Graphs of Linear Equations in Two Variables

Draw the graph of $ax + by = c$ from a table of values or from its intercepts, to a given scale and range; read values from it; tell whether a point lies on it; and draw it for a real setting.

Before you start

In Secondary 1 you drew straight-line graphs, such as the graph of $y = 2x + 1$ , from a table of values: work out $y$ for some values of $x$ , plot the points and join them with a ruler.
In lesson 2.1 the solutions of $2x + 3y = 12$ , plotted as points, lay on one straight line.

Drawing the Graph from a Table of Values

The graph of ax + by = c

The graph of a linear equation $ax + by = c$ is a straight line. Every point on the line is a solution of the equation, and every solution is a point on the line.

To draw it from a table of values:

(1) make $y$ the subject, $y = \frac{c - ax}{b}$ ;

(2) choose at least three values of $x$ across the given range, if possible ones that give whole-number values of $y$ , and work out $y$ for each;

(3) plot the points, draw a straight line through them with a ruler across the given range, and label it with its equation.

Example 1: Drawing 3x + 2y = 6 from a table

Draw the graph of $3x + 2y = 6$ for values of $x$ from $-2$ to $4$ .

Solution

Make $y$ the subject: $2y = 6 - 3x$ , so $y = \frac{6 - 3x}{2}$ .

An odd value of $x$ gives a half ( $x = 1$ gives $y = 1.5$ ), so choose the even values $-2$ , $0$ , $2$ and $4$ , which include both ends of the range:

$x = -2$ : $y = \frac{6 + 6}{2} = 6$ . $x = 0$ : $y = \frac{6}{2} = 3$ . $x = 2$ : $y = \frac{6 - 6}{2} = 0$ . $x = 4$ : $y = \frac{6 - 12}{2} = -3$ .

$x$	$-2$	$0$	$2$	$4$
$y$	$6$	$3$	$0$	$-3$

Plot the four points. They lie on one straight line, which checks the table. Draw the line from $x = -2$ to $x = 4$ .

The graph is the straight line through $(-2, 6)$ and $(4, -3)$ , passing through $(0, 3)$ and $(2, 0)$ .

Drawing the Graph from the Intercepts

The x-intercept and the y-intercept

The $x$ -intercept is the $x$ -coordinate of the point where the line crosses the $x$ -axis. Every point on the $x$ -axis has $y = 0$ , so put $y = 0$ and solve for $x$ .

The $y$ -intercept is the $y$ -coordinate of the point where the line crosses the $y$ -axis. Every point on the $y$ -axis has $x = 0$ , so put $x = 0$ and solve for $y$ .

Plot the two points, check with a third point, and draw the line.

Example 2: Drawing 2x + 3y = 12 from its intercepts

Find the $x$ -intercept and the $y$ -intercept of $2x + 3y = 12$ . Use them to draw its graph for $x$ from $-3$ to $9$ , and check with a third point.

Solution

$x$ -intercept: put $y = 0$ . $2x + 3(0) = 12$ , so $2x = 12$ and $x = 6$ . The line crosses the $x$ -axis at $(6, 0)$ .

$y$ -intercept: put $x = 0$ . $2(0) + 3y = 12$ , so $3y = 12$ and $y = 4$ . The line crosses the $y$ -axis at $(0, 4)$ .

Third point: put $x = 3$ . $6 + 3y = 12$ , so $3y = 6$ and $y = 2$ , giving $(3, 2)$ .

Ends of the range: $x = -3$ gives $-6 + 3y = 12$ , $y = 6$ ; $x = 9$ gives $18 + 3y = 12$ , $y = -2$ .

The check point $(3, 2)$ lies on the line through the two intercepts, so the line is right.

The $x$ -intercept is $6$ and the $y$ -intercept is $4$ ; the graph is the line through $(6, 0)$ and $(0, 4)$ .

The intercepts are quick to find, but they are not always good points to plot.

When two intercepts are not enough

The line passes through the origin ( $c = 0$ , as in $3x - 2y = 0$ ): both intercepts are $0$ , so they give the one point $(0, 0)$ . You need other points from a table.

The intercepts are awkward fractions, as in $2x + 3y = 7$ : points with whole-number coordinates from a table are easier to plot accurately.

$a = 0$ or $b = 0$ , as in $y = 3$ or $x = -2$ : the line crosses only one axis (lesson 2.4).

