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Year 9 Maths Linear Relations

Introduction to Linear Relations

20 practice questions 2 video lessons Theory + worked examples
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  • Introduction to Linear Relations - Video - Plotting Points on the Coordinate Plane Watch
  • Introduction to Linear Relations - Video - Write a Linear Equation Given a Table Watch
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Theory

A linear relation links two variables so that equal steps in one give equal steps in the other, and its points lie on a straight line. This Year 9 maths page revises the number plane, builds a table of values from a rule, finds the rule from a table, and plots the points to draw the line.

The number plane (Cartesian plane) has a horizontal \(x\)-axis and a vertical \(y\)-axis that cross at the origin \((0, 0)\). The axes split the plane into four quadrants.

A point is written as \((x, y)\). The \(x\)-coordinate comes first and says how far across (right is positive, left is negative). The \(y\)-coordinate says how far up or down (up is positive).

A rule is an equation that links two variables, such as \(y = 2x + 1\). Each \(x\)-value you put in gives one \(y\)-value out. A table of values lists some of these pairs; each pair is a point to plot.

A relation is linear when \(y\) changes by the same amount every time \(x\) goes up by \(1\). Then all its points lie on a straight line. In the rule \(y = mx + c\), that change is \(m\), the number multiplying \(x\), and \(c\) is the value of \(y\) when \(x = 0\).

Every point on the line fits the rule, including points between and beyond the ones in the table. That is why you can extend the line to read off new values.

Four points on the number planeThe points A(3, 2), B(-2, 4), C(-4, -1) and D(2, -3), one in each quadrant. Dashed lines show the path to A: 3 across to the right, then 2 up. x y βˆ’5 βˆ’4 βˆ’3 βˆ’2 βˆ’1 1 2 3 4 5 βˆ’4 βˆ’3 βˆ’2 1 2 3 4 5 A(3, 2) B(βˆ’2, 4) C(βˆ’4, βˆ’1) D(2, βˆ’3)
Across first, then up or down. \(A(3, 2)\) is \(3\) to the right and \(2\) up; \(C(-4, -1)\) is \(4\) to the left and \(1\) down.
Points from a table of values for y = 2x - 1The points (-1, -3), (0, -1), (1, 1), (2, 3) and (3, 5) from a table of values lie on one straight line. A dashed step from (1, 1) goes 1 across and 2 up to (2, 3). x y βˆ’2 βˆ’1 1 3 4 2 4 6 βˆ’4 1 2 y = 2x βˆ’ 1
For \(y = 2x - 1\), each step of \(1\) across is a step of \(2\) up, so the points line up. The \(2\) in the rule is that step.

When the \(x\)-values go up in steps of \(1\):

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

where \(m\) is the change in \(y\) for each step of \(1\) in \(x\), and \(c\) is the value of \(y\) when \(x = 0\).

Pattern in the tableRule
\(y\) goes up \(3\) each step; \(y = 1\) when \(x = 0\)\(y = 3x + 1\)
\(y\) goes down \(4\) each step; \(y = 9\) when \(x = 0\)\(y = 9 - 4x\)
\(y\) is always \(4\) more than \(x\)\(y = x + 4\)
Steps that are not 1. If \(x\) goes up in steps of \(2\), divide the change in \(y\) by \(2\). For \(x = 0, 2, 4\) and \(y = 5, 11, 17\), \(y\) goes up \(6\) for each step of \(2\), which is \(3\) for each \(1\). The rule is \(y = 3x + 5\).
Negative and decimal inputs. Put a negative value in brackets when you substitute. For \(y = 4x - 3\): at \(x = -2\), \(y = 4 \times (-2) - 3 = -11\); at \(x = 1.5\), \(y = 4 \times 1.5 - 3 = 3\).
Is a point on the line? Substitute its \(x\)-value into the rule. If you get its \(y\)-value, the point is on the line. For \(y = 2x - 1\), the point \((5, 9)\) is on the line because \(2 \times 5 - 1 = 9\).
Graphs of real situations. The horizontal axis is usually time, and the vertical axis is the amount. A line that rises shows an amount growing at a steady rate; a line that falls shows it shrinking. Where a falling line meets the horizontal axis, the amount has run out.

How to complete a table of values

  1. Substitute each \(x\)-value into the rule, using brackets for negatives.
  2. Work out \(y\) and write it under its \(x\)-value.
  3. Check that the \(y\)-values change by the same amount each step.

How to find the rule from a table

  1. Find the step: how much \(y\) changes when \(x\) goes up by \(1\). This is the number multiplying \(x\) (negative if \(y\) goes down).
  2. Find the start: the value of \(y\) when \(x = 0\). Work backwards one step if the table does not show \(x = 0\).
  3. Write \(y = (\text{step})\,x + (\text{start})\) and test it on another pair.

How to graph the relation

  1. Plot each pair as a point \((x, y)\), across first.
  2. Rule a straight line through the points and extend it past both ends.
Example 1 β€” Completing a table
Complete the table of values for \(y = 3x - 4\).
\(x\)\(-2\)\(-1\)\(0\)\(1\)\(3\)
\(y\)\(a\)\(b\)\(c\)\(d\)\(e\)
Solution

Substitute each \(x\)-value. Put the negative values in brackets.

