Simultaneous Equations: Finding the Meeting Points (0606 Additional Maths)
Hello future A-Maths expert! This chapter on Simultaneous Equations is about finding values that satisfy two or more equations at the same time. Think of each equation as a path on a map; solving them simultaneously means finding the exact spot (the coordinates!) where those paths intersect.
In basic IGCSE Maths, you mainly dealt with two straight lines. In Additional Maths, we step up the game! We often solve systems involving a straight line and a curve, or sometimes two curves, which requires careful algebraic manipulation.
1. The Concept: What Are We Solving For?
A simultaneous equation system in two unknowns (usually \(x\) and \(y\)) is a set of two equations that must both be true for the same pair of \(x\) and \(y\) values.
Visualizing the Solution:
If you graph the two equations on the same axes, the solution is simply the point(s) of intersection.
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Two straight lines: Usually one solution (one intersection point).
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A line and a parabola (Quadratic curve): Can have two solutions, one solution (tangent), or no solutions.
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Two complex curves: Can have multiple solutions, depending on their shapes.
Key Takeaway: When solving, we are looking for the coordinate pair \((x, y)\) that makes both equations true.
2. Prerequisite Review: Linear Systems
While you learned Elimination and Substitution in core Maths, Substitution is the technique you will rely on almost exclusively in A-Maths simultaneous equations.
Review: The Substitution Method
The substitution method works by isolating one variable in one equation and replacing it in the second equation, reducing the system to a single equation with one variable.
Example:
Equation 1: \(x + 2y = 10\)
Equation 2: \(3x - y = 9\)
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Isolate: Rearrange Equation 1 to make \(x\) the subject: \(x = 10 - 2y\).
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Substitute: Replace \(x\) in Equation 2 with \((10 - 2y)\): \(3(10 - 2y) - y = 9\).
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Solve: Simplify and solve for \(y\). \(30 - 6y - y = 9 \implies 30 - 7y = 9 \implies 21 = 7y\), so \(y = 3\).
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Find the other variable: Substitute \(y=3\) back into the rearranged equation: \(x = 10 - 2(3) = 4\).
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Solution: \((x, y) = (4, 3)\).
Memory Aid: The Substitution Chain
Isolate \(\rightarrow\) Substitute \(\rightarrow\) Solve (for 1 variable) \(\rightarrow\) Substitute back (for the 2nd variable) \(\rightarrow\) Check!
3. The A-Maths Challenge: Linear and Non-Linear
This is the most common type of simultaneous equation in the 0606 syllabus. You will have one linear equation (degree 1) and one non-linear equation (often a quadratic, degree 2).
The Golden Rule: ALWAYS use Substitution. Attempting elimination usually fails because of the squared terms or multiplied terms (\(xy\)) in the non-linear equation.
Step-by-Step Guide for Linear-Quadratic Systems
Let's use the syllabus example:
(1) \(y - x + 3 = 0\)
(2) \(x^2 - 3xy + y^2 + 19 = 0\)
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Make a subject from the Linear Equation (1).
It's easiest to isolate \(y\):
\(y = x - 3\) -
Substitute this expression into the Non-Linear Equation (2).
Replace every instance of \(y\) in (2) with \((x - 3)\).
\(x^2 - 3x(x - 3) + (x - 3)^2 + 19 = 0\) -
Expand and Simplify to form a standard Quadratic Equation.
Be careful with expanding brackets, especially the squared term!
\((x - 3)^2 = x^2 - 6x + 9\)
The equation becomes:
\(x^2 - (3x^2 - 9x) + (x^2 - 6x + 9) + 19 = 0\)
\(x^2 - 3x^2 + 9x + x^2 - 6x + 9 + 19 = 0\)
Collect like terms (\(x^2\), \(x\), and constants):
\((1 - 3 + 1)x^2 + (9 - 6)x + (9 + 19) = 0\)
\(-x^2 + 3x + 28 = 0\) -
Solve the resulting Quadratic Equation.
Multiply by \(-1\) to make the leading coefficient positive: \(x^2 - 3x - 28 = 0\)
(Using factorization: \((x - 7)(x + 4) = 0\))
Solutions for \(x\): \(x = 7\) or \(x = -4\). -
Find the corresponding \(y\) values.
You must substitute *both* \(x\) values back into the simple linear equation \(y = x - 3\).
If \(x = 7\): \(y = 7 - 3 = 4\). Solution 1: \((7, 4)\).
If \(x = -4\): \(y = -4 - 3 = -7\). Solution 2: \((-4, -7)\).
!!! Common Mistake Alert !!!
Students often forget to find the second variable, \(y\), or they substitute the \(x\) values back into the wrong equation. Always use the simplest linear rearrangement (Step 1) to find the \(y\) values quickly and accurately.