Example 3: Fractional intercepts

(a) Find the intercepts of $5x - 4y = 10$ . (b) Find the intercepts of $2x + 3y = 7$ , then draw its graph for $x$ from $-1$ to $5$ .

Solution

(a) $y = 0$ : $5x = 10$ , so $x = 2$ . $x = 0$ : $-4y = 10$ , so $y = -\frac{10}{4} = -2.5\text{.}$ The intercepts are $2$ and $-2.5$ .

(b) $y = 0$ : $2x = 7$ , so $x = 3.5$ . $x = 0$ : $3y = 7$ , so $y = \frac{7}{3} = 2\frac{1}{3}$ .

The point $(0, 2\frac{1}{3})$ is hard to plot exactly, so make a table instead. $y = \frac{7 - 2x}{3}$ is a whole number when $x = -1$ , $2$ or $5$ :

$x$	$-1$	$2$	$5$
$y$	$3$	$1$	$-1$

The line crosses the axes where the intercepts said: at $3.5$ on the $x$ -axis and between $2$ and $3$ on the $y$ -axis.

(a) $x$ -intercept $2$ , $y$ -intercept $-2.5$ . (b) $x$ -intercept $3.5$ , $y$ -intercept $2\frac{1}{3}$ ; the graph is the line through $(-1, 3)$ , $(2, 1)$ and $(5, -1)$ .

Example 4: A line through the origin

Draw the graph of $3x - 2y = 0$ for $x$ from $-2$ to $4$ .

Solution

Putting $y = 0$ gives $x = 0$ , and putting $x = 0$ gives $y = 0$ . Both intercepts give the origin, one point only, so make a table.

Make $y$ the subject: $-2y = -3x$ , so $y = \frac{3x}{2}$ . Even values of $x$ give whole numbers:

$x$	$-2$	$0$	$2$	$4$
$y$	$-3$	$0$	$3$	$6$

The graph is the straight line through the origin, $(2, 3)$ and $(4, 6)$ , drawn from $(-2, -3)$ to $(4, 6)$ .

Drawing to a Given Scale, and Reading the Graph

Examination questions give the scale and the range. "2 cm to 1 unit on the $x$ -axis" means that each unit of $x$ is 2 cm long; on 2 mm graph paper that is 10 small squares. The two axes may have different scales, so always read the numbers on an axis, not the squares.

To read from the graph, draw a guide line from the given value on one axis to the line, then across (or down) to the other axis.

Example 5: Exam style

Using a scale of 2 cm to 1 unit on the $x$ -axis and 1 cm to 1 unit on the $y$ -axis, draw the graph of $2x + y = 4$ for $-1 \le x \le 3$ . Use your graph to find (a) the value of $y$ when $x = 1.5$ , (b) the value of $x$ when $y = 3$ .

Solution

Make $y$ the subject: $y = 4 - 2x$ .

$x$	$-1$	$0$	$1$	$2$	$3$
$y$	$6$	$4$	$2$	$0$	$-2$

So the $x$ -axis runs from $-1$ to $3$ and the $y$ -axis from $-2$ to $6$ . In the drawing each square is 1 cm: one unit of $x$ is two squares, one unit of $y$ is one square.

(a) From $x = 1.5$ (three squares to the right of $0$ ) go up to the line, then across to the $y$ -axis: $y = 1$ . Check: $2(1.5) + 1 = 3 + 1 = 4\text{.}$

(b) From $y = 3$ go across to the line, then down to the $x$ -axis: one square to the right of $0$ , so $x = 0.5$ . Check: $2(0.5) + 3 = 1 + 3 = 4\text{.}$

(a) $y = 1$ (b) $x = 0.5$

Does a point lie on the line?

A point lies on the line $ax + by = c$ exactly when its coordinates make the equation true. Substitute them and see.

On a drawn graph, a point that is on the line sits on it; a point above or below the line is not a solution.

Example 6: Points on 2x + 3y = 12

Which of the points $P(1.5, 3)$ , $Q(4, 1)$ and $R(7.5, -1)$ lie on the line $2x + 3y = 12$ ? Show them on the graph.

Solution

$P$ : $2(1.5) + 3(3) = 3 + 9 = 12\text{.}$ On the line.

$Q$ : $2(4) + 3(1) = 8 + 3 = 11\text{.}$ Not $12$ , so not on the line.

