Linear Inequalities – SS1 Further Mathematics Lesson Note (First Term)

A linear inequality compares two expressions using <, >, ≤ or ≥, and it is solved like a linear equation except that multiplying or dividing by a negative number reverses the sign.

Behavioural objectives

By the end of the lesson, students should be able to:

1. solve linear inequalities in one variable and state the solution set;
2. solve quadratic inequalities in one variable by factorising and checking the signs of the factors;
3. use the parabola of a quadratic function to solve a quadratic inequality;
4. solve inequalities that contain absolute values.

What are the inequality symbols?

There are four inequality symbols, and each one compares two expressions.

How do you solve linear inequalities in one variable?

You solve a linear inequality by collecting the variable on one side, using the same rules as for a linear equation.

• Add or subtract: adding or subtracting the same number on both sides keeps the sign.
• Multiply or divide by a positive number: the sign stays the same.
• Multiply or divide by a negative number: the sign reverses. For example, -2x > 6 gives x < -3.

How do you show a solution on a number line?

• Open circle: used for < and >, because the end value is not included.
• Closed circle: used for ≤ and ≥, because the end value is included.
• Arrow direction: points right for “greater than” and left for “less than”.

Worked example

Find the solution set of each inequality and represent it on a number line.

(a) 2x – 3 < x + 7 (b) 3x + 4 > 1 – 2x

Solution to (a)

1. Start with 2x – 3 < x + 7.
2. Add 3 to both sides: 2x < x + 10.
3. Subtract x from both sides: x < 10.
4. Number line: open circle at 10, arrow pointing left.

Solution to (b)

1. Start with 3x + 4 > 1 – 2x.
2. Subtract 4 from both sides: 3x > -3 – 2x.
3. Add 2x to both sides: 5x > -3.
4. Divide both sides by 5: x > -3/5.
5. Number line: open circle at -3/5, arrow pointing right.

How do you solve quadratic inequalities in one variable?

A quadratic inequality has the form ax² + bx + c ≥ 0 or ax² + bx + c ≤ 0, and you solve it by factorising and checking the signs of the factors.

What are the sign rules for a product a.b?

What are the steps?

1. Move every term to one side so the other side is 0.
2. Factorise the quadratic expression.
3. Decide whether the product must be positive or negative.
4. Write the two cases for the signs of the factors.
5. Solve each case and reject any case with no common values.
6. State the solution set.

Worked example 1

Find the solution set of x² + x – 6 > 0.

1. Factorise: (x – 2)(x + 3) > 0.
2. The product is positive when both factors are positive or both are negative.
3. Both positive: x – 2 > 0 and x + 3 > 0, so x > 2 and x > -3. This gives x > 2.
4. Both negative: x – 2 < 0 and x + 3 < 0, so x < 2 and x < -3. This gives x < -3.
5. The solution set is x < -3 or x > 2.

Worked example 2

Show the solution set of x² + 3x – 4 ≤ 0.

1. Factorise: (x – 1)(x + 4) ≤ 0.
2. The product is negative or zero when the two factors have opposite signs, or one factor is zero.
3. First case: x – 1 ≤ 0 and x + 4 ≥ 0, so x ≤ 1 and x ≥ -4. This gives -4 ≤ x ≤ 1.
4. Second case: x – 1 ≥ 0 and x + 4 ≤ 0, so x ≥ 1 and x ≤ -4. No number satisfies both, so this case gives no solution.
5. The solution set is -4 ≤ x ≤ 1.
6. Number line: closed circles at -4 and 1, with the line drawn between them.

How do you use the graph of a quadratic function to solve an inequality?

The graph of f(x) = ax² + bx + c is a parabola, and the places where it crosses the x-axis show where the expression changes sign.

What does the discriminant tell you?

The discriminant is D = b² – 4ac, and it shows how the parabola meets the x-axis.

• D > 0: the parabola crosses the x-axis at two distinct points.
• D = 0: the parabola touches the x-axis at one point (equal roots).
• D < 0: the parabola does not meet the x-axis.

What are the steps for the parabola method?

1. Write the inequality with 0 on one side.
2. Solve the related equation to find where the curve cuts the x-axis.
3. Sketch the parabola. It opens upwards if a > 0 and downwards if a < 0.
4. For “≥ 0” or “> 0”, read the x-values where the curve is above the x-axis.
5. For “≤ 0” or “< 0”, read the x-values where the curve is below the x-axis.

Worked example 1

Determine the solution set of x² – x – 10 < 2.

1. Subtract 2 from both sides: x² – x – 12 < 0.
2. Factorise: (x + 3)(x – 4) < 0.
3. The curve cuts the x-axis where (x + 3)(x – 4) = 0, so x = -3 and x = 4.
4. The product is negative when one factor is positive and the other is negative.
5. Case 1: x + 3 < 0 and x – 4 > 0, so x < -3 and x > 4. No number satisfies both.
6. Case 2: x + 3 > 0 and x – 4 < 0, so x > -3 and x < 4.
7. The solution set is -3 < x < 4. On the graph, the curve is below the x-axis between -3 and 4.

Worked example 2

Find the solution of x² – 2x – 3 ≥ 0.

1. Factorise: (x + 1)(x – 3) ≥ 0.
2. The curve cuts the x-axis at x = -1 and x = 3.
3. Both factors positive or zero: x ≥ -1 and x ≥ 3. This gives x ≥ 3.
4. Both factors negative or zero: x ≤ -1 and x ≤ 3. This gives x ≤ -1.
5. The solution set is x ≤ -1 or x ≥ 3.

What is an absolute value?

The absolute value of a number x, written |x|, is its size without its sign, so it is never negative.

• If x ≥ 0: |x| = x. For example, |5| = 5.
• If x < 0: |x| = -x. For example, |-5| = 5.

How do you solve inequalities with absolute values?

• |x| ≥ a (a > 0): x ≥ a or x ≤ -a.
• |x| < a (a > 0): -a < x < a.
• |f(x)| < |g(x)|: square both sides, because both sides are never negative.

Worked example 1

Solve |2x – 3| ≥ 4.

1. First case: 2x – 3 ≥ 4.
2. Add 3: 2x ≥ 7, so x ≥ 7/2, that is x ≥ 3½.
3. Second case: -(2x – 3) ≥ 4.
4. Expand: -2x + 3 ≥ 4, so -2x ≥ 1.
5. Divide by -2 and reverse the sign: x ≤ -½.
6. The solution set is x ≤ -½ or x ≥ 3½.

Worked example 2

Find the solution set of |x – 2| < |x + 3|.

1. Square both sides: (x – 2)² < (x + 3)².
2. Expand: x² – 4x + 4 < x² + 6x + 9.
3. Cancel x² and collect terms: -4x – 6x < 9 – 4.
4. Simplify: -10x < 5.
5. Divide by -10 and reverse the sign: x > -5/10.
6. The solution set is x > -½.

Evaluation

1. State the rule that changes the sense of an inequality when you solve it.
2. When is the product of two numbers a and b positive? When is it negative?
3. What does the discriminant D tell you about where the parabola of f(x) = ax² + bx + c meets the x-axis?
4. What is the absolute value of a number?
5. Solve the inequality 3x – 5 > 4 – 2x.

Assignment

1. Find the range of values of x for which 7x – 12 ≥ x².
2. For what values of x is 2x² – 11x + 12 positive?
3. Find the values of x satisfying |3x – 2| ≥ 3|x – 1|.
4. Find the value of the constant k for which the equation 2x² + (k + 3)x + 2k = 0 has equal roots.