Year 11 Maths - Methods (Unit 1 and Unit 2) Linear relations and equations

Simultaneous Equations 3X3

Q: How do you solve a 3x3 system of equations?

Use elimination. Remove one variable from two of the equations to get a two by two system in the other two unknowns, solve that, then back-substitute to find the third unknown.

Q: How do I know if a 3x3 system has no solution?

When elimination leads to a false statement such as zero equals seven, the equations are inconsistent and there is no solution.

Q: How can a system have infinitely many solutions?

If elimination reduces an equation to zero equals zero, the equations are dependent, so there are infinitely many solutions rather than a single point.

Q: Do I have to eliminate x first?

No. You may eliminate whichever variable is easiest, but you must remove the same variable from both of the other equations to reach a two by two system.

Q: How do I check my answer?

Substitute the triple x, y, z back into all three original equations. Each one must be satisfied.

20 practice questions 0 video lessons Theory + worked examples

Master simultaneous equations in three unknowns for Victorian Year 11 Mathematical Methods (VCAA) — three linear equations solved together to find the single point that satisfies all three.

You will learn to solve these systems by elimination and back-substitution, decide whether a system has one solution, none or infinitely many, and set up and answer worded problems involving three unknowns.

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Practice questions

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Theory

In Year 11 Mathematical Methods (Unit 1), a $3\times 3$ system is three linear equations in the three unknowns $x$, $y$ and $z$. This page shows how to solve one by elimination — combine equations to remove a variable, drop to a $2\times 2$ system, solve it, then back-substitute — and how to tell whether a system has a unique solution, no solution, or infinitely many solutions.

A $3\times 3$ system of linear equations is three equations in the three unknowns $x$, $y$ and $z$, each of the form $ax+by+cz=d$. A solution is an ordered triple $(x,\,y,\,z)$ that satisfies all three equations at once.

Geometrically each equation is a plane in space, so solving the system means finding the point (or points) common to all three planes.

The standard technique is elimination: add suitable multiples of one equation to another to cancel a chosen variable. Doing this twice reduces the $3\times 3$ system to a $2\times 2$ system in two unknowns, which you solve, then back-substitute to recover the third unknown.

A $3\times 3$ system has exactly one of three outcomes:

a unique solution — the three planes meet at a single point;
no solution — the equations are inconsistent (elimination produces a false statement such as $0=5$);
infinitely many solutions — the equations are dependent (elimination produces $0=0$).

Eliminate the same variable twice. Use one equation to remove \(x\) from the other two; that leaves a \(2\times 2\) system in \(y\) and \(z\).

Unique solution: the three planes meet at one point $P(x,y,z)$.

No solution: two planes are parallel, so the three share no common point — the system is inconsistent.

A general $3\times 3$ linear system:

\[a_1x+b_1y+c_1z=d_1,\quad a_2x+b_2y+c_2z=d_2,\quad a_3x+b_3y+c_3z=d_3\]

a_{1} x + b_{1} y + c_{1} z = d_{1}

Elimination uses a linear combination of two equations to cancel a variable. To remove $x$ from equations $(1)$ and $(2)$:

\[a_2\times(1)-a_1\times(2)\ \Rightarrow\ \text{a new equation in }y,z\ \text{only}\]

a_{2} \times (1) - a_{1} \times (2)

Classifying the outcome. After elimination: a normal line like \(z=3\) gives a unique solution; a false line like \(0=-12\) means no solution; and \(0=0\) means infinitely many solutions.

How to solve a $3\times 3$ system by elimination

Label the three equations $(1)$, $(2)$, $(3)$.
Eliminate one variable twice. Use equation $(1)$ to remove $x$ from $(2)$ and from $(3)$, giving two new equations $(4)$ and $(5)$ in $y$ and $z$ only.
Solve the $2\times 2$ system $(4)$,$(5)$ for $y$ and $z$ (eliminate again, or substitute).
Back-substitute $y$ and $z$ into an original equation to find $x$.
Check the triple $(x,\,y,\,z)$ in all three original equations.

Watch the classification. If a step gives a false statement (e.g. \(0=7\)) stop — there is no solution; if it gives \(0=0\), the system has infinitely many solutions.

Example 1 — Unique solution by elimination

Solve the system $x+y+z=4$, $x-y+z=6$, $2x+y-z=0$ for $x$, $y$ and $z$.

