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Year 9 Maths Core Straight Line

Solving Linear Equations Graphically

20 practice questions 1 video lesson Theory + worked examples

An equation such as \(2x - 1 = 5\) can be solved with graphs: draw \(y = 2x - 1\) and \(y = 5\), and the \(x\)-coordinate where they cross is the solution. This NSW Year 9 Core maths subtopic revises and extends the Stage 4 graphing work under NESA outcome MA4‑LIN‑C‑01.

You will split an equation into two lines, one for each side, choose the right pair of lines from a list, and read the solution as the \(x\)-coordinate of the crossing point, not the \(y\)-coordinate. You will solve equations where two sloping lines cross, as in \(2x - 3 = 6 - x\), and where the crossing falls between grid lines, see why parallel lines mean an equation has no solution, compare two phone plans to find when they cost the same, and find an unknown gradient. It is part of Year 9 Core Straight Line.

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Theory

To solve a linear equation graphically, graph each side as a straight line. The two lines cross at one point, and the \(x\)-coordinate of that point of intersection is the solution. This Year 9 maths page revises the earlier method, with sloping and horizontal lines, answers between grid lines, parallel lines and real cost comparisons.

An equation such as \(3x - 2 = x + 4\) asks: for which \(x\) do the two sides have the same value?

Graph the left-hand side as one line, \(y = 3x - 2\), and the right-hand side as another, \(y = x + 4\). Every point on a line satisfies its equation.

Where the lines cross, the point of intersection, both lines have the same \(y\)-value. So the \(x\)-coordinate of that point is the solution of the equation. The \(y\)-coordinate is just the value both sides equal there.

If one side is a plain number, such as \(8\), its line is horizontal: \(y = 8\).

If the two lines are parallel, they never cross, and the equation has no solution.

Solving x + 1 = 4 with two graphsThe line y = x + 1 and the horizontal line y = 4 cross at (3, 4). A dashed red line drops from the crossing point to x = 3 on the x-axis. x y βˆ’2 βˆ’1 1 2 4 5 6 βˆ’2 βˆ’1 1 2 3 5 6 y = x + 1 y = 4 (3, 4) x = 3
To solve \(x + 1 = 4\), graph \(y = x + 1\) and \(y = 4\). They cross at \((3, 4)\), so the solution is the \(x\)-coordinate: \(x = 3\).
Two parallel lines that never crossThe lines y = x + 2 and y = x - 1 have the same gradient, so they are parallel and never meet. x y βˆ’3 βˆ’2 βˆ’1 1 2 3 4 βˆ’4 βˆ’3 βˆ’2 βˆ’1 1 2 3 4 5 6 y = x + 2 y = x βˆ’ 1
Lines with the same gradient never cross. So \(x + 2 = x - 1\) has no solution.

To solve \(\text{left side} = \text{right side}\), graph:

\[y = \text{left side} \qquad \text{and} \qquad y = \text{right side}\]
y=left side,y=right side
\[\text{solution} = x\text{-coordinate of the point of intersection}\]
solution = x-coordinate of the intersection
Same gradient, no solution. \(y = 3x + 1\) and \(y = 3x - 4\) have the same gradient and different intercepts, so they are parallel. The equation \(3x + 1 = 3x - 4\) has no solution.
A crossing point used backwards. If two lines are known to cross where \(x = 2\), both equations give the same \(y\) there. For \(y = 4x - 3\) and \(y = kx + 1\): \(4 \times 2 - 3 = 5\), so \(2k + 1 = 5\) and \(k = 2\).
Verify. The intersection point must satisfy both equations. Substitute its \(x\) into each side; both must give its \(y\).

How to solve an equation graphically

  1. Split the equation: \(y = \) left side and \(y = \) right side. A plain number gives a horizontal line.
  2. Graph both lines on the same axes. Use a table of values or the gradient and \(y\)-intercept.
  3. Find the point where the lines cross.
  4. Read its \(x\)-coordinate. That is the solution. It may be a decimal such as \(2.5\).
  5. Check by substituting the solution into both sides of the original equation.
Example 1 β€” Choosing the lines
Which two lines would you graph to solve \(5x - 2 = 8\)? Use the graph to solve it.
Solution

Each side of the equation becomes a line, copied exactly. The right-hand side is a number, so its line is horizontal.

