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Solving three equations in three unknowns

Covers AS91587

What a $3 \times 3$ system is

A system of three linear equations in three unknowns looks like

$\begin{cases} a_1x + b_1y + c_1z = d_1 \\ a_2x + b_2y + c_2z = d_2 \\ a_3x + b_3y + c_3z = d_3 \end{cases}$

A solution is a triple $(x, y, z)$ satisfying all three equations at once.
Each equation is a plane in three-dimensional space, and the solutions are the points where all three planes meet.
Three unknowns generally need three independent equations. Fewer cannot pin down a point; more can only agree or contradict.

The strategy: reduce $3 \times 3$ to $2 \times 2$

This is the whole method, and it is worth stating before you start writing:
- Eliminate one variable — say $x$ — from two different pairs of equations.
- You now have two equations in the two remaining unknowns.
- Solve that $2 \times 2$ system the usual way.
- Back-substitute into any original equation to find the third unknown.
- Check in the equation you did not use for the back-substitution.
Use two DIFFERENT pairs. Combining (1)&(2) and then (1)&(2) again produces the same information twice and leaves you stuck.

The elimination step

To eliminate a variable from a pair of equations:
- Multiply one or both equations so that the coefficients of that variable match in size.
- Subtract if the signs are the same, add if they are opposite.
- Label the result as a new equation.
Choose the easiest variable to eliminate, not necessarily $x$ $x$ :
- Look for a variable with a coefficient of 1, which needs no multiplication.
- Look for coefficients that are multiples of each other.
- Look for a variable missing from one equation — that equation is already half-eliminated.

Worked through — a complete solve

Solve

$\begin{cases} x + 2y - z = 3 & (1) \\ 2x - y + 3z = 6 & (2) \\ x + y + z = 4 & (3) \end{cases}$

Choose to eliminate $x$ , since (1) and (3) both have coefficient 1.
$(2) - 2 \times (1)$ :
- $(2x - y + 3z) - (2x + 4y - 2z) = 6 - 6$
- $-5y + 5z = 0$ , so $y = z$ — call this (4)
$(3) - (1)$ :
- $(x + y + z) - (x + 2y - z) = 4 - 3$
- $-y + 2z = 1$ — call this (5)
Solve (4) and (5). Substituting $y = z$ into (5): $-z + 2z = 1$ , so $z = 1$ , and $y = 1$ .
Back-substitute into (1): $x + 2(1) - 1 = 3$ , so $x = 2$ .
Check in (2), the equation not used for the back-substitution: $2(2) - 1 + 3(1) = 4 - 1 + 3 = 6$ ✓

$\textbf{Solution: } x = 2, \ y = 1, \ z = 1$

Subtracting negatives

This is where most errors happen. Write the subtraction out in full, with brackets:

$(2x - y + 3z) - (2x + 4y - 2z) = 2x - y + 3z - 2x - 4y + 2z = -5y + 5z$

Every sign inside the second bracket flips. Doing it in your head produces an answer that looks right and is not.

Checking your solution

Substitute the full triple into ALL THREE original equations, not just the one you back-substituted into.
A solution that satisfies two equations and fails the third is not a solution — and it is exactly what an arithmetic slip produces.
The check takes three lines and is worth doing every time.

Tips and Tricks

Label every equation, including the ones you create. Writing "(2) − 2(1) → (4)" makes your method visible and lets you refer back without re-deriving.

Eliminate from two different pairs. Using (1)&(2) twice gives the same equation again and wastes a page. A good habit is (1)&(2) then (1)&(3).

Pick the easiest variable, not the first. If $z$ is missing from one equation, eliminating $z$ from the other two is one step instead of three.

Write subtractions with brackets. $(2x - y + 3z) - (2x + 4y - 2z)$ set out in full is the single most effective error-prevention habit in this topic.

Check in all three original equations. Back-substitution proves nothing about the equation you back-substituted into.

Keep fractions exact. $z = \frac{7}{3}$ is a better intermediate value than 2.33, and the final check will not balance if you round early.

