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Learning Resources

Linear Graphs & Simultaneous Eq

Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.

Form 2 Pathway: N/A

First Principles

Objective: Plot linear graphs, determine slopes, and solve simultaneous equations graphically and algebraically.

Interactive Linear Graph Explorer

Adjust the slope (\(m\)) and \(y\)-intercept (\(c\)) of the two lines to see how their point of intersection changes dynamically on the Cartesian plane.

Line 1: \(y = m_1x + c_1\)

Line 2: \(y = m_2x + c_2\)

Intersection: (2.00, 3.00)

(a) Concrete Scenario

Imagine two Kenyan entrepreneurs starting vegetable kiosks in Nairobi. Kiosk A starts with KSh 100 in reserve and earns a steady net profit of KSh 50 per day. Kiosk B starts with KSh 250 in capital but earns KSh 20 per day. Initially, Kiosk B has more total cash. However, because Kiosk A grows at a steeper rate, their total money eventually meets and Kiosk A takes the lead. The exact day and total cash value where they equal each other represents the point of intersection of two straight lines.

(b) Geometric Insight

Every straight line on a Cartesian plane is represented by the linear equation \(y = mx + c\), where \(m\) is the gradient (rate of change or steepness) and \(c\) is the \(y\)-intercept (initial value at \(x = 0\)).

  • Intersecting Lines: If the slopes are different (\(m_1 \neq m_2\)), the lines cross at exactly one unique point \((x, y)\). This point satisfies both equations at the same time.
  • Parallel Lines: If \(m_1 = m_2\) but \(c_1 \neq c_2\), the lines never meet (no solution).
  • Coincident Lines: If \(m_1 = m_2\) and \(c_1 = c_2\), the lines lie on top of each other (infinitely many solutions).

Key Concept: The slope \(m\) measures the vertical change (rise) per unit horizontal change (run): \[ m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} \]

(c) Algebraic Methods

We solve a system of two linear equations simultaneously using three primary approaches:

  • Graphical Method: Draw both lines accurately on squared graph paper and locate the intersection \((x, y)\).
  • Substitution Method: Express one variable in terms of the other from one equation, then substitute it into the other equation.
  • Elimination Method: Multiply one or both equations by suitable constants so that adding or subtracting eliminates one variable entirely.

Key Formulas

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \] — Gradient / Slope: The vertical change divided by the horizontal change between points \((x_1, y_1)\) and \((x_2, y_2)\).
\[ y = mx + c \] — Slope-Intercept Form: Where \(m\) is the gradient and \(c\) is the \(y\)-intercept.
\[ y - y_1 = m(x - x_1) \] — Point-Slope Form: Equation of a line through point \((x_1, y_1)\) with gradient \(m\).
\[ m_1x + c_1 = m_2x + c_2 \implies x = \frac{c_2 - c_1}{m_1 - m_2} \quad (m_1 \neq m_2) \] — Point of Intersection (x-coordinate): Equating \(y\)-values of two linear functions.
\[ \Delta = a_1 b_2 - a_2 b_1 \] — System Determinant: For linear system \(a_1x + b_1y = d_1\) and \(a_2x + b_2y = d_2\). If \(\Delta \neq 0\), a unique solution exists.
\[ x = \frac{d_1 b_2 - d_2 b_1}{a_1 b_2 - a_2 b_1}, \quad y = \frac{a_1 d_2 - a_2 d_1}{a_1 b_2 - a_2 b_1} \] — Algebraic Solution by Determinants (Cramer's Rule).

Worked Examples

Problem 1 (Easy): Find the point of intersection of the lines \(y = 2x + 1\) and \(y = -x + 4\).

Step-by-Step Solution:
  1. Equate the expressions for \(y\): \[ 2x + 1 = -x + 4 \]
  2. Collect like terms: Add \(x\) to both sides: \[ 2x + x + 1 = 4 \implies 3x + 1 = 4 \]
  3. Solve for \(x\): Subtract 1 from both sides and divide by 3: \[ 3x = 3 \implies x = 1 \]
  4. Calculate \(y\): Substitute \(x = 1\) into \(y = 2x + 1\): \[ y = 2(1) + 1 = 3 \]
  5. Verification: Check using the second equation: \(y = -(1) + 4 = 3\). Both match.
    Intersection Point: \((1, 3)\).
Problem 2 (Medium): Solve the simultaneous equations using the elimination method: \[ 3x + 2y = 12 \] \[ x - y = 1 \]

