Topic 8: Linear Equations and Inequalities

🟢 Lite — Quick Review (1h–1d)

A linear equation is an equation in which the highest power of any variable is 1, producing a straight line when graphed. The general form of a linear equation in one variable is ax + b = 0, where a ≠ 0. Solving such equations involves finding the value of x that makes the equation true. A linear inequality, such as 3x - 7 > 5, similarly relates expressions but with inequality signs (<, >, ≤, ≥) rather than an equals sign. The critical distinction is that while a linear equation typically has exactly one solution, a linear inequality has a range of solutions satisfying the condition.

The process of solving linear equations follows a fundamental principle: whatever operation you perform on one side of the equation, you must perform on the other side to maintain equality. This is the “balance method”—imagine a balance scale where both sides must remain level. For inequalities, a similar principle applies, but with one crucial exception: when multiplying or dividing both sides by a negative number, the inequality sign reverses direction. For instance, if -2x > 8, dividing both sides by -2 gives x < -4 (notice the sign reversal).

Key Facts:

• Linear equation: highest power of variable is 1 (degree = 1)
• To solve ax + b = c: subtract b from both sides, then divide by a
• Linear inequality signs: < (less than), > (greater than), ≤ (less than or equal), ≥ (greater than or equal)
• When multiplying/dividing inequality by a negative: reverse the inequality sign
• Solution of inequality: expressed as a range, e.g., x > 3 or -2 ≤ x < 5
• Graphical representation: linear inequalities shade regions on coordinate axes

⚡ Exam Tip: In WASSCE questions with fractional coefficients, multiply the entire equation by the LCM of denominators to clear fractions first—this prevents sign errors.

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🟡 Standard — Regular Study (2d–2mo)

Solving Linear Equations

Solve: 4(2x - 3) = 3(x + 4) + 7

Step 1 — Expand brackets: 8x - 12 = 3x + 12 + 7

Step 2 — Simplify right side: 8x - 12 = 3x + 19

Step 3 — Collect variable terms on left: 8x - 3x = 19 + 12

Step 4 — Simplify: 5x = 31

Step 5 — Divide by coefficient: x = 31/5 = 6.2

Equations with Fractions

Solve: (2x + 3)/4 - (x - 1)/3 = 5/6

Step 1 — Find LCM of denominators (4, 3, 6) = 12 Step 2 — Multiply entire equation by 12: 12 × [(2x + 3)/4] - 12 × [(x - 1)/3] = 12 × (5/6) 3(2x + 3) - 4(x - 1) = 10

Step 3 — Expand: 6x + 9 - 4x + 4 = 10

Step 4 — Simplify: 2x + 13 = 10

Step 5 — Solve: 2x = -3 → x = -3/2 = -1.5

Linear Inequalities

Solve and graph: 5 - 2x ≤ 3(x - 2) + 4

5 - 2x ≤ 3x - 6 + 4 5 - 2x ≤ 3x - 2 5 + 2 ≤ 3x + 2x 7 ≤ 5x x ≥ 7/5 or x ≥ 1.4

On a number line: closed circle at 1.4, shading to the right.

Inequalities with Negative Coefficients

Solve: 7 - 3x < 2x - 8

7 + 8 < 2x + 3x 15 < 5x 3 < x or x > 3

(Notice we did NOT reverse the sign because we added 3x to both sides, not divide.)

Solve: -4x + 3 ≥ 2x - 9

3 + 9 ≥ 2x + 4x 12 ≥ 6x 2 ≥ x or x ≤ 2

(Here we divided by positive 6, so no sign reversal. But check: if we divided by -6, we’d get -2 ≤ x, which is equivalent.)

