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Year 8 Maths Graphing Linear Equations

Solving Equations Graphically

20 practice questions 1 video lesson Theory + worked examples
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Theory

Every point on a line fits its rule, so a graph can solve equations. To solve \(1 - 3x = 7\), graph \(y = 1 - 3x\), draw the horizontal line \(y = 7\), and read the \(x\)-value where they cross: \(x = -2\). This Year 8 maths page shows how.

Finding \(y\) from \(x\). Start at the \(x\)-value on the \(x\)-axis, go straight up or down to the line, then straight across to the \(y\)-axis. (Or draw the vertical line through that \(x\)-value and read where it meets the graph.)

Finding \(x\) from \(y\). Start at the \(y\)-value, go straight across to the line, then up or down to the \(x\)-axis.

Extend the line past the table points so values outside the table can still be read.

Is a point on the line? Substitute its \(x\)-value into the rule. If the rule gives its \(y\)-value, it is on the line.

Solving an equation. The left side is the rule, already drawn as a line. The right side is a number, so draw that horizontal line. Read the \(x\)-value where the two lines cross. To solve \(2x + 2 = -2\), draw \(y = -2\).

The \(x\)-axis is \(y = 0\), so where the line crosses it solves “rule \(= 0\)”. A crossing halfway between \(2\) and \(3\) gives \(x = 2.5\).

Always check by substituting the answer into the rule: a graph can only be read approximately.

Solving 2x + 2 = βˆ’2 The line y = 2x + 2 and the horizontal line y = -2. They cross at (-2, -2), so x = -2 solves 2x + 2 = -2. y = 2x + 2 xy −4−3−2−11234−4−3−2−1123456780 y = βˆ’2 (βˆ’2, βˆ’2)
The horizontal line \(y = -2\) meets \(y = 2x + 2\) at \((-2, -2)\), so \(x = -2\). Check: \(2 \times (-2) + 2 = -2\).

Table of values for \(y = 2x + 2\):

\(x\)\(-1\)\(0\)\(1\)\(2\)
\(y\)\(0\)\(2\)\(4\)\(6\)
To solveDraw
\(2x + 2 = -2\)\(y = -2\)
\(2x + 2 = 7\)\(y = 7\)
\(2x + 2 = 0\)nothing: use the \(x\)-axis
The solution is the x-value. The crossing point is \((-2, -2)\), but the answer to \(2x + 2 = -2\) is \(x = -2\).
The horizontal line comes from the right side. To solve \(2x + 2 = 7\), draw \(y = 7\), not \(x = 7\).

Solving an equation with a graph

  1. Graph the left side as a line \(y = \ldots\).
  2. Draw the horizontal line \(y = \) (right side).
  3. Read the \(x\)-value of the crossing point.
  4. Check by substituting into the rule.
Example 1 — y from x
Graph \(y = 2x + 1\) and use it to find \(y\) when \(x = -1\).
y = 2x + 1 and x = βˆ’1 The line y = 2x + 1 and the vertical line x = -1, crossing at (-1, -1). xy −3−2−1123−3−2−112340 (βˆ’1, βˆ’1)
Solution

Plot \((0, 1)\) and \((1, 3)\) and join them. Draw the vertical line \(x = -1\); it meets the graph at \((-1, -1)\).

\(2 \times (-1) + 1\)\(=\)\(-1\)
y =−1

\(y = -1\)

Example 2 — Solving with a horizontal line
Solve \(1 - 3x = 7\) graphically.
1 βˆ’ 3x = 7 The line y = 1 - 3x and the horizontal line y = 7, crossing at (-2, 7). xy −3−2−1123−2−1123456780 (βˆ’2, 7)
Solution

Graph \(y = 1 - 3x\) through \((0, 1)\) and \((-1, 4)\). Draw \(y = 7\). The lines cross at \((-2, 7)\).

\(1 - 3 \times (-2)\)\(=\)\(7\)
x =−2

\(x = -2\)

Example 3 — Reading both ways
The graph of \(y = 4 - x\) is drawn.
(i) Find \(y\) when \(x = 6\).
(ii) Find \(x\) when \(y = 3\).
y = 4 βˆ’ x The line y = 4 - x. At x = 6 the line is at y = -2; at y = 3 it is at x = 1. xy −2−11234567−3−2−11234560
Solution
\(4 - 6\)\(=\)\(-2\)
\(4 - 1\)\(=\)\(3\)
y =−2 x =1

(i) \(y = -2\) (go down from \(6\) to the line, then across)
(ii) \(x = 1\) (go across from \(3\) to the line, then down)

Example 4 — Choosing the second line
The graph of \(y = 4x + 2\) is drawn.
(i) Which horizontal line solves \(4x + 2 = -6\)?
(ii) Solve \(4x + 2 = -6\).
Solution

The right side is \(-6\), so draw \(y = -6\). It crosses the graph at \((-2, -6)\).

\(4 \times (-2) + 2\)\(=\)\(-6\)
y =−6 x =−2

(i) \(y = -6\)
(ii) \(x = -2\)

Example 5 — The lemonade stall
A stall starts with \(40\) L of lemonade and sells \(8\) L each hour, so \(L = 40 - 8h\).
(i) Use the graph to solve \(40 - 8h = 12\).
(ii) When does the lemonade run out?
Lemonade left A line from 40 litres at h = 0 down to 0 litres at h = 5, and a horizontal line L = 12 that meets it at h = 3.5. 0 20 40 0 1 2 3 4 5 6 h (hours) L (litres)
Solution
\(40 - 8 \times 3.5\)\(=\)\(12\)
\(40 - 8 \times 5\)\(=\)\(0\)
h =3.55 hours

(i) \(h = 3.5\) hours (halfway between \(3\) and \(4\))
(ii) After \(5\) hours, where the line meets the \(h\)-axis

Example 6 — Three questions, one graph
The graph of \(y = 2x - 6\) is drawn.
(i) Solve \(2x - 6 = 3\).
(ii) Solve \(2x - 6 = 0\).
(iii) Is \((5, 5)\) on the line?
Solution
\(2 \times 4.5 - 6\)\(=\)\(3\)
\(2 \times 3 - 6\)\(=\)\(0\)
\(2 \times 5 - 6\)\(=\)\(4\)
x =4.5 x =3 no

(i) \(x = 4.5\) (\(y = 3\) meets the line halfway between \(4\) and \(5\))
(ii) \(x = 3\), where the line meets the \(x\)-axis
(iii) No: the line passes through \((5, 4)\).

Common pitfalls

Giving the point instead of the x-value. The solution of \(2x + 2 = -2\) is \(x = -2\), not \((-2, -2)\).
Drawing the wrong extra line. For rule \(= 7\), draw \(y = 7\).
Reading by eye. Go straight along the grid lines, then substitute to check.
Not extending the line. An answer may lie beyond the table points.

Frequently asked questions

How do I solve an equation using a graph?

Graph the left side as a line, draw the horizontal line y = (right side), and read the x-value where they cross.

How do I find y for a given x on a graph?

Go from the x-value straight up or down to the line, then across to the y-axis.

What does the x-axis crossing tell me?

It solves rule = 0, because every point on the x-axis has y = 0.

How do I know if a point is on a line?

Substitute its x-value into the rule. It is on the line only if the rule gives its y-value.

Why should I check a graphical answer?

Graphs can only be read approximately, so substitute the answer into the rule.