Algebra

Simultaneous Equations

Year 10 · Year 11

✓Solve two linear simultaneous equations using both elimination and substitution methods.
✓Solve simultaneous equations where one equation is linear and the other is quadratic (Higher Tier content).
✓Interpret the solutions of simultaneous equations graphically.
✓Formulate and solve simultaneous equations from given problem contexts.

Key concepts

What are Simultaneous Equations?

Simultaneous equations are a set of two or more equations that share common variables. The goal is to find the values of these variables that satisfy *all* equations in the set at the same time. Graphically, the solutions represent the points where the graphs of the equations intersect.

Solving Two Linear Simultaneous Equations: Method 1 - Elimination

The elimination method involves manipulating the equations so that one of the variables has the same (or opposite) coefficient in both equations. Then, by adding or subtracting the equations, that variable is 'eliminated', leaving a single equation with one variable which can then be solved. This method is often efficient when coefficients are already the same or easily made so.

Solving Two Linear Simultaneous Equations: Method 2 - Substitution

The substitution method involves rearranging one of the equations to make one variable the subject (e.g., y = ... or x = ...). This expression is then substituted into the *other* equation, resulting in a single equation with one variable, which can then be solved. This method is particularly useful when one variable is already isolated or easily isolated.

Solving Linear and Quadratic Simultaneous Equations (Higher Tier)

When one equation is linear and the other is quadratic, the substitution method is typically used. Rearrange the linear equation to make one variable the subject, and then substitute this expression into the quadratic equation. This will result in a quadratic equation in one variable, which can be solved using factorisation, the quadratic formula, or completing the square. There can be zero, one (tangent), or two pairs of solutions, corresponding to zero, one, or two points of intersection between the line and the curve.

Key facts to remember

1Simultaneous equations require finding values for variables that satisfy all equations in the set.
2The two main methods for solving linear simultaneous equations are elimination and substitution.
3For linear and quadratic simultaneous equations, the substitution method is generally used.
4Two linear equations typically have one unique solution (one point of intersection).
5A linear and a quadratic equation can have zero, one (tangent), or two solutions (points of intersection).
6Always check your solutions by substituting them back into *both* original equations.
7Solutions to simultaneous equations represent the coordinates of the intersection points of their graphs.

Worked examples

Example 1

Solve the following simultaneous equations: 2x + y = 11 (1) 3x - y = 4 (2)

INotice that the 'y' coefficients are already the same magnitude but opposite signs. We can add the equations to eliminate 'y'.

II(1) + (2): (2x + y) + (3x - y) = 11 + 4

III5x = 15

IVx = 3

VSubstitute x = 3 into equation (1):

VI2(3) + y = 11

VII6 + y = 11

VIIIy = 11 - 6

9y = 5

10Check with equation (2): 3(3) - 5 = 9 - 5 = 4. This is correct.

Answer

x = 3, y = 5

This example demonstrates the elimination method when coefficients are already suitable for adding.

Example 2

Solve the following simultaneous equations: y = 2x - 3 (1) 4x + 3y = 21 (2)

IEquation (1) already has 'y' as the subject. Substitute (1) into (2).

II4x + 3(2x - 3) = 21

III4x + 6x - 9 = 21

IV10x - 9 = 21

V10x = 30

VIx = 3

VIISubstitute x = 3 back into equation (1):

VIIIy = 2(3) - 3

9y = 6 - 3

10y = 3

11Check with equation (2): 4(3) + 3(3) = 12 + 9 = 21. This is correct.

Answer

x = 3, y = 3

This example demonstrates the substitution method, which is often efficient when one variable is already isolated.

Example 3

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

ISubstitute equation (1) into equation (2).

IIx^2 + (x + 1)^2 = 13

IIIx^2 + (x^2 + 2x + 1) = 13

IV2x^2 + 2x + 1 = 13

V2x^2 + 2x - 12 = 0

VIDivide by 2: x^2 + x - 6 = 0

VIIFactorise the quadratic: (x + 3)(x - 2) = 0

VIIISo, x + 3 = 0 or x - 2 = 0

9x = -3 or x = 2

10Now find the corresponding 'y' values using equation (1):

11If x = -3, y = -3 + 1, so y = -2

12If x = 2, y = 2 + 1, so y = 3

13Check solutions:

14For (-3, -2): (-3)^2 + (-2)^2 = 9 + 4 = 13. Correct.

15For (2, 3): (2)^2 + (3)^2 = 4 + 9 = 13. Correct.

Answer

x = -3, y = -2 and x = 2, y = 3

Remember to find both 'x' and 'y' values for each solution. This type of problem can have two pairs of solutions.

Common mistakes

✗**Sign Errors**: Incorrectly adding or subtracting equations, especially with negative coefficients, leading to incorrect elimination.
✗**Incomplete Multiplication**: When scaling an equation for elimination, forgetting to multiply *every* term (including the constant) by the chosen factor.
✗**Algebraic Errors**: Mistakes when expanding brackets, collecting like terms, or rearranging equations, particularly when dealing with quadratic expressions.
✗**Forgetting the Second Variable**: Only finding the value for one variable (e.g., 'x') and forgetting to substitute back to find the other variable (e.g., 'y').
✗**Incorrect Substitution**: Substituting an expression back into the *same* equation it came from, rather than the *other* equation, which will lead to an identity (e.g., 0=0) or an incorrect result.

Exam tips

★**Choose Wisely**: For linear equations, decide whether elimination or substitution is more efficient. Substitution is almost always best for linear and quadratic equations.
★**Show All Working**: Present your steps clearly and logically, labelling equations (e.g., (1), (2)) to make your method easy to follow for the examiner.
★**Check Your Answers**: Substitute your final 'x' and 'y' values into *both* original equations to ensure they satisfy both. This is a crucial step to verify your solution and avoid losing marks.
★**Be Meticulous with Negatives**: Pay close attention to negative signs throughout your calculations, as these are a common source of error. Use brackets to help manage negative numbers during substitution.

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