Solving simultaneous equations graphically/algebraically/with technology | Year 12

01 / LEARN

Understand the idea.

Satisfy both conditions

A simultaneous solution is one pair $(x, y)$ satisfying both equations. On a graph, that pair is an intersection: its coordinates lie on both lines, rather than giving one point from each line.

Choose elimination when matching coefficients can cancel a variable. Choose substitution when one variable is already isolated. In either method, preserve equality at every step, then use the recovered variable to find the other.

Check a graph or calculator result

Graphing technology can suggest an intersection, but a poor window may hide it and displayed coordinates may be rounded. Substitute the proposed pair into both original equations to verify it.

Distinct parallel lines have no common point. Equivalent equations describe the same line and have infinitely many solutions. A contradiction or an identity in algebra explains these cases more reliably than a calculator message alone.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Elimination

Solve $2 x + y = 11$ and $x - y = 1$ .

Worked solution

1
Add the two equations.
$\begin{aligned} (2 x + y) + (x - y) & = 11 + 1 \\ 3 x & = 12 \end{aligned}$
The coefficients of $y$ are opposites, so their terms cancel.
2
Find the first coordinate.
$x = \frac{12}{3} = 4$
Divide both sides by 3 to isolate $x$ .
3
Recover the second coordinate.
$\begin{aligned} 4 - y & = 1 \\ - y & = - 3 \\ y & = 3 \end{aligned}$
The pair must also satisfy the original equation $x - y = 1$ .
4
Verify both conditions.
$\begin{aligned} 2 (4) + 3 & = 11 \\ 4 - 3 & = 1 \end{aligned}$
Both equalities hold, so the simultaneous solution is $(4, 3)$ .

Answer

$(4, 3)$ .

EXAMPLE 02

An intersection

Find where $y = 2 x + 1$ meets $y = - x + 7$ .

A coordinate grid runs from zero to eight on x and zero to ten on y. The two supplied lines are y=2x+1 and x+y=7. The intersection is not labelled.

Worked solution

1
Equate the two expressions for height.
$2 x + 1 = - x + 7$
At an intersection the two lines have the same $y$ -value for the same $x$ .
2
Solve for the horizontal coordinate.
$\begin{aligned} 3 x & = 6 \\ x & = 2 \end{aligned}$
Adding $x$ and subtracting 1 preserves the equality.
3
Find and check the height.
$\begin{aligned} 2 (2) + 1 & = 5 \\ - (2) + 7 & = 5 \end{aligned}$
Both original rules give $y = 5$ , so the intersection is $(2, 5)$ .

Answer

$(2, 5)$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Solve $x + y = 9$ and $x - y = 3$ .

Show answer & reasoning

$(6, 3)$ .

Adding the equations gives $2 x = 12$ , so $x = 6$ . Then $6 + y = 9$ gives $y = 3$ , and $6 - 3 = 3$ verifies the second equation.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Solve $x + y = 13$ and $x - y = 5$ .
2 marks
Solve $3 x + 2 y = 16$ and $x - y = 2$ .
2 marks
Find where $y = 4 x - 3$ meets $y = x + 6$ .
2 marks
How many solutions do $x + 2 y = 6$ and $2 x + 4 y = 15$ have?
2 marks
How many solutions do $3 x - y = 8$ and $6 x - 2 y = 16$ have?
2 marks
A solver returns $(2, 5)$ for $4 x - y = 3$ and $x + 3 y = 17$ . Verify it.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Solve $x + y = 13$ and $x - y = 5$ .
2 marks
Adding gives $2 x = 18$ ; then $y = 4$ .
$(9, 4)$ .
Solve $3 x + 2 y = 16$ and $x - y = 2$ .
2 marks
Substitute $x = y + 2$ : $5 y + 6 = 16$ .
Thus $y = 2$ , $x = 4$ .
$(4, 2)$ .
Find where $y = 4 x - 3$ meets $y = x + 6$ .
2 marks
$4 x - 3 = x + 6$ gives $x = 3$ ; then $y = 9$ .
$(3, 9)$ .
How many solutions do $x + 2 y = 6$ and $2 x + 4 y = 15$ have?
2 marks
Doubling the first requires $2 x + 4 y = 12$ , which contradicts 15.
None.
How many solutions do $3 x - y = 8$ and $6 x - 2 y = 16$ have?
2 marks
The second equation is twice the first, so they describe the same line.
Infinitely many.
A solver returns $(2, 5)$ for $4 x - y = 3$ and $x + 3 y = 17$ . Verify it.
2 marks
$4 (2) - 5 = 3$ and $2 + 3 (5) = 17$ .
Correct.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 General Mathematics: Modelling with Linear Relationships, a compulsory school-assessed topic. The external examination scope differs from the full course.

Official curriculum source ↗

In this lesson

Understand the idea.

Satisfy both conditions

Check a graph or calculator result

See the method in action.

Elimination

Add the two equations.

Find the first coordinate.

Recover the second coordinate.

Verify both conditions.

An intersection

Equate the two expressions for height.

Solve for the horizontal coordinate.

Find and check the height.

Try it yourself.

Check your understanding

Keep practising