Welcome to Simultaneous Equations!
Welcome to one of the most rewarding and powerful topics in CCEA GCSE Further Mathematics (Unit 1: Pure Mathematics): Simultaneous Equations. In standard GCSE maths, you solved pairs of linear equations. In Further Mathematics, we level up! You will master solving systems with three linear equations (three unknowns) as well as pairs containing one linear and one non-linear equation (such as quadratics, circles, or hyperbolas).
Why are they called "simultaneous"? The word "simultaneous" means happening at the exact same time. We are looking for values of the variables that make all the given equations true at the exact same time.
Real-World Link: Engineers use simultaneous equations to calculate electrical currents across circuits. Business analysts use them to find break-even pricing points. Video game developers use them to detect when two objects collide on screen!
Topic 1: Linear Equations in Two Unknowns (Quick Refresher)
Before jumping into three variables or non-linear systems, let's briefly review the two main methods for solving two linear equations: Elimination and Substitution.
Method 1: Elimination
This method works best when equations are arranged in the standard form \(ax + by = c\). You multiply one or both equations by suitable numbers so that the coefficients of one variable become equal (or opposites), then add or subtract the equations to eliminate that variable.
Method 2: Substitution
This method is essential for non-linear equations later! You rearrange one equation to make one variable the subject (e.g., \(y = \dots\)), and substitute this expression into the other equation.
Quick Review Box:
• Same signs: Subtract the equations to eliminate the term.
• Different signs: Add the equations to eliminate the term.
• Memory Trick: SSS = Same Sign Subtract!
Topic 2: Linear Equations in Three Unknowns
In Further Mathematics, you will solve systems with three variables, typically \(x\), \(y\), and \(z\). To find unique values for three unknowns, you need three independent linear equations.
The Step-by-Step Strategy
Don't worry if this looks intimidating at first! The secret is simple: reduce the 3-variable problem into a familiar 2-variable problem.
Step 1: Label your equations (1), (2), and (3).
Step 2: Choose one variable that looks easiest to eliminate (for example, look for coefficients of \(1\) or \(-1\)).
Step 3: Pair equation (1) with equation (2) to eliminate that chosen variable. Label this new result equation (4).
Step 4: Pair equation (1) with equation (3) (or equation (2) with (3)) to eliminate the same variable. Label this new result equation (5).
Step 5: You now have equations (4) and (5), which only contain two variables! Solve this 2-variable system as normal.
Step 6: Substitute your two known values back into any of the original equations (1), (2), or (3) to find the final variable.
Step 7: Always check your complete set of answers in all three original equations!
Worked Example: Solving a 3-Variable System
Solve the following simultaneous equations:
(1) \(x + 2y - z = 6\)
(2) \(2x - y + z = 3\)
(3) \(3x + y + 2z = 13\)
Solution:
Step A: Choose a variable to eliminate.
Notice that the variable \(z\) has coefficients of \(-1\) in (1) and \(+1\) in (2). Adding (1) and (2) will eliminate \(z\) effortlessly!
Step B: Form Equation (4).
Add equation (1) and equation (2):
\((x + 2y - z) + (2x - y + z) = 6 + 3\)
\(3x + y = 9\) --- (4)
Step C: Form Equation (5).
We must eliminate \(z\) again using a different pair. Let's multiply equation (2) by \(2\), then subtract equation (3):
\(2 \times \text{(2)} \implies 4x - 2y + 2z = 6\)
Subtract equation (3):
\((4x - 2y + 2z) - (3x + y + 2z) = 6 - 13\)
\(x - 3y = -7\) --- (5)
Step D: Solve the 2-variable system of (4) and (5).
From (5), rearrange for \(x\):
\(x = 3y - 7\)
Substitute this into (4):
\(3(3y - 7) + y = 9\)
\(9y - 21 + y = 9\)
\(10y - 21 = 9\)
\(10y = 30 \implies y = 3\)
Now find \(x\) using \(x = 3y - 7\):
\(x = 3(3) - 7 = 9 - 7 \implies x = 2\)
Step E: Find the third variable (\(z\)).
Substitute \(x = 2\) and \(y = 3\) into equation (1):
\((2) + 2(3) - z = 6\)
\(2 + 6 - z = 6\)
\(8 - z = 6 \implies z = 2\)
Final Solution:
\(x = 2\), \(y = 3\), \(z = 2\)
Check in Equation (3): \(3(2) + 3 + 2(2) = 6 + 3 + 4 = 13\). It works!
Key Takeaway for 3 Unknowns
Eliminate the same variable twice using two different pairs of equations. This creates two equations with two unknowns, which you can easily solve.
Topic 3: One Linear and One Non-Linear Equation
In this section, you will solve pairs of equations where one equation is linear (contains only single powers like \(y = mx + c\) or \(ax + by = d\)) and the other is non-linear, such as:
• A parabola: \(y = ax^2 + bx + c\)
• A circle centered at the origin: \(x^2 + y^2 = r^2\)
• A reciprocal curve: \(xy = k\) or \(y = \frac{k}{x}\)
The Golden Rule: Always Use Substitution!
You cannot easily use elimination when powers do not match. Substitution is your best friend here.
Step 1: Rearrange the linear equation to make one variable the subject (e.g., \(y = \dots\) or \(x = \dots\)). Pick whichever variable has a coefficient of \(1\) or \(-1\) to avoid fractions where possible.
Step 2: Substitute this expression into the non-linear equation. Use brackets carefully!