$R$ : $2(7.5) + 3(-1) = 15 - 3 = 12\text{.}$ On the line.

On the graph, $Q$ is just below the line: at $x = 4$ the line has $y = \frac{12 - 8}{3} = 1\frac{1}{3}\text{.}$

$P$ and $R$ lie on the line; $Q$ does not.

A Graph from a Real Setting

In a real setting the letters stand for amounts, and only some of their values make sense. Draw only the part of the line that makes sense, and say what each reading from it means.

Example 7: Mixing coffee beans

A coffee shop mixes $x$ kg of beans costing $ $20$ per kg with $y$ kg of beans costing $ $25$ per kg. The beans cost $ $200$ altogether. (a) Show that $4x + 5y = 40$ . (b) Draw the graph of $4x + 5y = 40$ for the values that make sense. (c) Use your graph to find the mass of the $ $25$ beans when $2.5$ kg of the $ $20$ beans is used. (d) Use your graph to find $x$ when $y = 2$ .

Solution

(a) The $ $20$ beans cost $ $20x$ and the $ $25$ beans cost $ $25y$ , so $20x + 25y = 200$ . Divide every term by $5$ : $4x + 5y = 40$ .

(b) A mass cannot be negative, so $x \ge 0$ and $y \ge 0$ . Intercepts: $y = 0$ gives $4x = 40$ , so $x = 10$ ; $x = 0$ gives $5y = 40$ , so $y = 8$ . Draw the line from $(0, 8)$ to $(10, 0)$ only.

Third point: $x = 5$ gives $20 + 5y = 40$ , so $5y = 20$ and $y = 4$ , giving $(5, 4)$ , which lies on the line.

A mass need not be a whole number of kilograms, so every point of the line from $(0, 8)$ to $(10, 0)$ makes sense, and the line is drawn solid.

(c) From $x = 2.5$ , halfway between $2$ and $3$ , go up to the line, then across to the $y$ -axis: $y = 6$ . Check: $4(2.5) + 5(6) = 10 + 30 = 40\text{.}$

(d) From $y = 2$ go across to the line, then down to the $x$ -axis: halfway between $7$ and $8$ , so $x = 7.5$ . Check: $4(7.5) + 5(2) = 30 + 10 = 40\text{.}$

(a) $20x + 25y = 200$ , which is $4x + 5y = 40$ . (b) The line from $(0, 8)$ to $(10, 0)$ . (c) $6$ kg of the $ $25$ beans. (d) $x = 7.5$ : $7.5$ kg of the $ $20$ beans.

Common Mistakes

Mixing up the intercepts. Putting $x = 0$ gives the $y$ -intercept; putting $y = 0$ gives the $x$ -intercept.
Using the intercepts for a line through the origin. In $3x - 2y = 0$ both intercepts give $(0, 0)$ , one point only. Use a table.
Losing a sign when making $y$ the subject. From $x - 2y = 4$ : $-2y = 4 - x$ , and dividing both terms by $-2$ gives $y = \frac{x - 4}{2}$ , not $\frac{4 - x}{2}$ .
Counting squares instead of reading numbers. When the scales differ, one square may be half a unit on one axis and one unit on the other.
Drawing beyond the range, or beyond the values that make sense in a real setting, or not labelling the line with its equation.

Practice Questions

Draw each graph on graph paper before you open its solution. The questions get harder as you go.

Question 1

Find the $x$ -intercept and the $y$ -intercept of (a) $x + 2y = 8$ , (b) $3x - 4y = 12$ , (c) $4x + 6y = 9$ .

ShowHide solution

(a) $y = 0$ : $x = 8$ . $x = 0$ : $2y = 8$ , $y = 4$ .

(b) $y = 0$ : $3x = 12$ , $x = 4$ . $x = 0$ : $-4y = 12$ , $y = -3$ .

(a) $8$ and $4$ (b) $4$ and $-3$ (c) $2.25$ and $1.5$ ( $x$ -intercept first)

Question 2

Use the intercepts to draw the graph of $x - 2y = 4$ for $x$ from $-2$ to $6$ . Check with a third point.

ShowHide solution

$y = 0$ : $x = 4$ , giving $(4, 0)$ . $x = 0$ : $-2y = 4$ , $y = -2$ , giving $(0, -2)$ .

Third point: $x = 2$ gives $2 - 2y = 4$ , $-2y = 2$ , $y = -1$ , giving $(2, -1)$ .

