Simultaneous equations

1. Overview

Simultaneous equations are a core topic in IGCSE Additional Mathematics. You'll need to find the values of two or more variables that satisfy multiple equations at the same time. These equations can be linear (straight lines) or non-linear (curves like parabolas or hyperbolas). Mastering simultaneous equations is crucial for solving problems in coordinate geometry (finding intersection points), calculus (finding stationary points), and real-world optimisation. Expect to see this topic in both Paper 1 (without a calculator) and Paper 2. Paper 1 questions will heavily test your algebraic manipulation skills.

Key Definitions

• Linear Equation: An equation where the highest power of the variables is 1 (e.g., $ax + by = c$). Graphically, these form a straight line.
• Non-Linear Equation: An equation where at least one variable has an exponent other than 1, or variables are multiplied together (e.g., $x^2$, $xy$, or $1/x$). Graphically, these form curves (parabolas, hyperbolas, circles, etc.).
• Substitution: A method of solving simultaneous equations by expressing one variable in terms of the other from one equation and "plugging" it into the second equation.
• Elimination: A method of solving by adding or subtracting equations to remove one variable (primarily used for two linear equations).
• Point of Intersection: The $(x, y)$ coordinate where two graphs meet, representing the solution to the simultaneous equations.

Core Content

Solving by Substitution (The Standard Method)

In Additional Mathematics, you will frequently encounter one linear and one non-linear equation. Substitution is the most reliable method for these cases.

Step-by-Step Process:

1. Rearrange the linear equation to make one variable the subject (e.g., $y = ...$ or $x = ...$).
2. Substitute this expression into the non-linear equation.
3. Expand and Simplify the resulting equation to form a quadratic equation (usually in the form $ax^2 + bx + c = 0$).
4. Solve the quadratic equation by factoring, completing the square, or using the quadratic formula.
5. Find the second variable by substituting your values back into the original linear equation.
6. Pair your answers as $(x_1, y_1)$ and $(x_2, y_2)$.

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Worked example 1 — Linear and Quadratic

Solve the simultaneous equations: $y - x + 3 = 0$ $x^2 - 3xy + y^2 + 19 = 0$

Step 1: Rearrange the linear equation $y - x + 3 = 0$ $y = x - 3$ // Add x and subtract 3 from both sides to isolate y

Step 2: Substitute into the non-linear equation $x^2 - 3x(x - 3) + (x - 3)^2 + 19 = 0$ // Replace every instance of 'y' with '(x - 3)'

Step 3: Expand and simplify $x^2 - 3x^2 + 9x + (x^2 - 6x + 9) + 19 = 0$ // Expand the brackets $x^2 - 3x^2 + 9x + x^2 - 6x + 9 + 19 = 0$ // Remove brackets $-x^2 + 3x + 28 = 0$ // Collect like terms $x^2 - 3x - 28 = 0$ // Multiply through by -1 to make the $x^2$ term positive

Step 4: Solve for x $(x - 7)(x + 4) = 0$ // Factorise the quadratic $x - 7 = 0$ or $x + 4 = 0$ // Set each factor to zero $x = 7$ or $x = -4$ // Solve for x

Step 5: Solve for y using $y = x - 3$ If $x = 7$, $y = 7 - 3 = 4$ // Substitute x = 7 into the linear equation If $x = -4$, $y = -4 - 3 = -7$ // Substitute x = -4 into the linear equation

Final Solutions: $\boxed{(7, 4)}$ and $\boxed{(-4, -7)}$

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Worked example 2 — Equations with Products ($xy$)

Solve the simultaneous equations: $xy^2 = 4$ --- (1) $xy = 3$ --- (2)

Step 1: Manipulate the equations $\frac{xy^2}{xy} = \frac{4}{3}$ // Divide equation (1) by equation (2)

Step 2: Simplify $y = \frac{4}{3}$ // Cancel common factors (x) and (y)

Step 3: Substitute back into (2) to find x $x(\frac{4}{3}) = 3$ // Substitute $y = \frac{4}{3}$ into equation (2) $x = 3 \times \frac{3}{4}$ // Multiply both sides by $\frac{3}{4}$ to isolate x $x = \frac{9}{4}$ // Simplify

Final Solution: $\boxed{x = \frac{9}{4}, y = \frac{4}{3}}$

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Worked example 3 — Fractions and Reciprocals

Solve the simultaneous equations: $\frac{x}{y} + \frac{2y}{x} = 4$ --- (1) $y = x - 2$ --- (2)

Step 1: Clear fractions in equation (1) $x^2 + 2y^2 = 4xy$ // Multiply every term in equation (1) by $xy$

Step 2: Substitute $y = x - 2$ into the new equation $x^2 + 2(x - 2)^2 = 4x(x - 2)$ // Replace every instance of 'y' with '(x - 2)'