Solution

Label the three equations:

$x+y+z$	$=$	$4 \qquad (1)$
$x-y+z$	$=$	$6 \qquad (2)$
$2x+y-z$	$=$	$0 \qquad (3)$

Eliminate $x$ — subtract (1) from (2):

$(x-y+z)-(x+y+z)$	$=$	$6-4$
$-2y$	$=$	$2 \qquad (4)$

Eliminate $x$ — subtract $2\times$(1) from (3):

$(2x+y-z)-2(x+y+z)$	$=$	$0-8$
$-y-3z$	$=$	$-8 \qquad (5)$

Solve (4) for $y$:

$-2y$	$=$	$2$
$y$	$=$	$-1$

Substitute $y=-1$ into (5) and solve for $z$:

$-(-1)-3z$	$=$	$-8$
$1-3z$	$=$	$-8$
$-3z$	$=$	$-9$
$z$	$=$	$3$

Back-substitute $y=-1$, $z=3$ into (1) for $x$:

$x+(-1)+3$	$=$	$4$
$x+2$	$=$	$4$
$x$	$=$	$2$

Solution: $x=2$, $y=-1$, $z=3$.

x = 2, y = - 1, z = 3

Example 2 — Eliminate, then eliminate again

Solve $x+y+z=4$, $2x-y+z=0$, $3x+y-2z=11$.

Solution

Label the equations:

$x+y+z$	$=$	$4 \qquad (1)$
$2x-y+z$	$=$	$0 \qquad (2)$
$3x+y-2z$	$=$	$11 \qquad (3)$

Eliminate $x$ — (2) minus $2\times$(1):

$(2x-y+z)-2(x+y+z)$	$=$	$0-8$
$-3y-z$	$=$	$-8 \qquad (4)$

Eliminate $x$ — (3) minus $3\times$(1):

$(3x+y-2z)-3(x+y+z)$	$=$	$11-12$
$-2y-5z$	$=$	$-1 \qquad (5)$

Eliminate $z$ from (4),(5) using $5\times(4)-(5)$:

$5\times(4)$	$\Rightarrow$	$-15y-5z=-40$
$(5)$	$\Rightarrow$	$-2y-5z=-1$
$-13y$	$=$	$-39$
$y$	$=$	$3$

Substitute $y=3$ into (4) for $z$:

$-3(3)-z$	$=$	$-8$
$-9-z$	$=$	$-8$
$-z$	$=$	$1$
$z$	$=$	$-1$

Back-substitute into (1) for $x$:

$x+3+(-1)$	$=$	$4$
$x+2$	$=$	$4$
$x$	$=$	$2$

Solution: $x=2$, $y=3$, $z=-1$.

x = 2, y = 3, z = - 1

Example 3 — A worded application

A cinema sells adult, concession and child tickets at $\$x$, $\$y$ and $\$z$ each. Two adult, one concession and one child ticket cost $\$60$; one adult, two concession and two child tickets cost $\$66$; three adult, one concession and two child tickets cost $\$88$. Find each price.

Solution

Set up one equation per sale:

$2x+y+z$	$=$	$60 \qquad (1)$
$x+2y+2z$	$=$	$66 \qquad (2)$
$3x+y+2z$	$=$	$88 \qquad (3)$

Eliminate $x$ — $2\times(2)-(1)$:

$2(x+2y+2z)-(2x+y+z)$	$=$	$2(66)-60$
$3y+3z$	$=$	$72$
$y+z$	$=$	$24 \qquad (4)$

Eliminate $x$ — $2\times(3)-3\times(1)$:

$2(3x+y+2z)-3(2x+y+z)$	$=$	$2(88)-3(60)$
$-y+z$	$=$	$-4 \qquad (5)$

Add (4) and (5) to find $z$:

$(y+z)+(-y+z)$	$=$	$24+(-4)$
$2z$	$=$	$20$
$z$	$=$	$10$

Substitute $z=10$ into (4) for $y$:

$y+10$	$=$	$24$
$y$	$=$	$14$

Substitute into (1) for $x$:

$2x+14+10$	$=$	$60$
$2x$	$=$	$36$
$x$	$=$	$18$

Prices: adult $\$18$, concession $\$14$, child $\$10$.

x = 18, y = 14, z = 10

Example 4 — A system with no solution

Determine how many solutions the system $x+2y-z=3$, $2x-y+3z=4$, $3x+y+2z=8$ has.

Solution

Label the equations:

$x+2y-z$	$=$	$3 \qquad (1)$
$2x-y+3z$	$=$	$4 \qquad (2)$
$3x+y+2z$	$=$	$8 \qquad (3)$

Add (1) and (2) to combine two of the equations:

$(x+2y-z)+(2x-y+3z)$	$=$	$3+4$
$3x+y+2z$	$=$	$7 \qquad (4)$

Equation (3) has the same left-hand side $3x+y+2z$ as (4), but a different right-hand side:

Compare (3) and (4):

$3x+y+2z$	$=$	$8 \qquad (3)$
$3x+y+2z$	$=$	$7 \qquad (4)$

The one expression $3x+y+2z$ cannot equal both $8$ and $7$, so the system is inconsistent.

No solution — the three planes share no common point.

No solution

Common pitfalls

Eliminating a different variable each time. To reduce a $3\times 3$ system to a $2\times 2$ one, you must remove the same variable from both of the other equations — not $x$ from one pair and $y$ from another.