Step 3: Expand and simplify to obtain a quadratic equation in one variable: \(ax^2 + bx + c = 0\).
Step 4: Solve the quadratic equation by factorising or using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
Step 5: Substitute each value back into the rearranged linear equation to find the corresponding values of the second variable.
Step 6: Present your solutions clearly as matching pairs \((x_1, y_1)\) and \((x_2, y_2)\).
Worked Example 1: Line and Parabola
Solve the simultaneous equations:
(1) \(y - 2x = 1\)
(2) \(y = x^2 - x - 3\)
Solution:
Step 1: Rearrange the linear equation (1) to make \(y\) the subject:
\(y = 2x + 1\)
Step 2: Substitute this expression for \(y\) into equation (2):
\(2x + 1 = x^2 - x - 3\)
Step 3: Rearrange into standard quadratic form (\(= 0\)):
\(0 = x^2 - x - 2x - 3 - 1\)
\(x^2 - 3x - 4 = 0\)
Step 4: Factorise the quadratic:
\((x - 4)(x + 1) = 0\)
So, \(x = 4\) or \(x = -1\)
Step 5: Find corresponding \(y\)-values using \(y = 2x + 1\):
When \(x = 4\): \(y = 2(4) + 1 = 8 + 1 = 9\)
When \(x = -1\): \(y = 2(-1) + 1 = -2 + 1 = -1\)
Final Solution:
\(x = 4, y = 9\) and \(x = -1, y = -1\)
As coordinates: \((4, 9)\) and \((-1, -1)\)
Worked Example 2: Line and Circle
Solve the simultaneous equations:
(1) \(x + y = 5\)
(2) \(x^2 + y^2 = 13\)
Solution:
Step 1: Make \(y\) the subject of the linear equation:
\(y = 5 - x\)
Step 2: Substitute into the circle equation:
\(x^2 + (5 - x)^2 = 13\)
Step 3: Carefully expand brackets and collect like terms:
Remember: \((5 - x)^2 = (5 - x)(5 - x) = 25 - 10x + x^2\)
\(x^2 + 25 - 10x + x^2 = 13\)
\(2x^2 - 10x + 25 - 13 = 0\)
\(2x^2 - 10x + 12 = 0\)
Divide the whole equation by \(2\) to simplify:
\(x^2 - 5x + 6 = 0\)
Step 4: Factorise:
\((x - 2)(x - 3) = 0\)
So, \(x = 2\) or \(x = 3\)
Step 5: Find the matching \(y\)-values using \(y = 5 - x\):
When \(x = 2\): \(y = 5 - 2 = 3\)
When \(x = 3\): \(y = 5 - 3 = 2\)
Final Solution:
\(x = 2, y = 3\) and \(x = 3, y = 2\)
As coordinates: \((2, 3)\) and \((3, 2)\)
Topic 4: Geometric Interpretation
Understanding simultaneous equations graphically gives you great insight and helps you check your answers visually!
• Two linear equations: Represent two straight lines. The solution is the single point where the two lines intersect.
• A line and a non-linear curve: The solutions represent the points of intersection between the line and the curve.
- Two distinct solutions: The line cuts through the curve at 2 points (a secant line).
- One repeated solution: The line touches the curve at exactly 1 point (the line is a tangent to the curve).
- No real solutions: The line does not meet the curve at all (the discriminant \(b^2 - 4ac < 0\)).
Did You Know? The condition for a straight line to be a tangent to a curve is that when you combine their equations into a quadratic \(ax^2 + bx + c = 0\), the discriminant must equal zero: \(b^2 - 4ac = 0\)!
Topic 5: Setting Up Equations from Real-Life Problems
Sometimes CCEA exam questions describe a scenario in words, and you must form the simultaneous equations before solving them.
Strategy for Word Problems:
1. Define variables clearly: Let \(x =\) price of an adult ticket, \(y =\) price of a child ticket, etc.
2. Translate words to algebra: "Total cost of 3 adult and 2 child tickets is £28" \(\implies 3x + 2y = 28\).
3. Solve using the standard algebraic methods.
4. State your final answer in context: Include units (e.g., £, cm, kg) and ensure the numbers make sense in the real world (e.g., lengths and prices cannot be negative).
Common Mistakes to Avoid
• Squaring a bracket incorrectly: Writing \((x - 3)^2\) as \(x^2 + 9\) instead of \(x^2 - 6x + 9\). Always write it as \((x - 3)(x - 3)\) and expand using FOIL.
• Mixing up matching pairs: Pairing \(x_1\) with \(y_2\). Always substitute \(x_1\) to find its matching \(y_1\), then substitute \(x_2\) to find its matching \(y_2\).
• Sign errors when subtracting equations: Be extra careful with double negatives, e.g., \(2x - (-3x) = 5x\).
• Substituting into the wrong equation: When solving linear-nonlinear pairs, substitute the found \(x\)-values back into the linear equation, not the quadratic, to avoid extra work or confusion.
Chapter Summary & Key Takeaways
• 3 Variables, 3 Equations: Eliminate one variable twice using two different pairs of equations to produce a 2-variable system, then solve.
• Linear + Non-Linear: Rearrange the linear equation first, substitute into the non-linear equation, and solve the resulting quadratic equation.
• Solutions as Points: The algebraic solutions represent the coordinates \((x, y)\) where graphs intersect.
• Check your work: Always plug your final values back into the original equations to confirm they are 100% correct!