Comparison Table: Equation vs Inequality Rules

Operation	Equation: Result	Inequality: Result
Add same value to both sides	Equality preserved	Inequality preserved
Subtract same value	Equality preserved	Inequality preserved
Multiply both sides by positive	Equality preserved	Inequality preserved
Divide both sides by positive	Equality preserved	Inequality preserved
Multiply both sides by negative	N/A	Sign reverses
Divide both sides by negative	N/A	Sign reverses

Common Mistakes to Avoid:

1. Forgetting to reverse the inequality sign when multiplying/dividing by negatives
2. Making arithmetic errors when clearing fractions
3. Losing track of negative signs during expansion
4. Incorrectly identifying the direction of inequality after solving
5. Forgetting to express answers in simplest form

Problem-Solving Strategy:

1. Simplify both sides of the inequality independently
2. Collect variable terms on one side (usually left)
3. Collect constant terms on the other side
4. Divide by the coefficient of the variable, remembering to reverse if negative
5. Express answer in simplest form and represent on a number line if required
6. Check by substitution: pick a value within your solution range and verify it satisfies the original inequality

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🔴 Extended — Deep Study (3mo+)

Simultaneous Linear Equations

Solving two equations with two unknowns requires finding values of x and y that satisfy both equations simultaneously.

Method 1 — Substitution: Given: y = 2x + 3 … (1) 3x + y = 18 … (2)

Substitute (1) into (2): 3x + (2x + 3) = 18 5x = 15 x = 3

Substitute back: y = 2(3) + 3 = 9 Solution: (x, y) = (3, 9)

Method 2 — Elimination: Given: 2x + y = 8 … (1) x - y = 2 … (2)

Add (1) and (2): 3x = 10 → x = 10/3 Substitute: 10/3 - y = 2 → y = 10/3 - 2 = 4/3 Solution: (10/3, 4/3)

Method 3 — Graphical: The solution is the point where the two lines intersect.

Word Problems Leading to Linear Equations

“In a mathematics test, Ama scored twice as many marks as Kofi. Together they scored 99 marks. Find Kofi’s marks.”

Let Kofi’s marks = k Ama’s marks = 2k Total: k + 2k = 99 3k = 99 → k = 33

Linear Inequalities in Two Variables

A linear inequality in two variables, such as y ≤ 2x + 1, represents a region on the coordinate plane. The boundary line y = 2x + 1 is drawn as a solid line (for ≤ or ≥) or dashed line (for < or >). The region satisfying the inequality is then shaded.

To determine which side to shade:

1. Choose a test point not on the boundary (the origin (0,0) is convenient)
2. Substitute into the inequality
3. If the inequality is satisfied, shade the side containing the test point; otherwise, shade the opposite side

For y ≤ 2x + 1 at point (0,0): 0 ≤ 2(0) + 1 = 1 → TRUE → shade the region containing (0,0)

Systems of Linear Inequalities

Finding the solution region for multiple inequalities simultaneously:

• Graph each inequality separately
• The solution is the region where ALL inequalities overlap
• Vertices of this region are found by solving pairs of boundary equations

Absolute Value Equations and Inequalities

|x - 3| = 5 means x - 3 = 5 or x - 3 = -5 So x = 8 or x = 2

|x + 2| < 4 means -4 < x + 2 < 4 Subtracting 2: -6 < x < 2

|x - 1| ≥ 3 means x - 1 ≤ -3 or x - 1 ≥ 3 So x ≤ -2 or x ≥ 4

WASSCE Examination Patterns:

The WASSCE quantitative reasoning paper typically includes:

1. Simple linear equation solving (Objective)
2. Linear equations with brackets and fractions (Objective and Theory)
3. Linear inequalities and solution sets (Objective)
4. Simultaneous linear equations in two unknowns (Theory)
5. Word problems requiring formulation as linear equations (Theory)

⚡ Pro Exam Tip: In the WASSCE Theory paper, always verify your solution by substituting back into the original equation(s). For simultaneous equations, if one equation is multiplied by a constant before adding/subtracting, clearly show this step. For inequality graphs, always indicate which region is the solution by arrows or shading.

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