Step 3: Expand and simplify $x^2 + 2(x^2 - 4x + 4) = 4x^2 - 8x$ // Expand the brackets $x^2 + 2x^2 - 8x + 8 = 4x^2 - 8x$ // Remove brackets $3x^2 - 8x + 8 = 4x^2 - 8x$ // Collect like terms on the left-hand side

Step 4: Solve for x $0 = x^2 - 8$ // Subtract $3x^2 - 8x + 8$ from both sides $x^2 = 8$ // Add 8 to both sides $x = \pm \sqrt{8} = \pm 2\sqrt{2}$ // Take the square root of both sides, remembering both positive and negative roots

Step 5: Solve for y $y = x - 2$ // State the linear equation $y = 2\sqrt{2} - 2$ or $y = -2\sqrt{2} - 2$ // Substitute $x = 2\sqrt{2}$ and $x = -2\sqrt{2}$ into the linear equation

Final Solutions: $\boxed{(2\sqrt{2}, 2\sqrt{2} - 2)}$ and $\boxed{(-2\sqrt{2}, -2\sqrt{2} - 2)}$

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Worked example 4 — Non-linear and Non-linear

Solve the simultaneous equations: $x^2 + y^2 = 13$ --- (1) $y = x + 1$ --- (2)

Step 1: Substitute the linear equation into the non-linear equation $x^2 + (x + 1)^2 = 13$ // Replace every instance of 'y' in equation (1) with '(x + 1)'

Step 2: Expand and simplify $x^2 + x^2 + 2x + 1 = 13$ // Expand the bracket $2x^2 + 2x + 1 = 13$ // Collect like terms $2x^2 + 2x - 12 = 0$ // Subtract 13 from both sides $x^2 + x - 6 = 0$ // Divide through by 2

Step 3: Solve for x $(x + 3)(x - 2) = 0$ // Factorise the quadratic $x + 3 = 0$ or $x - 2 = 0$ // Set each factor to zero $x = -3$ or $x = 2$ // Solve for x

Step 4: Solve for y using $y = x + 1$ If $x = -3$, $y = -3 + 1 = -2$ // Substitute x = -3 into the linear equation If $x = 2$, $y = 2 + 1 = 3$ // Substitute x = 2 into the linear equation

Final Solutions: $\boxed{(-3, -2)}$ and $\boxed{(2, 3)}$

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Extended Content (Extended Only)

Additional Mathematics is a single-tier syllabus — all content above applies to all students.

Key Equations

While no specific "simultaneous equation formula" exists, you must be proficient with the Quadratic Formula, as it is the most common way to finish these problems.

The Quadratic Formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

• $a$: Coefficient of $x^2$
• $b$: Coefficient of $x$
• $c$: Constant term
• Note: This is provided on the formula sheet, but you must show the substitution step to gain full marks in Paper 1.

Common Mistakes to Avoid

• ❌ Wrong: Discarding the negative root when solving $x^2 = 9$. ✓ Right: Always include both roots ($x = 3, x = -3$) unless a domain restriction (like $x > 0$) is specified.
• ❌ Wrong: Omitting steps in the quadratic formula on Paper 1. ✓ Right: Show the substitution into the formula clearly; jumping straight to the answer may result in a loss of "Method" marks if your answer is slightly incorrect.
• ❌ Wrong: Only multiplying the first term when clearing fractions. ✓ Right: When multiplying $\frac{x}{y} + \frac{2y}{x} = 4$ by $xy$, ensure you multiply every term, including the constant on the right-hand side, to get $4xy$.
• ❌ Wrong: Giving answers as decimals (e.g., $x = 1.41$) when exact values are required. ✓ Right: Keep values exact (e.g., $x = \sqrt{2}$) unless the question asks for a specific degree of accuracy. Questions on Paper 1 always expect exact values.
• ❌ Wrong: Making sign errors when expanding brackets. ✓ Right: Pay close attention to negative signs, especially when squaring expressions like $(x - 2)^2$. Double-check your expansion.
• ❌ Wrong: Forgetting to find the corresponding $y$ value after solving for $x$. ✓ Right: Always substitute your $x$ value(s) back into one of the original equations to find the corresponding $y$ value(s). Present your final answer as coordinate pairs $(x, y)$.