Ends: $x = -2$ gives $-2 - 2y = 4$ , $y = -3$ ; $x = 6$ gives $6 - 2y = 4$ , $y = 1$ .

The line through $(4, 0)$ and $(0, -2)$ ; the check point $(2, -1)$ lies on it.

Question 3

Make $y$ the subject of $2x + 5y = 10$ , and draw its graph for $-5 \le x \le 10$ from a table of values.

ShowHide solution

$5y = 10 - 2x$ , so $y = \frac{10 - 2x}{5}$ . Multiples of $5$ give whole numbers:

$x$	$-5$	$0$	$5$	$10$
$y$	$4$	$2$	$0$	$-2$

The line from $(-5, 4)$ to $(10, -2)$ , through $(0, 2)$ and $(5, 0)$ .

Question 4

Explain why the intercepts alone cannot be used to draw the graph of $4x + 3y = 0$ . Then draw it for $x$ from $-3$ to $3$ .

ShowHide solution

Putting $y = 0$ gives $x = 0$ and putting $x = 0$ gives $y = 0$ : both intercepts give the single point $(0, 0)$ , and one point does not fix a line.

Make $y$ the subject: $3y = -4x$ , $y = -\frac{4x}{3}$ . Multiples of $3$ give whole numbers: $x = -3$ , $y = 4$ ; $x = 0$ , $y = 0$ ; $x = 3$ , $y = -4$ .

The line passes through the origin, so both intercepts are the same point. The graph is the line through $(-3, 4)$ , $(0, 0)$ and $(3, -4)$ .

Question 5

Using a scale of 1 cm to 1 unit on the $x$ -axis and 1 cm to 2 units on the $y$ -axis, draw the graph of $3x - y = 2$ for $-2 \le x \le 4$ . Use it to find (a) $y$ when $x = 2.5$ , (b) $x$ when $y = -5$ .

ShowHide solution

$y = 3x - 2$ .

$x$	$-2$	$0$	$2$	$4$
$y$	$-8$	$-2$	$4$	$10$

(a) Up from $x = 2.5$ to the line and across: one square on the $y$ -axis is $2$ units, and the line is reached three quarters of a square above $4$ , so $y = 4 + 1.5 = 5.5$ . Check: $3(2.5) - 5.5 = 7.5 - 5.5 = 2\text{.}$

(b) Across from $y = -5$ (halfway between $-4$ and $-6$ ) to the line and up: $x = -1$ . Check: $3(-1) - (-5) = -3 + 5 = 2\text{.}$

(a) $y = 5.5$ (b) $x = -1$

Question 6

Which of the points $A(1, 2)$ , $B(-2, 5)$ , $C(4, -1)$ and $D(5, -4)$ lie on the line $3x + 2y = 7$ ? (a) Find out by substituting. (b) Draw the line for $-2 \le x \le 6$ and mark the four points. Does the graph agree?

ShowHide solution

(a) $A$ : $3 + 4 = 7$ . Yes.

$B$ : $-6 + 10 = 4$ . No.

$C$ : $12 - 2 = 10$ . No.

$D$ : $15 - 8 = 7$ . Yes.

(b) Make $y$ the subject: $y = \frac{7 - 3x}{2}$ . When $x = -2$ , $y = 6.5$ ; when $x = 6$ , $y = -5.5$ . Join $(-2, 6.5)$ and $(6, -5.5)$ .

Yes: $A$ and $D$ are on the line. $B$ is below it (at $x = -2$ the line has $y = 6.5$ ), and $C$ is above it (at $x = 4$ the line has $y = -2.5$ ).

$A$ and $D$ lie on the line; $B$ and $C$ do not.

Question 7

Copy and complete the table of values for $x + 2y = 6$ . Using a scale of 1 cm to 1 unit on both axes, draw the graph for $-2 \le x \le 8$ . Use your graph to find (a) $y$ when $x = 3$ , (b) $x$ when $y = 3.5$ .

$x$	$-2$	$0$	$2$	$4$	$6$	$8$
$y$	$4$	?	$2$	?	$0$	?

ShowHide solution

$2y = 6 - x$ , so $y = \frac{6 - x}{2}$ : $x = 0$ gives $y = 3$ ; $x = 4$ gives $y = 1$ ; $x = 8$ gives $y = \frac{-2}{2} = -1$ .