\(a\)\(=\)\(3 \times (-2) - 4 = -10\)
\(b\)\(=\)\(3 \times (-1) - 4 = -7\)
\(c\)\(=\)\(3 \times 0 - 4 = -4\)
\(d\)\(=\)\(3 \times 1 - 4 = -1\)
\(e\)\(=\)\(3 \times 3 - 4 = 5\)
y=3x−4
\(x\)\(-2\)\(-1\)\(0\)\(1\)\(3\)
\(y\)\(-10\)\(-7\)\(-4\)\(-1\)\(5\)

From \(x = -2\) to \(x = 1\) the \(y\)-values go up \(3\) each step, as the rule says. The jump from \(x = 1\) to \(x = 3\) is two steps, so \(y\) goes up \(6\).

Example 2 β€” Finding the rule
The table follows a pattern.
\(x\)\(0\)\(1\)\(2\)\(3\)\(4\)
\(y\)\(7\)\(4\)\(1\)\(-2\)\(-5\)

(i) Find the rule linking \(x\) and \(y\).
(ii) Find \(y\) when \(x = 10\).
Solution

(i) Each time \(x\) goes up \(1\), \(y\) goes down \(3\), so the number multiplying \(x\) is \(-3\). When \(x = 0\), \(y = 7\).

\(y\)\(=\)\(-3x + 7\)
\(\)\(=\)\(\boldsymbol{7 - 3x}\)

Test it on another pair: \(x = 3\) gives \(7 - 9\) \(= -2\ \checkmark\)

(ii) Substitute \(x = 10\).

\(y\)\(=\)\(7 - 3 \times 10\)
\(\)\(=\)\(\mathbf{-23}\)
y=7−3x
Example 3 β€” Plotting and reading a graph

(i) Complete a table of values for \(y = 4 - x\) for \(x = 0\) to \(5\), and graph the line.
(ii) Where does the line cross each axis?
(iii) Does the point \((6, -2)\) lie on the line?
Solution

(i) Substitute each \(x\)-value: \(4 - 0 = 4\), \(4 - 1 = 3\), and so on.

\(x\)\(0\)\(1\)\(2\)\(3\)\(4\)\(5\)
\(y\)\(4\)\(3\)\(2\)\(1\)\(0\)\(-1\)
Graph of y = 4 - x from a table of valuesThe points (0, 4), (1, 3), (2, 2), (3, 1), (4, 0) and (5, -1) lie on the straight line y = 4 - x. The line crosses the y-axis at (0, 4) and the x-axis at (4, 0). The point (6, -2) is also on the line. x y βˆ’2 βˆ’1 1 2 3 7 βˆ’3 1 2 3 6 (0, 4) (4, 0) (6, βˆ’2)

(ii) The line crosses the \(y\)-axis at \(\mathbf{(0, 4)}\) and the \(x\)-axis at \(\mathbf{(4, 0)}\). Both points are in the table.

(iii) Substitute \(x = 6\).

\(y\)\(=\)\(4 - 6 = -2\ \checkmark\)
y=4−x

The rule gives \(-2\), the \(y\)-coordinate of the point, so \((6, -2)\) is on the line. Extending the graph shows the same thing.

Example 4 β€” A real situation
Jai has $15 saved and adds $10 every week. Let \(S\) be his savings in dollars after \(w\) weeks.
(i) Make a table for \(w = 0\) to \(4\) and write a rule for \(S\).
(ii) After how many weeks will he have $95?
Solution

(i) Start at \(15\) and add \(10\) each week.

\(w\)\(0\)\(1\)\(2\)\(3\)\(4\)
\(S\)\(15\)\(25\)\(35\)\(45\)\(55\)

\(S\) goes up \(10\) each week and starts at \(15\), so the rule is \(\boldsymbol{S = 10w + 15}\).

(ii) Put \(S = 95\) and solve.

\(10w + 15\)\(=\)\(95\)
\(10w\)\(=\)\(80\)
\(w\)\(=\)\(\mathbf{8}\)
S=10w+15
Savings against weeksJai's savings: a straight line from $15 at week 0 rising $10 each week, through the plotted points for weeks 0 to 4. A dashed guide shows that the line reaches $95 at week 8. w S 1 2 3 4 5 6 7 9 20 40 60 80 100 (8, 95) 8

Jai has $95 after \(8\) weeks. The graph gives the same answer: across from $95 to the line, then down to week \(8\).

Common pitfalls

Across first, then up. The point \((3, 2)\) is \(3\) across and \(2\) up. Plotting it at \(2\) across and \(3\) up gives the point \((2, 3)\) instead.
Brackets for negatives. For \(y = 3x - 4\) at \(x = -2\), work out \(3 \times (-2) - 4 = -10\). Dropping the sign gives \(2\), which is wrong.
Check the step in x. If the \(x\)-values jump by \(2\), the change in \(y\) covers two steps. Halve it to get the number multiplying \(x\).
Test the rule on more than one pair. A rule that fits the first pair may not fit the others. \(y = x + 6\) and \(y = 8 - x\) both give \(y = 7\) at \(x = 1\), but only one of them fits a whole table.

Frequently asked questions

What is a linear relation?

A relation between two variables where \(y\) changes by the same amount each time \(x\) goes up by \(1\). Its points lie on a straight line, and its rule can be written as \(y = mx + c\).

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

Find how much \(y\) changes when \(x\) goes up by \(1\); that is the number multiplying \(x\). Then find \(y\) when \(x = 0\); that is the number added on. Test the rule on another pair.

How do you plot the point (x, y)?

Start at the origin. Move across by the \(x\)-coordinate (right if positive, left if negative), then up or down by the \(y\)-coordinate.

How can you tell if a table gives a straight line?

Check that the \(x\)-values go up in equal steps and the \(y\)-values change by the same amount each step. If they do, the points lie on a straight line.

How many points do you need to draw a straight line?

Two points fix a straight line. Plot a third as a check: if it is not in line with the other two, there is a mistake.