Worked ExampleA fertiliser blend

A supplier makes three fertiliser blends. Each 100 kg of Blend A contains 20 kg nitrogen, 10 kg phosphorus and 5 kg potassium; Blend B contains 10, 20 and 10; Blend C contains 5, 10 and 30. A farmer needs a mix containing exactly 275 kg of nitrogen, 300 kg of phosphorus and 425 kg of potassium.

How many hundred-kilogram units of each blend should be ordered?

Step 1 — Define the variables

Step 2 — Form one equation per nutrient

Nitrogen — 20 kg per unit of A, 10 per unit of B, 5 per unit of C, total 275:

$20a + 10b + 5c = 275 \qquad (1)$

Phosphorus:

$10a + 20b + 10c = 300 \qquad (2)$

Potassium:

$5a + 10b + 30c = 425 \qquad (3)$

Step 3 — Simplify before eliminating

Divide (1) by 5:

$4a + 2b + c = 55 \qquad (1')$

Divide (2) by 10:

$a + 2b + c = 30 \qquad (2')$

Divide (3) by 5:

$a + 2b + 6c = 85 \qquad (3')$

Step 4 — Eliminate $a$ from the first pair

$(1') - 4 \times (2')$ :

$(4a + 2b + c) - (4a + 8b + 4c) = 55 - 120$

Expand the subtraction in full:

$4a + 2b + c - 4a - 8b - 4c = -65$

$-6b - 3c = -65 \qquad (4)$

Step 5 — Eliminate $a$ from a DIFFERENT pair

$(3') - (2')$ :

$(a + 2b + 6c) - (a + 2b + c) = 85 - 30$

$5c = 55$

$c = 11 \qquad (5)$

Step 6 — Substitute back to find $b$

Into (4):

$-6b - 3(11) = -65$

$-6b - 33 = -65$

$-6b = -32$

$b = \frac{32}{6} = \frac{16}{3} = 5.333\ldots$

Step 7 — Find $a$

Into (2′):

$a + 2\left(\frac{16}{3}\right) + 11 = 30$

$a + \frac{32}{3} = 19$

$a = 19 - \frac{32}{3} = \frac{57 - 32}{3} = \frac{25}{3} = 8.333\ldots$

Step 8 — Check in ALL THREE original equations

Nitrogen (1):

$20\left(\frac{25}{3}\right) + 10\left(\frac{16}{3}\right) + 5(11) = \frac{500}{3} + \frac{160}{3} + 55 = \frac{660}{3} + 55 = 220 + 55 = 275 \ \checkmark$

Phosphorus (2):

$10\left(\frac{25}{3}\right) + 20\left(\frac{16}{3}\right) + 10(11) = \frac{250}{3} + \frac{320}{3} + 110 = 190 + 110 = 300 \ \checkmark$

Potassium (3):

$5\left(\frac{25}{3}\right) + 10\left(\frac{16}{3}\right) + 30(11) = \frac{125}{3} + \frac{160}{3} + 330 = 95 + 330 = 425 \ \checkmark$

All three balance, so the solution is correct.

Step 9 — Interpret in context

All three values are positive, so the mix is physically possible — a negative value would have meant the nutrient targets could not be met with these three blends.

A practical note. The answers for A and B are not whole units. Fertiliser is measured continuously, so 833 kg is a perfectly orderable quantity — but if the blends were only sold in sealed 100 kg sacks, no exact solution would exist and the farmer would have to choose the nearest whole combination and accept a small nutrient error.

Solving three equations in three unknowns

Covers AS91587

What a $3 \times 3$ system is

A system of three linear equations in three unknowns looks like

$\begin{cases} a_1x + b_1y + c_1z = d_1 \\ a_2x + b_2y + c_2z = d_2 \\ a_3x + b_3y + c_3z = d_3 \end{cases}$

A solution is a triple $(x, y, z)$ satisfying all three equations at once.
Each equation is a plane in three-dimensional space, and the solutions are the points where all three planes meet.
Three unknowns generally need three independent equations. Fewer cannot pin down a point; more can only agree or contradict.