$x$	$-2$	$0$	$2$	$4$	$6$	$8$
$y$	$4$	$3$	$2$	$1$	$0$	$-1$

(a) Up from $x = 3$ to the line, then across: halfway between $1$ and $2$ , so $y = 1.5$ . Check: $3 + 2(1.5) = 3 + 3 = 6\text{.}$

(b) Across from $y = 3.5$ , halfway between $3$ and $4$ , to the line, then down: $x = -1$ . Check: $-1 + 2(3.5) = -1 + 7 = 6\text{.}$

The missing values of $y$ are $3$ , $1$ and $-1$ . (a) $y = 1.5$ (b) $x = -1$

Question 8

A rectangle is $x$ cm long and $y$ cm wide, and its perimeter is $20$ cm. (a) Show that $x + y = 10$ . (b) Draw the graph of $x + y = 10$ for $1 \le x \le 9$ . (c) Use your graph to find the width when the length is $6.5$ cm. (d) Explain why the graph is not drawn for $x = 10$ or more.

ShowHide solution

(a) The perimeter is $x + y + x + y = 2x + 2y\text{,}$ so $2x + 2y = 20$ . Divide every term by $2$ : $x + y = 10$ .

(b) $y = 10 - x$ : $x = 1$ gives $y = 9$ ; $x = 5$ gives $y = 5$ ; $x = 9$ gives $y = 1$ . Draw the line from $(1, 9)$ to $(9, 1)$ .

(c) Up from $x = 6.5$ , halfway between $6$ and $7$ , to the line, then across: $y = 3.5$ . Check: $2(6.5) + 2(3.5) = 13 + 7 = 20\text{.}$

(d) When $x = 10$ , $y = 0$ , and a rectangle cannot have a width of $0$ . When $x$ is more than $10$ , $y$ is negative, and a width cannot be negative.

(a) $2x + 2y = 20$ , which is $x + y = 10$ . (c) $3.5$ cm (d) The width would be $0$ or negative.

Question 9

Mrs Tan spends exactly $ $60$ on $x$ kg of prawns at $ $15$ per kg and $y$ kg of fish at $ $10$ per kg. (a) Show that $3x + 2y = 12$ . (b) Using a scale of 2 cm to 1 kg on the $x$ -axis and 1 cm to 1 kg on the $y$ -axis, draw the graph for the values that make sense. (c) Use your graph to find how much fish she buys with $1$ kg of prawns. (d) Use your graph to find the mass of prawns she buys with $1.5$ kg of fish.

ShowHide solution

(a) The prawns cost $ $15x$ and the fish $ $10y$ , so $15x + 10y = 60$ . Divide every term by $5$ : $3x + 2y = 12$ .

(b) Neither mass can be negative. Intercepts: $y = 0$ gives $3x = 12$ , $x = 4$ ; $x = 0$ gives $2y = 12$ , $y = 6$ . Third point: $x = 2$ gives $6 + 2y = 12$ , $y = 3$ . Draw the line from $(0, 6)$ to $(4, 0)$ ; a mass need not be a whole number, so the line is solid.

(c) From $x = 1$ (two squares to the right of $0$ ) go up to the line, then across: halfway between $4$ and $5$ , so $y = 4.5$ . Check: $3(1) + 2(4.5) = 3 + 9 = 12\text{.}$

(d) From $y = 1.5$ , halfway between $1$ and $2$ , go across to the line, then down: six squares, which is $3$ kg, so $x = 3$ . Check: $3(3) + 2(1.5) = 9 + 3 = 12\text{.}$

(a) $15x + 10y = 60$ , which is $3x + 2y = 12$ . (c) $4.5$ kg of fish (d) $3$ kg of prawns

Key Points

The graph of $ax + by = c$ is a straight line; its points are exactly the solutions of the equation.
From a table: make $y$ the subject, choose values of $x$ that give whole numbers, plot at least three points.
The $x$ -intercept: put $y = 0$ . The $y$ -intercept: put $x = 0$ .
Two intercepts fix the line only when they are two different points; for a line through the origin, or awkward fractions, use a table. Always check with a third point.
With a given scale, read the numbers on each axis, not the squares.
To read the graph, go from the given value on one axis to the line, then across (or down) to the other axis.
A point is on the line exactly when its coordinates make the equation true.
In a real setting, draw only the part of the line that makes sense.