Solving Simultaneous Equations
Finding where two equations meet
What are Simultaneous Equations?
Simultaneous equations are two or more equations that share the same variables and must be solved together. The solution is the set of values that satisfies ALL equations at the same time.
For two linear equations, the solution represents the point where the two lines intersect on a graph.
Methods of Solving Simultaneous Equations
There are three main methods for solving simultaneous equations:
1. Elimination Method
Add or subtract equations to eliminate one variable, then solve for the remaining variable.
Best when: Coefficients are easy to match
2. Substitution Method
Express one variable in terms of the other, then substitute into the second equation.
Best when: One equation has a coefficient of 1
Method 1: Elimination
Example 1: Elimination by Addition
Solve the simultaneous equations:
\[2x + 3y = 13 \quad \text{...(1)}\]
\[5x - 3y = 8 \quad \text{...(2)}\]
\[(2x + 3y) + (5x - 3y) = 13 + 8\]
\[7x = 21\]
\[x = \frac{21}{7} = 3\]
\[2(3) + 3y = 13\]
\[6 + 3y = 13\]
\[3y = 7\]
\[y = \frac{7}{3}\]
Example 2: Elimination by Making Coefficients Equal
Solve:
\[3x + 2y = 12 \quad \text{...(1)}\]
\[2x + 5y = 19 \quad \text{...(2)}\]
Equation (1) × 2: \(6x + 4y = 24\) ...(3)
Equation (2) × 3: \(6x + 15y = 57\) ...(4)
\[(6x + 15y) - (6x + 4y) = 57 - 24\]
\[11y = 33\]
\[y = 3\]
\[3x + 2(3) = 12\]
\[3x + 6 = 12\]
\[3x = 6\]
\[x = 2\]
Method 2: Substitution
Example 3: Substitution Method
Solve:
\[y = 2x + 1 \quad \text{...(1)}\]
\[3x + 2y = 12 \quad \text{...(2)}\]
\[3x + 2(2x + 1) = 12\]
\[3x + 4x + 2 = 12\]
\[7x + 2 = 12\]
\[7x = 10\]
\[x = \frac{10}{7}\]
\[y = 2\left(\frac{10}{7}\right) + 1 = \frac{20}{7} + \frac{7}{7} = \frac{27}{7}\]
Interactive Simultaneous Equations Solver
Graphical Solver
Enter coefficients for two linear equations in the form \(ax + by = c\)
Equation 1
Equation 2
Solution
Enter coefficients and click "Solve & Graph"
Word Problems
Example 4: Age Problem
Problem: John is 4 years older than Mary. In 6 years, the sum of their ages will be 40. Find their current ages.
Let \(j\) = John's current age
Let \(m\) = Mary's current age
\(j = m + 4\) ...(John is 4 years older)
\((j + 6) + (m + 6) = 40\) ...(sum in 6 years)
Simplifying: \(j + m = 28\) ...(2)
\((m + 4) + m = 28\]
\[2m + 4 = 28\]
\[2m = 24\]
\[m = 12\]
\[j = 12 + 4 = 16\]
Practice Problems
Question 1: Solve the simultaneous equations:
\[4x + 3y = 17\]
\[2x - y = 3\]
Show Solution
From equation (2): \(y = 2x - 3\)
Substitute into (1): \(4x + 3(2x - 3) = 17\)
\(4x + 6x - 9 = 17\)
\(10x = 26\), so \(x = 2.6\)
\(y = 2(2.6) - 3 = 2.2\)
Solution: x = 2.6, y = 2.2
Question 2: The sum of two numbers is 15 and their difference is 3. Find the numbers.
Show Solution
Let the numbers be \(x\) and \(y\).
\(x + y = 15\) ...(1)
\(x - y = 3\) ...(2)
Adding: \(2x = 18\), so \(x = 9\)
From (1): \(y = 15 - 9 = 6\)
The numbers are 9 and 6
Question 3: 3 books and 2 pens cost $21. 2 books and 5 pens cost $20. Find the cost of one book and one pen.
Show Solution
Let \(b\) = cost of a book, \(p\) = cost of a pen
\(3b + 2p = 21\) ...(1)
\(2b + 5p = 20\) ...(2)
Multiply (1) by 5: \(15b + 10p = 105\) ...(3)
Multiply (2) by 2: \(4b + 10p = 40\) ...(4)
Subtract (4) from (3): \(11b = 65\), so \(b = \frac{65}{11} \approx 5.91\)
From (1): \(p = \frac{21 - 3(5.91)}{2} \approx 1.64\)
Book: $5.91, Pen: $1.64
Key Tips for CSEC Exams
- Always label your equations (1), (2), etc.
- Show all working clearly - marks are given for method
- Check your answer by substituting back into BOTH original equations
- For word problems, define your variables clearly first
- Use elimination when coefficients are easy to match
- Use substitution when one equation is already solved for a variable