Key Takeaway: Linear/Non-Linear systems produce a single-variable quadratic equation, leading to two possible pairs of solutions.
4. Advanced Non-Linear Systems (A-Maths Special Cases)
Sometimes, you are given two equations that are both non-linear, and they require a bit of clever algebraic work first.
Case 4.1: Multiplicative Terms (\(xy\))
Consider the syllabus example:
(1) \(xy^2 = 4\)
(2) \(xy = 3\)
We cannot easily isolate a single variable and create a quadratic using standard substitution because the powers are too high. However, look closely at how the terms relate!
We can rewrite (1) using the parts of (2):
\(xy^2\) is the same as \((xy) \times y\).
Step-by-Step Solution:
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Rewrite (1) using parts of (2):
\( (xy) \times y = 4 \) -
Substitute the known value from (2):
We know \(xy = 3\), so substitute 3 into the new equation:
\( 3 \times y = 4 \)
\( y = \frac{4}{3} \) -
Find the corresponding \(x\) value:
Substitute \(y = \frac{4}{3}\) back into the simplest equation, (2):
\( x \left(\frac{4}{3}\right) = 3 \)
\( x = 3 \times \frac{3}{4} = \frac{9}{4} \) -
Solution: \(\left(\frac{9}{4}, \frac{4}{3}\right)\). (In this specific case, there was only one solution.)
Tip: Look for ways to rearrange equations into factors. If you see \(x^2y\) in one equation and \(xy\) in another, think: \(x^2y = x(xy)\).
Case 4.2: Fractional Systems
Sometimes, equations appear intimidating due to fractions, but substitution still works, leading to a standard quadratic after clearing denominators.
Example from syllabus requiring substitution and manipulation:
(1) \(\frac{2}{y} + \frac{1}{x} = 4\)
(2) \(y = x - 2\)
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Substitute (2) into (1):
Replace \(y\) in (1) with \((x - 2)\):
\(\frac{2}{x - 2} + \frac{1}{x} = 4\) -
Clear the denominators:
Multiply the entire equation by the Lowest Common Denominator (LCD), which is \(x(x - 2)\).
\( x(x - 2) \left(\frac{2}{x - 2}\right) + x(x - 2) \left(\frac{1}{x}\right) = 4 x(x - 2) \)
\( 2x + (x - 2) = 4x^2 - 8x \) -
Form and Solve the Quadratic:
\( 3x - 2 = 4x^2 - 8x \)
Move all terms to one side to set the equation to zero:
\( 0 = 4x^2 - 11x + 2 \) -
Solve for \(x\) (using formula or calculator/factorization):
This may require the quadratic formula (\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)). Find the two solutions for \(x\). -
Find the corresponding \(y\) values:
Use \(y = x - 2\) for each \(x\) solution.
Did you know?
The number of solutions you find often corresponds to the maximum number of times the two graphs can cross. A line and a quadratic curve can cross at most twice, which is why your substitution process usually yields a quadratic equation with two roots!
5. Essential Tips and Common Pitfalls
1. Always Check Your Answer Pairs
Once you find your solutions (e.g., \((4, 3)\) and \((2, 5)\)), quickly substitute them back into both original equations. If the left-hand side equals the right-hand side for both equations, your solution is correct. This is a crucial step for securing full marks!
2. Be FEARLESS when dealing with Fractions and Negatives
Mistake to Avoid: Forgetting to distribute the negative sign when substituting. If you substitute \(y = 3 - 2x\) into \(x^2 - 4y\), make sure you write \(x^2 - 4(3 - 2x)\), which is \(x^2 - 12 + 8x\).
3. Dealing with No Real Roots
If, after substitution, you end up with a quadratic equation like \(2x^2 + 5x + 10 = 0\), and the discriminant (\(b^2 - 4ac\)) is negative, there are no real solutions. This means the graphs do not intersect. You must state this conclusion clearly in your exam answer.
4. Organizing Your Work
Simultaneous equations can generate lots of algebraic lines. Use numbering (1) and (2) for your original equations and clearly label your substitution steps to avoid getting lost. This helps examiners follow your logic, securing you method marks even if a minor arithmetic error occurs.
Quick Review Box: The Simultaneous Workflow
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Goal: Find \((x, y)\) pairs.
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Method: Substitution (Isolate a variable from the linear/simpler equation).
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Key Result: A single-variable quadratic equation (\(ax^2 + bx + c = 0\)).
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Final Step: Solve the quadratic for two \(x\) values, then use the linear equation to find the two corresponding \(y\) values. Write answers as ordered pairs \((x_1, y_1)\) and \((x_2, y_2)\).