Chapter 5: Simultaneous Equations
5.1 Linear Simultaneous Equations
Definition:
A system of linear simultaneous equations consists of two or more linear equations in two (or more) variables. For O Level Additional Mathematics, we focus on two equations in two unknowns.
General form:
\[
\begin{cases}
a_1 x + b_1 y = c_1 \\
a_2 x + b_2 y = c_2
\end{cases}
\]
Methods of solution:
| Method |
Procedure |
| Elimination |
Multiply equations to make coefficients of one variable equal, then add or subtract to eliminate that variable |
| Substitution |
Solve one equation for one variable, substitute into the other equation |
Proof: Why elimination works
Given:
\[
a_1 x + b_1 y = c_1 \quad \text{(1)}
\]
\[
a_2 x + b_2 y = c_2 \quad \text{(2)}
\]
Multiply (1) by \(b_2\) and (2) by \(b_1\):
\[
a_1 b_2 x + b_1 b_2 y = c_1 b_2
\]
\[
a_2 b_1 x + b_1 b_2 y = c_2 b_1
\]
Subtract:
\[
(a_1 b_2 - a_2 b_1)x = c_1 b_2 - c_2 b_1
\]
This gives \(x\) directly. The method works because adding or subtracting equal quantities from both sides preserves equality.
Proof: Why substitution works
If we solve (1) for \(x\): \(x = \dfrac{c_1 - b_1 y}{a_1}\) (assuming \(a_1 \neq 0\)).
Substituting this into (2) gives an equation in \(y\) only:
\[
a_2 \left(\frac{c_1 - b_1 y}{a_1}\right) + b_2 y = c_2
\]
This is a linear equation in \(y\), which can be solved. Substituting back gives \(x\). The method works because we are replacing equal quantities with equal quantities.
Worked Example 5.1 (Elimination method – non-calculator)
Solve:
\[
\begin{cases}
3x + 2y = 13 \\
5x - 2y = 11
\end{cases}
\]
Solution:
Add the two equations (\(2y\) and \(-2y\) cancel):
\[
(3x + 2y) + (5x - 2y) = 13 + 11
\]
\[
8x = 24 \Rightarrow x = 3
\]
Substitute \(x = 3\) into the first equation:
\[
3(3) + 2y = 13 \Rightarrow 9 + 2y = 13 \Rightarrow 2y = 4 \Rightarrow y = 2
\]
Answer: \(x = 3\), \(y = 2\)
Worked Example 5.2 (Substitution method – non-calculator)
Solve:
\[
\begin{cases}
2x - y = 5 \\
3x + 4y = 24
\end{cases}
\]
Solution:
From the first equation: \(y = 2x - 5\)
Substitute into the second:
\[
3x + 4(2x - 5) = 24
\]
\[
3x + 8x - 20 = 24
\]
\[
11x = 44 \Rightarrow x = 4
\]
Substitute \(x = 4\) into \(y = 2x - 5\):
\[
y = 2(4) - 5 = 8 - 5 = 3
\]
Answer: \(x = 4\), \(y = 3\)
Worked Example 5.3 (Mauritian context – shopping problem)
At a market in Port Louis, 2 kg of mangoes and 3 kg of lychees cost Rs 280.
5 kg of mangoes and 1 kg of lychees cost Rs 440.
Find the cost per kg of each fruit.
Solution:
Let \(x\) = cost per kg of mangoes, \(y\) = cost per kg of lychees.
\[
\begin{cases}
2x + 3y = 280 \\
5x + y = 440
\end{cases}
\]
From the second equation: \(y = 440 - 5x\)
Substitute into the first:
\[
2x + 3(440 - 5x) = 280
\]
\[
2x + 1320 - 15x = 280
\]
\[
-13x = -1040
\]
\[
x = 80
\]
Then \(y = 440 - 5(80) = 440 - 400 = 40\)
Answer: Mangoes cost Rs 80 per kg, lychees cost Rs 40 per kg.
