U. B^{\prime} retains 6,7,8, and intersecting with A removes 8.

Question 20Challenge4 marks

Phone plan A costs £9 plus 8 pence per minute. Plan B costs £17 plus 3 pence per minute. For a whole number of minutes, a customer wants A to cost less than B and A to cost no more than £21. What is the greatest possible number of minutes?

Hint

Write one inequality for each condition, using the same currency unit throughout.

Worked solution

Phone plan A costs £9 plus 8 pence per minute. Plan B costs £17 plus 3 pence per minute. For a whole number of minutes, a customer wants A to cost less than B and A to cost no more than £21. What is the greatest possible number of minutes?

  1. For m minutes, A costs 9+0.08m pounds and B costs 17+0.03m pounds.
  2. A being cheaper gives 9+0.08m<17+0.03m, so 0.05m<8 and m<160.
  3. The spending limit gives 9+0.08m\le21, so m\le150.
  4. Both conditions are satisfied up to 150 minutes. At this value A costs £21 and B costs £21.50, so the greatest number is 150.

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