Sign slips when subtracting equations. Subtract every term: $(2x-y+z)-2(x+y+z)$ gives $-3y-z$, because $-y-2y=-3y$ and $z-2z=-z$.

Forgetting to find the third unknown. After solving the $2\times 2$ system for $y$ and $z$, you must back-substitute into an original equation to get $x$ — the answer is a full triple $(x,\,y,\,z)$.

Misreading the classification. A false line such as $0=-12$ means no solution, while $0=0$ means infinitely many — do not report a number when elimination has collapsed a whole equation.

Frequently asked questions

How do you solve a 3x3 system of equations?

Use elimination: pick one variable and remove it from two of the equations, leaving a $2\times 2$ system in the other two unknowns. Solve that, then back-substitute to find the third unknown.

What does it mean to eliminate a variable?

You add a suitable multiple of one equation to another so that one variable cancels, producing a new equation with fewer unknowns.

How do I know if a 3x3 system has no solution?

When elimination leads to a false statement such as $0=7$ or $0=-12$, the equations are inconsistent and there is no solution.

How can a system have infinitely many solutions?

If elimination reduces an equation to $0=0$, the equations are dependent, so there are infinitely many solutions rather than a single point.

Do I have to eliminate x first?

No — you may eliminate whichever variable is easiest, but you must remove the same variable from both of the other equations to reach a $2\times 2$ system.

How do I check my answer?

Substitute the triple $(x,\,y,\,z)$ back into all three original equations; each one must be satisfied.

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Linear Graphs and their Equations

\(x+y+z\)	\(=\)	\(4 \qquad (1)\)
\(x-y+z\)	\(=\)	\(6 \qquad (2)\)
\(2x+y-z\)	\(=\)	\(0 \qquad (3)\)

\((x-y+z)-(x+y+z)\)	\(=\)	\(6-4\)
\(-2y\)	\(=\)	\(2 \qquad (4)\)

\((2x+y-z)-2(x+y+z)\)	\(=\)	\(0-8\)
\(-y-3z\)	\(=\)	\(-8 \qquad (5)\)

\(-2y\)	\(=\)	\(2\)
\(y\)	\(=\)	\(-1\)

\(-(-1)-3z\)	\(=\)	\(-8\)
\(1-3z\)	\(=\)	\(-8\)
\(-3z\)	\(=\)	\(-9\)
\(z\)	\(=\)	\(3\)

\((2x-y+z)-2(x+y+z)\)	\(=\)	\(0-8\)
\(-3y-z\)	\(=\)	\(-8 \qquad (4)\)

\((3x+y-2z)-3(x+y+z)\)	\(=\)	\(11-12\)
\(-2y-5z\)	\(=\)	\(-1 \qquad (5)\)

\(5\times(4)\)	\(\Rightarrow\)	\(-15y-5z=-40\)
\((5)\)	\(\Rightarrow\)	\(-2y-5z=-1\)
\(-13y\)	\(=\)	\(-39\)
\(y\)	\(=\)	\(3\)

\(-3(3)-z\)	\(=\)	\(-8\)
\(-9-z\)	\(=\)	\(-8\)
\(-z\)	\(=\)	\(1\)
\(z\)	\(=\)	\(-1\)

\(2x+y+z\)	\(=\)	\(60 \qquad (1)\)
\(x+2y+2z\)	\(=\)	\(66 \qquad (2)\)
\(3x+y+2z\)	\(=\)	\(88 \qquad (3)\)

\(2(x+2y+2z)-(2x+y+z)\)	\(=\)	\(2(66)-60\)
\(3y+3z\)	\(=\)	\(72\)
\(y+z\)	\(=\)	\(24 \qquad (4)\)

\(2(3x+y+2z)-3(2x+y+z)\)	\(=\)	\(2(88)-3(60)\)
\(-y+z\)	\(=\)	\(-4 \qquad (5)\)

\((y+z)+(-y+z)\)	\(=\)	\(24+(-4)\)
\(2z\)	\(=\)	\(20\)
\(z\)	\(=\)	\(10\)

\(x+2y-z\)	\(=\)	\(3 \qquad (1)\)
\(2x-y+3z\)	\(=\)	\(4 \qquad (2)\)
\(3x+y+2z\)	\(=\)	\(8 \qquad (3)\)

\((x+2y-z)+(2x-y+3z)\)	\(=\)	\(3+4\)
\(3x+y+2z\)	\(=\)	\(7 \qquad (4)\)

\(y+10\)	\(=\)	\(24\)
\(y\)	\(=\)	\(14\)

\(2x+14+10\)	\(=\)	\(60\)
\(2x\)	\(=\)	\(36\)
\(x\)	\(=\)	\(18\)

Simultaneous Equations 3X3

📖 Theory

How to solve a \(3\times 3\) system by elimination

Common pitfalls

Frequently asked questions

Theory