5.2 One Linear and One Quadratic Equation
System:
\[
\begin{cases}
y = mx + c \quad \text{(linear)} \\
y = ax^2 + bx + c \quad \text{(quadratic)}
\end{cases}
\]
Method:
Substitute the linear equation into the quadratic equation to get a quadratic in \(x\):
\[
mx + c = ax^2 + bx + c \Rightarrow ax^2 + (b - m)x + (c - c) = 0
\]
Number of solutions (intersection points):
| Discriminant \(\Delta\) |
Number of intersection points |
| \(\Delta > 0\) |
Two points (line cuts curve) |
| \(\Delta = 0\) |
One point (line is tangent) |
| \(\Delta < 0\) |
No intersection (line misses curve) |
Proof: Discriminant determines intersection type
After substitution, we obtain a quadratic equation in \(x\). The solutions to this equation give the x-coordinates of the intersection points. The number of real solutions is determined by the discriminant:
- \(\Delta > 0\): two distinct real x-values \(\rightarrow\) two intersection points
- \(\Delta = 0\): one repeated x-value \(\rightarrow\) one intersection point (tangent)
- \(\Delta < 0\): no real x-values \(\rightarrow\) no intersection
This follows directly from the quadratic formula.
Worked Example 5.4 (Line and parabola – two intersections – non-calculator)
Solve:
\[
\begin{cases}
y = 2x + 1 \\
y = x^2 - 3x + 5
\end{cases}
\]
Solution:
Equate: \(2x + 1 = x^2 - 3x + 5\)
\[
0 = x^2 - 3x + 5 - 2x - 1
\]
\[
x^2 - 5x + 4 = 0
\]
\[
(x - 1)(x - 4) = 0
\]
So \(x = 1\) or \(x = 4\)
For \(x = 1\): \(y = 2(1) + 1 = 3 \Rightarrow (1, 3)\)
For \(x = 4\): \(y = 2(4) + 1 = 9 \Rightarrow (4, 9)\)
Answer: \((1, 3)\) and \((4, 9)\)
Worked Example 5.5 (Tangency – non-calculator)
Find the value of \(k\) such that the line \(y = 2x + k\) is tangent to the curve \(y = x^2 - x + 3\).
Solution:
Equate: \(2x + k = x^2 - x + 3\)
\[
0 = x^2 - x + 3 - 2x - k
\]
\[
x^2 - 3x + (3 - k) = 0
\]
For tangency, \(\Delta = 0\):
\[
(-3)^2 - 4(1)(3 - k) = 0
\]
\[
9 - 12 + 4k = 0
\]
\[
-3 + 4k = 0 \Rightarrow k = \frac{3}{4}
\]
Answer: \(k = \frac{3}{4}\)
Worked Example 5.6 (No intersection – finding condition)
Find the value of \(k\) for which the line \(y = x + k\) does not intersect the curve \(y = x^2 - 2x + 5\).
Solution:
Equate: \(x + k = x^2 - 2x + 5\)
\[
0 = x^2 - 2x + 5 - x - k
\]
\[
x^2 - 3x + (5 - k) = 0
\]
For no intersection, \(\Delta < 0\):
\[
(-3)^2 - 4(1)(5 - k) < 0
\]
\[
9 - 20 + 4k < 0
\]
\[
-11 + 4k < 0 \Rightarrow k < \frac{11}{4}
\]
Answer: \(k < \frac{11}{4}\)
5.3 Graphical Interpretation of Solutions
The solution to a system of simultaneous equations is the point(s) of intersection of their graphs.
| Type of system |
Graphical interpretation |
| Two linear equations |
Intersection of two straight lines |
| One linear, one quadratic |
Intersection of a line and a parabola |
Number of solutions:
- Two lines: usually one solution (if not parallel); no solution if parallel
- Line and quadratic: 0, 1, or 2 solutions (determined by discriminant)
5.4 Problems Leading to Simultaneous Equations
Key steps:
- Define variables for the unknown quantities
- Form equations from the information given
- Solve the system
- Interpret the answer in context
Worked Example 5.8 (Mauritian context – age problem)
A father is three times as old as his son. In 12 years' time, the father will be twice as old as the son. Find their current ages.