The strategy: reduce $3 \times 3$ to $2 \times 2$

This is the whole method, and it is worth stating before you start writing:
- Eliminate one variable — say $x$ — from two different pairs of equations.
- You now have two equations in the two remaining unknowns.
- Solve that $2 \times 2$ system the usual way.
- Back-substitute into any original equation to find the third unknown.
- Check in the equation you did not use for the back-substitution.
Use two DIFFERENT pairs. Combining (1)&(2) and then (1)&(2) again produces the same information twice and leaves you stuck.

The elimination step

To eliminate a variable from a pair of equations:
- Multiply one or both equations so that the coefficients of that variable match in size.
- Subtract if the signs are the same, add if they are opposite.
- Label the result as a new equation.
Choose the easiest variable to eliminate, not necessarily $x$ $x$ :
- Look for a variable with a coefficient of 1, which needs no multiplication.
- Look for coefficients that are multiples of each other.
- Look for a variable missing from one equation — that equation is already half-eliminated.

Worked through — a complete solve

Solve

$\begin{cases} x + 2y - z = 3 & (1) \\ 2x - y + 3z = 6 & (2) \\ x + y + z = 4 & (3) \end{cases}$

Choose to eliminate $x$ , since (1) and (3) both have coefficient 1.
$(2) - 2 \times (1)$ :
- $(2x - y + 3z) - (2x + 4y - 2z) = 6 - 6$
- $-5y + 5z = 0$ , so $y = z$ — call this (4)
$(3) - (1)$ :
- $(x + y + z) - (x + 2y - z) = 4 - 3$
- $-y + 2z = 1$ — call this (5)
Solve (4) and (5). Substituting $y = z$ into (5): $-z + 2z = 1$ , so $z = 1$ , and $y = 1$ .
Back-substitute into (1): $x + 2(1) - 1 = 3$ , so $x = 2$ .
Check in (2), the equation not used for the back-substitution: $2(2) - 1 + 3(1) = 4 - 1 + 3 = 6$ ✓

$\textbf{Solution: } x = 2, \ y = 1, \ z = 1$

Subtracting negatives

This is where most errors happen. Write the subtraction out in full, with brackets:

$(2x - y + 3z) - (2x + 4y - 2z) = 2x - y + 3z - 2x - 4y + 2z = -5y + 5z$

Every sign inside the second bracket flips. Doing it in your head produces an answer that looks right and is not.

Checking your solution

Substitute the full triple into ALL THREE original equations, not just the one you back-substituted into.
A solution that satisfies two equations and fails the third is not a solution — and it is exactly what an arithmetic slip produces.
The check takes three lines and is worth doing every time.

Tips and Tricks

Label every equation, including the ones you create. Writing "(2) − 2(1) → (4)" makes your method visible and lets you refer back without re-deriving.

Eliminate from two different pairs. Using (1)&(2) twice gives the same equation again and wastes a page. A good habit is (1)&(2) then (1)&(3).

Pick the easiest variable, not the first. If $z$ is missing from one equation, eliminating $z$ from the other two is one step instead of three.

Write subtractions with brackets. $(2x - y + 3z) - (2x + 4y - 2z)$ set out in full is the single most effective error-prevention habit in this topic.

Check in all three original equations. Back-substitution proves nothing about the equation you back-substituted into.

Keep fractions exact. $z = \frac{7}{3}$ is a better intermediate value than 2.33, and the final check will not balance if you round early.

Worked ExampleA fertiliser blend

How many hundred-kilogram units of each blend should be ordered?

