Simultaneous equations in three unknowns Revision Notes for AQA GCSE Further Maths

Simultaneous equations in three unknowns

What are simultaneous equations in three unknowns?

When you have three different variables (usually called $x$ , $y$ , and $z$ ) that appear in multiple equations, you're dealing with simultaneous equations in three unknowns. To find unique values for all three variables, you need exactly three separate equations that contain these variables.

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Before tackling these problems, you should already feel confident solving simultaneous equations with just two unknowns, as this skill forms the foundation for the more complex three-variable problems.

The four-step elimination method

There's a systematic approach to solving these equations that makes the process much more manageable. Here's the method broken down into clear steps:

Step 1: Choose one variable to eliminate first, then combine two of your equations to remove that variable completely.

Step 2: Take the same variable you eliminated in Step 1, and eliminate it again by combining a different pair of equations.

Step 3: You now have two equations with only two unknowns remaining. Solve this pair using the standard methods for two-variable simultaneous equations.

Step 4: Once you've found the values for two variables, substitute these back into any of your original three equations to find the value of the third variable.

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The key to success is being systematic and eliminating the same variable in both Steps 1 and 2.

Worked example: using multiplication and addition

Let's work through a complete example to see how this method works in practice.

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Worked Example: Multiplication and Addition Method

Problem: Solve the system:

$3x + 2y - 3z = -13$ (equation ①)
$2x - 3y + 4z = 24$ (equation ②)
$4x - 5y + 2z = 22$ (equation ③)

Step 1: Let's eliminate $z$ by combining equations ① and ②.

Multiply equation ① by 4: $12x + 8y - 12z = -52$
Multiply equation ② by 3: $6x - 9y + 12z = 72$
Add these together: $18x - y = 20$ (equation ④)

Step 2: Now eliminate $z$ again using a different pair.

Multiply equation ③ by 2: $8x - 10y + 4z = 44$
Keep equation ② as is: $2x - 3y + 4z = 24$
Subtract the second from the first: $6x - 7y = 20$ (equation ⑤)

Step 3: Solve the two-variable system of equations ④ and ⑤.

From equations ④ and ⑤: $18x - y = 20$ and $6x - 7y = 20$
Multiply equation ④ by 3: $18x - 21y = 60$
Subtract equation ④ from this: $-20y = -40$
Therefore: $y = -2$

Substitute $y = -2$ into equation ④: $18x - (-2) = 20$ , so $x = 1$

Step 4: Find $z$ by substituting $x = 1$ and $y = -2$ into equation ①.

$3(1) + 2(-2) - 3z = -13$
$3 - 4 - 3z = -13$
$-3z = -12$
$z = 4$

Solution: x = 1, y = -2, z = 4

Worked example: rearranging equations first

Sometimes the equations aren't in standard form, so you'll need to rearrange them first before applying the elimination method.

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Worked Example: Rearranging Equations First

Problem: Solve the system:

$z - 8 = 2x + 3y$
$3x - 21 = 2y - 2z$
$2x + y = 3z - 4$

Step 1: First, rearrange all equations into standard form:

$z = 2x + 3y + 8$ (equation ①)
$3x - 21 = 2y - 2z$ becomes $7x + 4y = 5$ (equation ④)
$2x + y = 3z - 4$ becomes $x + 2y = -5$ (equation ⑤)

Step 2: Now we can eliminate variables systematically. From the rearranged equations, we get equation ④: $7x + 4y = 5$ and equation ⑤: $x + 2y = -5$ .

Step 3: Solve this two-variable system:

From equation ⑤: $x = -2y - 5$
Substitute into equation ④: $7(-2y - 5) + 4y = 5$
$-14y - 35 + 4y = 5$
$-10y = 40$
$y = -4$

Find $x$ : $x = -2(-4) - 5 = 3$

Step 4: Find $z$ using equation ①:

$z = 2(3) + 3(-4) + 8 = 2$

Solution: x = 3, y = -4, z = 2

Common exam tips and problem-solving strategies

When approaching these problems in exams, remember these key strategies:

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Stay organised: Label your equations clearly (①, ②, ③, etc.) and number your steps to keep track of your progress.

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Choose wisely: Pick the variable that's easiest to eliminate first - look for coefficients that are already the same or simple multiples of each other.

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Check your work: Always substitute your final answers back into all three original equations to verify they're correct. This catches calculation errors and confirms your solution.

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Show working clearly: Examiners want to see your method even if you make a calculation error. Clear working can earn you partial marks.

The elimination method is systematic and reliable. Even if the algebra gets messy, following these four steps will always lead you to the solution.

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Key Points to Remember:

You need exactly three equations to solve for three unknowns uniquely
Always eliminate the same variable in Steps 1 and 2 of the method
The four-step process is: eliminate, eliminate, solve two-variable system, substitute back
Keep your working organised and clearly labelled throughout
Check your final answer by substituting back into all original equations

Simultaneous equations in three unknowns (AQA GCSE Further Maths): Revision Notes