Linear Equations & Inequalities
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Core Objective: Master the algebraic logic of solving linear equations and linear inequalities in one variable from first principles.
The Golden Rule of Balance
An algebraic equation is like a traditional balance scale at an open-air market in Gikomba or Karatina: whatever operation you perform on one pan (adding, subtracting, multiplying, or dividing), you must perform identically on the other pan to preserve equilibrium.
1. The Linear Nature: An equation is called linear because the unknown variable \(x\) is raised to the power of 1 (first degree). As a result, any change in \(x\) causes a strictly proportional, constant change in the value of the expression.
2. Inverses and Isolation: To find the unknown quantity, we undo operations in reverse order using additive inverses (subtracting to undo addition, adding to undo subtraction) and multiplicative inverses (dividing to undo multiplication, multiplying to undo division).
3. Inequalities & The Directional Flip: When dealing with inequalities (\(<, \le, >, \ge\)), multiplying or dividing both sides by a negative quantity reflects the values across zero on the number line, reversing their relative sizes and thus flipping the inequality sign.
Key Formulas
\[ax + b = c \implies ax = c - b \implies x = \frac{c - b}{a} \quad (a \neq 0)\]
\[ax + b = cx + d \implies ax - cx = d - b \implies x(a - c) = d - b \implies x = \frac{d - b}{a - c}\]
If \(k > 0\) and \(ax < b\), then \(x < \frac{b}{a}\).
If \(k < 0\) and \(-kx < b\), then \(x > \frac{b}{-k}\) (reversing the inequality sign is mandatory).
\[a \le x < b\] Represents the continuous set of real numbers greater than or equal to \(a\) and strictly less than \(b\).
Worked Examples
Example 1 (Easy): Basic Two-Step Linear Equation
Problem: Solve for \(x\): \[4x - 9 = 15\]
Step-by-step Solution:
- Add \(9\) to both sides to isolate the \(x\)-term: \[4x - 9 + 9 = 15 + 9 \implies 4x = 24\]
- Divide both sides by \(4\): \[x = \frac{24}{4} = 6\]
- Verification: \(4(6) - 9 = 24 - 9 = 15\). (Correct)
Final Answer: \(x = 6\)
Example 2 (Medium): Brackets and Variables on Both Sides
Problem: Solve for \(x\): \[3(x - 2) = 2x + 7\]
Step-by-step Solution:
- Expand the left-hand side bracket: \[3x - 6 = 2x + 7\]
- Subtract \(2x\) from both sides to collect variable terms on the left: \[3x - 2x - 6 = 7 \implies x - 6 = 7\]
- Add \(6\) to both sides: \[x = 7 + 6 = 13\]
- Verification: LHS = \(3(13 - 2) = 3(11) = 33\); RHS = \(2(13) + 7 = 26 + 7 = 33\). Both sides match.
Final Answer: \(x = 13\)
Example 3 (Hard): Multi-Step Inequality with Sign Reversal
Problem: Solve the linear inequality: \[4 - 2x > x + 13\] and find the greatest integer value of \(x\).
Step-by-step Solution:
- Subtract \(x\) from both sides: \[4 - 3x > 13\]
- Subtract \(4\) from both sides: \[-3x > 13 - 4 \implies -3x > 9\]
- Divide both sides by \(-3\) and reverse the inequality sign: \[x < \frac{9}{-3} \implies x < -3\]
- The integers satisfying \(x < -3\) are \(\{-4, -5, -6, \dots\}\). The greatest among them is \(-4\).
Final Answer: \(x < -3\) (Greatest integer \(= -4\))
Common Mistakes
Mistake: Solving \(-2x > 6\) as \(x > -3\).
Correction: \(x < -3\).
Why it happens: In ordinary equations, dividing by \(-2\) leaves the equality sign unchanged (\(=\)). However, multiplying or dividing by a negative reverses relative size on the number line. For example, \(4 > 2\), but multiplying both sides by \(-1\) gives \(-4 < -2\).
Mistake: Treating \(\frac{1}{2}x\) as \(\frac{1}{2x}\).
Correction: \(\frac{1}{2}x = \frac{x}{2}\). To eliminate the \(\frac{1}{2}\), multiply both sides by \(2\), do NOT multiply by \(2x\).
Why it happens: Visual confusion occurs when handwriting fractions next to a letter variable.
Mistake: Expanding \(3(2x - 5)\) as \(6x - 5\).
Correction: \(3(2x - 5) = 3(2x) - 3(5) = 6x - 15\).
Why it happens: Students frequently multiply only the first term inside the bracket and forget to multiply the constant term by the outside multiplier.
Real World
Practice