Linear Equations
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective: Solve linear equations involving brackets and fractions using algebraic balance.
Interactive Scale: Balance & Expand
Equation: 2(x + 3) = 14
The Golden Rule of Equations: An equation is an exact balance. Whatever mathematical operation you perform on one side, you must perform identically on the other side to maintain equilibrium.
1. Unpacking Brackets (The Distributive Law)
When terms are grouped in brackets such as \(a(x + b)\), think of them as bundled bags. To unpack the contents, distribute the multiplier outside to every term inside:
\[ a(x + b) = a \cdot x + a \cdot b = ax + ab \]2. Clearing Fractions (Multiplying by Common Denominators)
When an equation has fractions, such as \(\frac{x + 2}{3} = 5\), the division binds the numerator together. Multiply both sides by the denominator (or the LCM of all denominators) to eliminate fractions immediately:
\[ 3 \cdot \left(\frac{x + 2}{3}\right) = 3 \cdot 5 \implies x + 2 = 15 \]Key Formulas
\[ a(bx + c) = abx + ac \] \[ a(bx - c) = abx - ac \]
\[ \frac{ax + b}{c} = d \implies ax + b = c \cdot d \]
For \(\frac{x}{a} + \frac{x}{b} = c\), multiply every term by \(\text{LCM}(a, b)\): \[ \text{LCM}(a, b) \cdot \frac{x}{a} + \text{LCM}(a, b) \cdot \frac{x}{b} = \text{LCM}(a, b) \cdot c \]
\[ \frac{a(x + b)}{c} = \frac{d}{e} \implies e \cdot a(x + b) = c \cdot d \]
Worked Examples
Example 1 (Easy): Single Bracket Expansion
Solve for \(x\): \(4(x + 3) = 28\)
- Expand the bracket: Distribute \(4\) to both terms inside.
\[ 4 \cdot x + 4 \cdot 3 = 28 \implies 4x + 12 = 28 \] - Isolate the variable term: Subtract \(12\) from both sides.
\[ 4x = 28 - 12 \implies 4x = 16 \] - Solve for \(x\): Divide both sides by \(4\).
\[ x = \frac{16}{4} = 4 \]
Check: \(4(4 + 3) = 4(7) = 28\) (True).
Answer: \(x = 4\)
Example 2 (Medium): Brackets on Both Sides
Solve for \(y\): \(3(2y - 4) = 2(y + 6)\)
- Expand brackets on both sides:
\[ 6y - 12 = 2y + 12 \] - Collect variable terms on the left: Subtract \(2y\) from both sides.
\[ 6y - 2y - 12 = 12 \implies 4y - 12 = 12 \] - Collect constants on the right: Add \(12\) to both sides.
\[ 4y = 12 + 12 \implies 4y = 24 \] - Solve for \(y\): Divide by \(4\).
\[ y = \frac{24}{4} = 6 \]
Answer: \(y = 6\)
Example 3 (Hard): Fractions with Brackets
Solve for \(k\): \(\frac{3(k - 2)}{5} - \frac{k - 1}{2} = 1\)
- Find the LCM of denominators: \(\text{LCM}(5, 2) = 10\).
- Multiply every term by 10 to clear all fractions:
\[ 10 \cdot \frac{3(k - 2)}{5} - 10 \cdot \frac{k - 1}{2} = 10 \cdot 1 \] \[ 2 \cdot 3(k - 2) - 5(k - 1) = 10 \] \[ 6(k - 2) - 5(k - 1) = 10 \] - Carefully expand brackets (watch the negative sign!):
\[ 6k - 12 - 5k + 5 = 10 \] - Combine like terms:
\[ (6k - 5k) + (-12 + 5) = 10 \implies k - 7 = 10 \] - Isolate \(k\): Add 7 to both sides.
\[ k = 10 + 7 = 17 \]
Answer: \(k = 17\)
Common Mistakes
Misconception 1: Incomplete Expansion
\[ 3(x + 5) = 3 \cdot x + 3 \cdot 5 = 3x + 15 \]
Misconception 2: Negative Sign Errors with Fractions
Misconception 3: Forgetting to Multiply the Standalone Constant
\[ 3 \cdot \left(\frac{x}{3}\right) + 3 \cdot (2) = 3 \cdot (5) \implies x + 6 = 15 \implies x = 9 \]
Real World
1. Agribusiness: Bagging Maize in Kitale
A farmer packages maize into sacks. Each sack contains \(x\) kg of grade 1 maize and a bonus \(5\) kg of grade 2 maize. A distributor purchases \(8\) sacks weighing a total of \(760\) kg. The equation is:
\[ 8(x + 5) = 760 \implies 8x + 40 = 760 \implies 8x = 720 \implies x = 90\text{ kg} \]2. Matatu Fleet Revenue
A 14-seater matatu charges \(x\) KES per seat on a route. If passengers receive a subsidized discount voucher of \(30\) KES each, the operator receives \(14(x - 30)\) KES. If the total trip earnings equal \(2{,}380\) KES, finding the regular fare uses:
\[ 14(x - 30) = 2380 \implies x - 30 = 170 \implies x = 200\text{ KES} \]3. Sharing Electricity Tokens
Three roommates in Nairobi share a prepaid electricity bill equally after subtracting a fixed service charge of \(150\) KES from total units bought. If each person pays \(850\) KES, the total bill \(T\) is modelled by:
\[ \frac{T - 150}{3} = 850 \implies T - 150 = 2550 \implies T = 2700\text{ KES} \]Practice