\(\text{left side}\)\(:\)\(y = 5x - 2\)
\(\text{right side}\)\(:\)\(y = 8\)
y=5x−2,y=8
y = 5x - 2 and y = 8The line y = 5x - 2 and the horizontal line y = 8 cross at (2, 8). x y βˆ’1 1 3 βˆ’3 βˆ’1 1 3 5 7 9 (2, 8) x = 2 y = 8

The lines cross at \((2, 8)\). The solution is the \(x\)-coordinate: \(\mathbf{x = 2}\). Check: \(5 \times 2 - 2 = 8\).

Example 2 β€” Two sloping lines
The lines \(y = 3x - 2\) and \(y = x + 4\) are graphed. Use them to solve \(3x - 2 = x + 4\).
Solution
y = 3x - 2 and y = x + 4The lines y = 3x - 2 and y = x + 4 cross at (3, 7). A dashed line drops to x = 3. x y βˆ’1 1 5 βˆ’3 βˆ’1 1 3 5 9 (3, 7) y = 3x βˆ’ 2 y = x + 4 x = 3

The lines cross at \((3, 7)\). The solution is its \(x\)-coordinate, \(x = 3\). The \(7\) is the value both sides take there.

\(3 \times 3 - 2\)\(=\)\(7\)
\(3 + 4\)\(=\)\(7 \ \checkmark\)
x=3

The solution is \(\mathbf{x = 3}\).

Example 3 β€” Between the grid lines
The lines \(y = 2x - 5\) and \(y = 2\) cross halfway between \(x = 3\) and \(x = 4\). Solve \(2x - 5 = 2\).
Solution
y = 2x - 5 and y = 2 crossing between grid linesThe line y = 2x - 5 meets the horizontal line y = 2 halfway between x = 3 and x = 4, at x = 3.5. x y βˆ’1 1 2 5 βˆ’5 βˆ’4 βˆ’3 βˆ’2 βˆ’1 1 2 3 4 5 3.5 y = 2

Read the crossing as \(x = 3.5\), then confirm it with algebra.

\(2x - 5\)\(=\)\(2\)
\(2x\)\(=\)\(7\)
\(x\)\(=\)\(\mathbf{3.5}\)
x=3.5

A crossing that misses a grid corner is normal. Give it as a decimal or a fraction, \(\dfrac{7}{2}\).

Example 4 β€” Comparing two plans
Gym A charges $15 to join plus $4 a visit: \(C = 4n + 15\). Gym B charges $35 to join plus $2 a visit: \(C = 2n + 35\). After how many visits do they cost the same?
Solution

Graph both lines on one set of axes and find where they cross. That solves \(4n + 15 = 2n + 35\).

Two gym plans comparedGym A, cost 4n + 15, and Gym B, cost 2n + 35, against the number of visits n. The lines cross at n = 10, where both cost 55 dollars. 0 2 4 6 8 10 12 14 20 40 60 80 (10, 55) A B visits n cost (dollars)
\(4n + 15\)\(=\)\(2n + 35\)
\(2n\)\(=\)\(20\)
\(n\)\(=\)\(\mathbf{10}\)
n=10

After 10 visits both cost \(4 \times 10 + 15 = 55\) dollars. For fewer visits Gym A is cheaper; for more, Gym B is cheaper.

Common pitfalls

The answer is the x-coordinate. The \(y\)-coordinate of the crossing point is the value both sides equal, not the solution.
A number gives y = …, not x = …. For \(5x - 2 = 8\), the second line is \(y = 8\), a horizontal line. \(x = 8\) would be vertical.
Copy each side exactly. Do not move terms across before graphing. Graph the left side and the right side as they are written.
Watch out for parallel lines. Lines with the same gradient never meet, so there is no solution. Two lines with different gradients always meet exactly once.

Frequently asked questions

How do you solve an equation using a graph?

Graph the left-hand side and the right-hand side as two lines on the same axes. Find where they cross. The \(x\)-coordinate of the crossing point is the solution.

Is the solution the x or the y value of the intersection?

The \(x\)-value. The \(y\)-value only tells you what both sides of the equation are equal to at the solution.

What does it mean if two lines do not intersect?

The lines are parallel, with the same gradient. The equation made from them has no solution.

What lines do you graph to solve 5x - 2 = 8?

Graph \(y = 5x - 2\) and the horizontal line \(y = 8\). They cross at \((2, 8)\), so \(x = 2\).

What if the lines cross between grid lines?

Read the \(x\)-coordinate as carefully as you can, such as \(3.5\), and confirm it by solving the equation with algebra.