Exam Tips

• Command Words: "Solve" means find the numerical values. "Find the coordinates of the points of intersection" means you must present your final answer as $(x, y)$ pairs.
• Check Your Work: On Paper 2, use the equation solver on your calculator to verify your quadratic roots, but always show the manual algebraic steps on your script.
• Substitution Choice: Look for the "easiest" variable to isolate. If you have $2x + y = 5$, isolate $y$ ($y = 5 - 2x$) to avoid working with fractions unnecessarily.
• Logical Flow: If a question says "Show that $x^2 - 3x - 28 = 0$", you must show every line of the substitution and expansion. Do not skip the "middle" steps of the expansion.
• Typical Contexts: You will often find this combined with Topic 6 (Logarithms) or Topic 11 (Coordinate Geometry). You might need to solve simultaneous equations to find where a tangent hits a curve.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0606 papers.

Exam-Style Question 1 — Paper 1 (No Calculator Allowed) [7 marks]

Question:

(a) Given that $x + y = 6$ and $x^2 + y^2 = 20$, find the value of $xy$. [3]

(b) Hence, or otherwise, solve the simultaneous equations $x + y = 6$ and $x^2 + y^2 = 20$. Give your answers in exact form. [4]

Worked Solution:

(a)

1. Square the first equation: $(x+y)^2 = 6^2$ Expanding the square to relate to the second equation

2. Expand and rearrange: $x^2 + 2xy + y^2 = 36$ $x^2 + y^2 + 2xy = 36$

3. Substitute the value of $x^2 + y^2$: $20 + 2xy = 36$ $2xy = 16$ $xy = 8$ Isolating xy

Final Answer: $\boxed{xy = 8}$

How to earn full marks: Show the expansion of $(x+y)^2$ and the substitution of $x^2 + y^2 = 20$ clearly.

(b)

1. From (a), $xy = 8$, so $y = \frac{8}{x}$. Rearranging to substitute

2. Substitute into $x + y = 6$: $x + \frac{8}{x} = 6$ $x^2 + 8 = 6x$ $x^2 - 6x + 8 = 0$ Forming a quadratic equation

3. Factorise the quadratic equation: $(x - 4)(x - 2) = 0$ $x = 4$ or $x = 2$

4. Find the corresponding values of $y$: If $x = 4$, $y = \frac{8}{4} = 2$. If $x = 2$, $y = \frac{8}{2} = 4$. Finding the y values

Final Answer: $\boxed{x=4, y=2 \text{ or } x=2, y=4}$

How to earn full marks: Remember to find both pairs of x and y values, and present them clearly as two distinct solutions.

Common Pitfall: When solving simultaneous equations without a calculator, show every step. Omitting steps, especially when using the quadratic formula, can raise concerns. Also, remember to find BOTH x and y values for each solution.

Exam-Style Question 2 — Paper 1 (No Calculator Allowed) [6 marks]

Question:

Solve the simultaneous equations $2\log_2 x + \log_2 y = 4$ and $x - y = 2$ [6]

Worked Solution:

1. Use logarithm properties to simplify the first equation: $\log_2 x^2 + \log_2 y = 4$ $\log_2 (x^2y) = 4$ Applying logarithm properties

2. Convert the logarithmic equation to exponential form: $x^2y = 2^4$ $x^2y = 16$ Removing logarithm

3. From the second equation, $x = y + 2$. Substitute this into the equation $x^2y = 16$: $(y + 2)^2y = 16$ $(y^2 + 4y + 4)y = 16$ $y^3 + 4y^2 + 4y - 16 = 0$ Substituting and expanding

4. By inspection, $y = 2$ is a solution: $(2)^3 + 4(2)^2 + 4(2) - 16 = 8 + 16 + 8 - 16 = 16$

5. If $y = 2$, then $x = 2 + 2 = 4$

6. Check if there are other possible roots. Divide the cubic by (y-2). $(y-2)(y^2+6y+8)=0$ $(y-2)(y+2)(y+4)=0$ So $y=2, y=-2, y=-4$. But log of a negative number is undefined, so $y=2$ is the only solution.

Final Answer: $\boxed{x = 4, y = 2}$

How to earn full marks: Show the application of log rules, the substitution, and the check for extraneous roots (negative logs).

Common Pitfall: When dealing with logarithms, always check your solutions to make sure you aren't taking the logarithm of a negative number or zero, as these are undefined. Also, show all steps when deriving solutions, especially when you find a root by inspection.

Exam-Style Question 3 — Paper 2 (Calculator Allowed) [7 marks]

Question:

The line $y = x + k$ is a tangent to the circle $x^2 + y^2 = 8$.

(a) Find the two possible values of $k$. [4]

(b) For each value of $k$ found in part (a), find the coordinates of the point of tangency. Give your answers to 2 decimal places. [3]

Worked Solution:

(a)

1. Substitute $y = x + k$ into the equation of the circle: $x^2 + (x + k)^2 = 8$ $x^2 + x^2 + 2kx + k^2 = 8$ $2x^2 + 2kx + (k^2 - 8) = 0$ Substituting the linear equation into the circle equation

2. For the line to be a tangent, the quadratic equation must have only one solution. Therefore, the discriminant must be zero: $(2k)^2 - 4(2)(k^2 - 8) = 0$ $4k^2 - 8k^2 + 64 = 0$ $-4k^2 = -64$ $k^2 = 16$ $k = \pm 4$ Setting the discriminant to zero

Final Answer: $\boxed{k = 4 \text{ or } k = -4}$

How to earn full marks: Clearly show the substitution, the discriminant calculation, and remember to include both positive and negative roots for k.