Step 1 — Define the variables

Step 2 — Form one equation per nutrient

Nitrogen — 20 kg per unit of A, 10 per unit of B, 5 per unit of C, total 275:

$20a + 10b + 5c = 275 \qquad (1)$

Phosphorus:

$10a + 20b + 10c = 300 \qquad (2)$

Potassium:

$5a + 10b + 30c = 425 \qquad (3)$

Step 3 — Simplify before eliminating

Divide (1) by 5:

$4a + 2b + c = 55 \qquad (1')$

Divide (2) by 10:

$a + 2b + c = 30 \qquad (2')$

Divide (3) by 5:

$a + 2b + 6c = 85 \qquad (3')$

Step 4 — Eliminate $a$ from the first pair

$(1') - 4 \times (2')$ :

$(4a + 2b + c) - (4a + 8b + 4c) = 55 - 120$

Expand the subtraction in full:

$4a + 2b + c - 4a - 8b - 4c = -65$

$-6b - 3c = -65 \qquad (4)$

Step 5 — Eliminate $a$ from a DIFFERENT pair

$(3') - (2')$ :

$(a + 2b + 6c) - (a + 2b + c) = 85 - 30$

$5c = 55$

$c = 11 \qquad (5)$

Step 6 — Substitute back to find $b$

Into (4):

$-6b - 3(11) = -65$

$-6b - 33 = -65$

$-6b = -32$

$b = \frac{32}{6} = \frac{16}{3} = 5.333\ldots$

Step 7 — Find $a$

Into (2′):

$a + 2\left(\frac{16}{3}\right) + 11 = 30$

$a + \frac{32}{3} = 19$

$a = 19 - \frac{32}{3} = \frac{57 - 32}{3} = \frac{25}{3} = 8.333\ldots$

Step 8 — Check in ALL THREE original equations

Nitrogen (1):

$20\left(\frac{25}{3}\right) + 10\left(\frac{16}{3}\right) + 5(11) = \frac{500}{3} + \frac{160}{3} + 55 = \frac{660}{3} + 55 = 220 + 55 = 275 \ \checkmark$

Phosphorus (2):

$10\left(\frac{25}{3}\right) + 20\left(\frac{16}{3}\right) + 10(11) = \frac{250}{3} + \frac{320}{3} + 110 = 190 + 110 = 300 \ \checkmark$

Potassium (3):

$5\left(\frac{25}{3}\right) + 10\left(\frac{16}{3}\right) + 30(11) = \frac{125}{3} + \frac{160}{3} + 330 = 95 + 330 = 425 \ \checkmark$

All three balance, so the solution is correct.

Step 9 — Interpret in context

All three values are positive, so the mix is physically possible — a negative value would have meant the nutrient targets could not be met with these three blends.

Solving three equations in three unknowns

What a 3×33 \times 33×3 system is

The strategy: reduce 3×33 \times 33×3 to 2×22 \times 22×2

The elimination step

Worked through — a complete solve

Subtracting negatives

Checking your solution

Step 1 — Define the variables

Step 2 — Form one equation per nutrient

Step 3 — Simplify before eliminating

Step 4 — Eliminate aaa from the first pair

Step 5 — Eliminate aaa from a DIFFERENT pair

Step 6 — Substitute back to find bbb

Step 7 — Find aaa

Step 8 — Check in ALL THREE original equations

Step 9 — Interpret in context

Solving three equations in three unknowns

What a 3×33 \times 33×3 system is

The strategy: reduce 3×33 \times 33×3 to 2×22 \times 22×2

The elimination step

Worked through — a complete solve

Subtracting negatives

Checking your solution

Step 1 — Define the variables

Step 2 — Form one equation per nutrient

Step 3 — Simplify before eliminating

Step 4 — Eliminate aaa from the first pair

Step 5 — Eliminate aaa from a DIFFERENT pair

Step 6 — Substitute back to find bbb

Step 7 — Find aaa

Step 8 — Check in ALL THREE original equations

Step 9 — Interpret in context

What a $3 \times 3$ system is

The strategy: reduce $3 \times 3$ to $2 \times 2$

Step 4 — Eliminate $a$ from the first pair

Step 5 — Eliminate $a$ from a DIFFERENT pair

Step 6 — Substitute back to find $b$

Step 7 — Find $a$

What a $3 \times 3$ system is

The strategy: reduce $3 \times 3$ to $2 \times 2$

Step 4 — Eliminate $a$ from the first pair

Step 5 — Eliminate $a$ from a DIFFERENT pair

Step 6 — Substitute back to find $b$

Step 7 — Find $a$