Step-by-Step Solution:
  1. Align coefficients: Multiply the second equation by 2 so the coefficients of \(y\) are additive inverses: \[ 2(x - y) = 2(1) \implies 2x - 2y = 2 \]
  2. Add equations to eliminate \(y\): \[ (3x + 2y) + (2x - 2y) = 12 + 2 \] \[ 5x = 14 \implies x = \frac{14}{5} = 2.8 \]
  3. Substitute to find \(y\): Use \(x - y = 1\): \[ \frac{14}{5} - y = 1 \implies y = \frac{14}{5} - 1 = \frac{9}{5} = 1.8 \]
  4. Conclusion: \((x, y) = \left(\frac{14}{5}, \frac{9}{5}\right)\) or \((2.8, 1.8)\).
Problem 3 (Hard): Solve the system algebraically using substitution and verify geometrically: \[ 2x - 3y = 7 \] \[ 4x + y = 1 \]

Step-by-Step Solution:
  1. Express \(y\) in terms of \(x\) from equation (2): \[ y = 1 - 4x \]
  2. Substitute into equation (1): \[ 2x - 3(1 - 4x) = 7 \]
  3. Expand brackets carefully: \[ 2x - 3 + 12x = 7 \] \[ 14x - 3 = 7 \implies 14x = 10 \implies x = \frac{10}{14} = \frac{5}{7} \]
  4. Find \(y\): Substitute \(x = \frac{5}{7}\) back into \(y = 1 - 4x\): \[ y = 1 - 4\left(\frac{5}{7}\right) = \frac{7}{7} - \frac{20}{7} = -\frac{13}{7} \]
  5. Final Solution: \((x, y) = \left(\frac{5}{7}, -\frac{13}{7}\right)\).

Common Mistakes

Mistake 1 Circular substitution — solving for \(y\) in Equation 1 and substituting it right back into Equation 1, resulting in \(0 = 0\).
Correction Always substitute the expressed variable into the other (unused) equation.
Why it feels right You solved an equation correctly, so plugging it back into the same expression feels intuitive, but it provides no new constraint.
Mistake 2 Identifying slope and intercept without converting to standard \(y = mx + c\) form.
Example: In \(2y = 6x + 8\), claiming the slope is 6 and intercept is 8.
Correction Divide every term by the coefficient of \(y\): \(y = 3x + 4\), so slope \(m = 3\) and intercept \(c = 4\).
Why it feels right The numbers 6 and 8 are explicitly written, so learners assume they represent \(m\) and \(c\) directly.
Mistake 3 Sign errors when subtracting simultaneous equations.
Example: Subtracting \((2x - 3y)\) from \((5x + y)\) and writing \(3x - 2y\) instead of \(3x + 4y\).
Correction Always distribute the negative sign: \(y - (-3y) = y + 3y = 4y\).
Why it feels right Minus signs easily get blended together during quick mental subtraction.
Mistake 4 Assuming parallel lines have a solution.
Correction If two lines have the same gradient \(m_1 = m_2\) and different intercepts \(c_1 \neq c_2\), they are parallel and will never intersect (no solution).

Real World

Boda Boda: Charges a flat entry fee of KSh 50 plus KSh 30 per kilometre: \[ C_1(x) = 30x + 50 \]
Tuk-tuk / Matatu: Charges a base boarding fee of KSh 110 plus KSh 15 per kilometre: \[ C_2(x) = 15x + 110 \]
For trips shorter than 4 km, the Boda Boda is cheaper.
At exactly 4 km, both options cost \(30(4) + 50 = \text{KSh } 170\).
For long trips beyond 4 km, the Matatu is cheaper.

Practice

A straight line passes through the points (2, 5) and (6, 13). What is the slope of the line? (Type only the number, e.g., 2)
Review the concepts above.
What is the slope of the line that passes through the points (-2, 5) and (3, -10)? (Type only the number, e.g., -3)
Review the concepts above.
The line through points (2, 5) and (k, 9) has a slope of 2. What is the value of k? (Type only the number, e.g., 4)
Review the concepts above.
The lines y = 2x + 3 and y = -x + k intersect at a point where x = 1. Find the value of k. (Type only the number, e.g., 6)
Review the concepts above.
The system of equations is 2x - 3y = 13 and 4x + y = 5. What is the value of x? (Type only the number, e.g., 2)
Review the concepts above.
Find the x-coordinate of the point where the lines 2x + 3y = 9 and 4x - y = 11 intersect. (Type only the number, e.g., 3)
Review the concepts above.