Solution:
Let \(f\) = father's current age, \(s\) = son's current age.
\[
\text{Equation 1: } f = 3s
\]
\[
\text{Equation 2: } f + 12 = 2(s + 12)
\]
Substitute \(f = 3s\) into equation 2:
\[
3s + 12 = 2s + 24
\]
\[
3s - 2s = 24 - 12
\]
\[
s = 12
\]
Then \(f = 3(12) = 36\)
Answer: Father is 36, son is 12.
Worked Example 5.9 (Mauritian context – mixture problem)
A shop sells two types of tea: Premium at Rs 120 per kg and Standard at Rs 80 per kg. A customer buys 5 kg of a mixture that costs Rs 460. How many kg of each type are in the mixture?
Solution:
Let \(p\) = kg of Premium, \(s\) = kg of Standard.
\[
\text{Equation 1 (total mass): } p + s = 5
\]
\[
\text{Equation 2 (total cost): } 120p + 80s = 460
\]
From equation 1: \(s = 5 - p\)
Substitute into equation 2:
\[
120p + 80(5 - p) = 460
\]
\[
120p + 400 - 80p = 460
\]
\[
40p = 60 \Rightarrow p = 1.5
\]
Then \(s = 5 - 1.5 = 3.5\)
Answer: 1.5 kg of Premium, 3.5 kg of Standard.
Chapter 5 Summary
| Concept |
Key points |
| Linear equations |
Elimination or substitution |
| Line and quadratic |
Substitute line into quadratic \(\rightarrow\) use discriminant |
| Number of solutions |
\(\Delta > 0\): two; \(\Delta = 0\): one (tangent); \(\Delta < 0\): none |
| Graphical interpretation |
Solutions are intersection points |
| Word problems |
Define variables, form equations, solve, interpret |
Exercises – Chapter 5
Easy (Drill – Non-Calculator)
- Solve:
(a) \(x + y = 7, x - y = 3\)
(b) \(2x + 3y = 12, 4x - y = 10\)
- Solve:
(a) \(y = 3x - 2, y = x^2\)
(b) \(y = x + 4, y = x^2 + 2\)
- Solve: \(x + 2y = 6, y = x^2 - 4x + 3\)
Medium (Examination Style)
- Find the points of intersection of the line \(y = 2x - 1\) and the curve \(y = x^2 - 3x + 4\).
- Find the value of \(k\) such that the line \(y = 3x + k\) is tangent to the curve \(y = x^2 + 2x + 1\).
- Mauritian context: A student buys some exercise books at Rs 25 each and some pens at Rs 15 each. The total number of items is 12 and the total cost is Rs 240. How many of each did the student buy?
- The line \(y = mx + 3\) intersects the curve \(y = x^2 + 2x + 5\) at two distinct points. Find the range of values of \(m\).
Hard (Challenge for A*)
- The line \(y = mx + c\) is tangent to the curve \(y = x^2 - 4x + 5\) at the point where \(x = 2\). Find \(m\) and \(c\).
- Mauritian context: A small business produces two products. The profit from product A is Rs 200 per unit and from product B is Rs 150 per unit. The total number of units produced is 50, and the total profit is Rs 8500.
(a) How many units of each product are produced?
(b) The business wants to increase total profit to Rs 9500 by reducing the number of product B units by 10 and increasing product A units. How many of each should they produce?
- The line \(y = 2x + c\) intersects the curve \(y = x^2 - 3x + 2\) at two points whose x-coordinates have a product of 3. Find the value of \(c\).