Solving linear inequalities Look at the following inequality, x + 3 ≥ 7 What values of x would make this inequality true? Any value of x greater or equal to 4 would solve this inequality. We could have solved this inequality as follows, x + 3 ≥ 7 When we talk about solving an inequality we are talking about finding the values of x that make the inequality true. Start by finding the solution by trial and improvement. For example, if x = 2 we have 2 + 3 ≥ 7. This is untrue and so x cannot be equal to 2. If x = 4 we have 4 + 3 ≥ 7. This is true and so x is a solution. Discuss the fact that any number less then 4 would not satisfy the inequality. Any number greater than 4 would satisfy the inequality and so the solution is x ≥ 4. Point out that when we solve an inequality we line up the inequality sign in the same way as solving an equation. x + 3 – 3 ≥ 7 – 3 subtract 3 from both sides: x ≥ 4 The solution has one letter on one side of the inequality sign and a number on the other.

Solving linear inequalities Like an equation, we can solve an inequality by adding or subtracting the same value to both sides of the inequality sign. We can also multiply or divide both sides of the inequality by a positive value. For example, Solve 4x – 7 > 11 – 2x add 7 to both sides: 4x > 18 – 2x add 2x to both sides: 6x > 18 divide both sides by 6: x > 3 How could we check this solution?

Checking solutions To verify that x > 3 is the solution to 4x – 7 > 11 – 2x substitute a value just above 3 into the inequality and then substitute a value just below 3. If we substitute x = 4 into the inequality we have 4 × 4 – 7 > 11 – 2 × 4 16 – 7 > 11 – 8 9 > 3 This is true. Explain that if x > 3 is the solution then any value of x above 3 will make the original inequality true. Any value of below 3 will make the inequality untrue. If we substitute x = 2 into the inequality we have, 4 × 2 – 7 > 11 – 2 × 2 8 – 7 > 11 – 4 1 > 7 This is not true.

Vertical regions We can represent all the points where the x-coordinate is equal to 2 with the line x = 2. The region where x > 2 does not include points where x = 2 and so we draw this as a dotted line. 1 2 3 4 5 6 –1 –2 –3 –4 –5 –6 y x x < 2 x > 2 The region to the right of the line x = 2 contains every point where x > 2. A common mistake is to think that the line x = 2 will be horizontal because the x-axis is horizontal. Show that the line x = 2 is as shown by asking pupils to tell you the coordinates of any point lying on the line. The region to the left of the line x = 2 contains every point where x < 2.

Horizontal regions The region where y ≤ 3 includes points where y = 3 and so we draw y = 3 as a solid line. 1 2 3 4 5 6 –1 –2 –3 –4 –5 –6 y x The region below the line y = 3 contains every point where y ≤ 3. y ≥ 3 The region above the line y = 3 contains every point where y ≥ 3. A common mistake is to think that the line y = 3 will be vertical because the y-axis is vertical. Show that the line y = 3 is as shown by asking pupils to tell you the coordinates of any point lying on the line. y ≤ 3

Inequalities in two variables We can represent all the points where the x-coordinate and the y-coordinate add up to 3 with the line x + y = 3. The region where x + y < 3 does not include points where x + y = 3 and so we draw this as a dotted line. 1 2 3 4 5 6 –1 –2 –3 –4 –5 –6 y x x + y > 3 The region below the line x + y = 3 contains every point where x + y < 3. x + y < 3 The region above the line x + y = 3 contains every point where x + y > 3.