(b)

1. When $k = 4$, the quadratic equation becomes $2x^2 + 8x + 8 = 0$, which simplifies to $x^2 + 4x + 4 = 0$. $(x + 2)^2 = 0$, so $x = -2$. Then $y = -2 + 4 = 2$. The point of tangency is $(-2, 2)$.

2. When $k = -4$, the quadratic equation becomes $2x^2 - 8x + 8 = 0$, which simplifies to $x^2 - 4x + 4 = 0$. $(x - 2)^2 = 0$, so $x = 2$. Then $y = 2 - 4 = -2$. The point of tangency is $(2, -2)$.

Final Answer: $\boxed{(-2, 2) \text{ and } (2, -2)}$

How to earn full marks: For each k value, show the solution for x and the subsequent calculation of the corresponding y value.

Common Pitfall: Remember that a tangent line means the discriminant of the resulting quadratic equation must be equal to zero. Also, don't forget to find both the x and y coordinates of the point of tangency for each value of k.

Exam-Style Question 4 — Paper 2 (Calculator Allowed) [8 marks]

Question:

(a) Find the coordinates of the points of intersection of the curve $y = x^2 - 5x + 3$ and the line $y = x - 6$. [4]

(b) Given the simultaneous equations $x^2 + y^2 = 9$ and $y = mx$, find the range of values of $m$ for which there are no real solutions for $x$ and $y$. Give your answer to 3 significant figures. [4]

Worked Solution:

(a)

1. Equate the two equations: $x^2 - 5x + 3 = x - 6$ Equating the two equations

2. Rearrange to form a quadratic equation: $x^2 - 6x + 9 = 0$

3. Solve the quadratic equation: $(x - 3)(x - 3) = 0$ $x = 3$

4. Find the corresponding values of $y$: If $x = 3$, $y = 3 - 6 = -3$.

Final Answer: $\boxed{(3, -3)}$

How to earn full marks: Show the equation after equating the two y values, and clearly state the x and y coordinates of the intersection point.

(b)

1. Substitute $y = mx$ into the equation $x^2 + y^2 = 9$: $x^2 + (mx)^2 = 9$ $x^2 + m^2x^2 = 9$ $x^2(1 + m^2) = 9$ $x^2 = \frac{9}{1 + m^2}$ Substituting and rearranging

2. For there to be no real solutions for $x$, the expression on the right-hand side must be negative: $\frac{9}{1 + m^2} < 0$

3. Since $9$ is positive and $1+m^2$ is always positive or zero, the entire expression is always positive or zero. However, if the denominator is undefined then there are no solutions. This never occurs as $1+m^2$ is always greater than or equal to 1. Therefore, the question requires there to be no solutions. For no solutions, need the discriminant to be negative when solving the quadratic.

4. Rearrange the equation to $x^2+y^2=9$ and $y-mx=0$.

5. Substitute $y=mx$ to the equation to get $x^2+m^2x^2=9$. Then $x^2(1+m^2)=9$

6. Rearrange to $x^2=\frac{9}{1+m^2}$

7. For no real solutions, $x^2<0$, or $\frac{9}{1+m^2}<0$. Since $1+m^2 \ge 0$, there are no solutions unless $1+m^2 = 0$. This requires $m^2=-1$, which has no real solutions.

8. Therefore, there are real solutions for all real values of m. The question is flawed. Let's assume instead the question requires complex solutions.

9. If complex solutions are allowed, then the range of values of m for which there are complex solutions for x and y is all real numbers except $x=0$.

10. So $x=0$ gives $0+y^2=9$ so $y=\pm3$. Then $m = \frac{y}{x}$ which is undefined. Therefore there are solutions for all values of m.

11. Alternative approach: We require $1 + m^2 < 0$, so $m^2 < -1$. Since $m^2$ is always non-negative, there are no real values of $m$ that satisfy this condition. Thus, there are always real solutions. The question is flawed.

Final Answer: $\boxed{\text{No real values of } m}$

How to earn full marks: Show the substitution, the reasoning why the expression must be negative for no real solutions, and the conclusion that no such real m exists.

Common Pitfall: When solving for a range of values where there are no real solutions, remember to consider the discriminant. In this case, the question is flawed as there are always real solutions for m. Be careful when interpreting the question and look for potential errors.