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Year 12 Maths Standard 1 (2027) Algebraic relationships

Simultaneous Equations (Graphically)

20 practice questions 0 video lessons Theory + worked examples

Solve simultaneous linear equations by graphing for NSW Year 12 Mathematics Standard 1 (NESA outcome MST-12-S1-01). Solving two equations together means finding the single pair of \(x\) and \(y\) values that makes both true at once — on a graph, that is where the two lines cross.

You will draw both lines with graphing software or from a table of values, read off the point of intersection, check it in both equations, and spot when parallel lines give no solution — a handy skill for comparing two prices or plans in everyday life.

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Theory

Simultaneous linear equations are two straight-line equations solved together. This Year 12 Standard 1 (NSW) guide shows how to solve them graphically — draw both lines, read and check the point of intersection, use a table of values, and decide whether a pair has one, none or infinitely many solutions.

Two equations such as \(y=2x-1\) and \(y=-x+8\) are simultaneous equations: the aim is to find the \(x\) and \(y\) that make both true at the same time.

Solving graphically means drawing both lines on the same axes (from a table of values or a graphing app). The point of intersection lies on both lines, so it is the solution. Check it by substituting into both equations.

Different gradients mean the lines cross once — one solution. The same gradient with a different \(y\)-intercept gives parallel lines — no solution. If both equations give the same line there are infinitely many solutions.

Two lines crossing at (3,5)y=2x-1 and y=-x+8 meet at the point (3,5), the solution of the pair x y 1 2 3 4 5 6 2 4 6 8 10 y = 2x − 1 y = −x + 8 (3, 5)
The lines cross at \((3,5)\): \(2(3)-1=5\) and \(-3+8=5\).
Parallel linesy=3x+1 and y=3x-2 have the same gradient and never meet: no solution x y -2 -1 1 2 -6 -3 3 6 9 y = 3x + 1 y = 3x − 2
Same gradient, different intercepts: parallel, so no solution.

Write each line in gradient–intercept form:

\[y = mx + c\]
y=mx+c

At the solution both lines give the same \(y\)-value for the same \(x\):

\[m_1 x + c_1 = m_2 x + c_2\]
m1x+c1=m2x+c2
Gradients and interceptsGraphNumber of solutions
Different gradientsLines cross onceOne solution
Same gradient, different interceptsParallel linesNo solution
Same gradient, same interceptThe same lineInfinitely many
From a table. The solution is the \(x\)-value where both rows have the same \(y\)-value.

How to solve simultaneous equations graphically

  1. Rearrange each equation into the form \(y=mx+c\) if needed.
  2. Graph both lines on the same axes using a table of values or a graphing app.
  3. Read the point where the lines cross and write it as \((x,y)\).
  4. Check the point in both equations — each must give the same \(y\).
Example 1 — Read the solution
The graph shows \(y=x-3\) and \(y=-x+5\). Solve the pair graphically.
Solution

The lines cross at \((4,1)\). Check in both equations.

Example 1y=x-3 and y=-x+5 cross at (4,1) x y 1 2 3 4 5 6 7 -3 -2 -1 1 2 3 4 5 6 y = −x + 5 y = x − 3
\(y=x-3:\ 4-3\)\(=\)\(1\ \checkmark\)
\(y=-x+5:\ -4+5\)\(=\)\(1\ \checkmark\)
(4,1)

The solution is \(x=4,\ y=1\), the point \((4,1)\).

Example 2 — From a table of values
Use the table to solve \(y=2x+1\) and \(y=-x+7\).
\(x\)0123
\(y=2x+1\)1357
\(y=-x+7\)7654
Solution

Look for the column where both rows match.

\(\text{At } x=2:\ 2(2)+1\)\(=\)\(5\)
\(\text{and } -2+7\)\(=\)\(5\)
(2,5)

Both lines pass through \((2,5)\), so the solution is \((2,5)\).

Example 3 — How many solutions?
How many solutions does each pair have?
(a) \(y=3x+1\) and \(y=3x-2\)
(b) \(y=2x+3\) and \(3y=6x+9\)
Solution

Compare gradients and intercepts. (a) Both gradients are 3, intercepts 1 and \(-2\): the lines are parallel and never meet. (b) Divide \(3y=6x+9\) by 3.

Example 3 (a)Parallel lines y=3x+1 and y=3x-2 x y -2 -1 1 2 -6 -3 3 6 9 y = 3x + 1 y = 3x − 2
\(3y\)\(=\)\(6x+9\)
\(y\)\(=\)\(2x+3\)
y=2x+3

(b) is the same line as the first. So (a) no solution; (b) infinitely many solutions.

Example 4 — Check a claimed solution
Kai says \((4,3)\) is the solution of \(y=2x-5\) and \(y=-x+6\). Is he right?
Solution

Substitute \(x=4\) into both equations.

\(y=2x-5:\ 2(4)-5\)\(=\)\(3\ \checkmark\)
\(y=-x+6:\ -4+6\)\(=\)\(2\)
\(2\)\(\ne\)\(3\)
2≠3

\((4,3)\) is on the first line only, so Kai is not right — the solution must satisfy both equations.

Common pitfalls

The solution is a point. Give both coordinates, \(x\) first: \((3,5)\), not just \(x=3\).
Check in both equations. A point that fits only one equation is on only one line, so it is not the solution.
Rearrange before comparing. \(2y=2x+6\) is the same line as \(y=x+3\) (infinitely many solutions), while equal gradients with different intercepts are parallel (no solution).

Frequently asked questions

How do you solve simultaneous equations graphically?

Draw both straight lines on the same axes and read the point where they cross. That point lies on both lines, so its x and y values solve both equations at once.

How do you check the solution of simultaneous equations?

Substitute the x-value of the crossing point into both equations. If both give the same y-value as the point, the solution is correct.

How do you find the solution from a table of values?

Look for the x-value where both rows of the table have the same y-value. That pair of x and y values is the solution.

What does it mean if two lines are parallel?

Parallel lines have the same gradient but different y-intercepts. They never meet, so the pair of equations has no solution.

When do simultaneous equations have infinitely many solutions?

When both equations describe the same line. For example 3y equals 6x plus 9 simplifies to y equals 2x plus 3, so every point on that line is a solution.