Solving Simultaneous Equations by Graphs

What simultaneous equations are, why the point where two lines meet is their solution, and how to solve a pair of linear equations by drawing their graphs, to a given scale, including when the answer can only be estimated.

Before you start

To draw $ax + by = c$ , find two or three points from a table of values or the intercepts, and join them with a ruler (lesson 2.2).
Each point of the line is a solution of its equation, and each solution is a point of the line.
A vertical line $x = h$ and a horizontal line $y = k$ meet at $(h, k)$ (lesson 2.4).

What Are Simultaneous Equations?

One equation in $x$ and $y$ , such as $x + y = 5$ , has infinitely many solutions. If a second fact must also hold, such as $x - y = 1$ , only the pairs that make both equations true at the same time will do.

Simultaneous equations

Two equations in the same two unknowns that must both be true at the same time are called simultaneous equations.

Their solution is the pair of values of $x$ and $y$ that satisfies both equations. To check a pair, substitute it into each equation.

Example 1: Checking a pair in both equations

Which of the pairs $(4, 1)$ , $(3, 2)$ and $(2, 1)$ is the solution of the simultaneous equations $x + y = 5$ and $x - y = 1$ ?

Solution

$(4, 1)$ : $4 + 1 = 5$ , true; $4 - 1 = 3$ , not $1$ . It satisfies the first equation only.

$(3, 2)$ : $3 + 2 = 5$ , true; $3 - 2 = 1$ , true. It satisfies both.

$(2, 1)$ : $2 + 1 = 3$ , not $5$ ; $2 - 1 = 1$ , true. It satisfies the second equation only.

$(3, 2)$ is the solution: $x = 3$ and $y = 2$ .

The Graphical Method

Draw the graphs of both equations on the same axes. Every point of the first line satisfies the first equation, and every point of the second line satisfies the second. The point where the lines meet is on both lines, so it satisfies both equations: its coordinates are the solution. Two different straight lines that cross meet at exactly one point, so there is one solution. (Lesson 2.9 looks at lines that never meet, and at two equations that give the same line.)

Solving simultaneous equations by drawing graphs

(1) For each equation, find at least three points (a table of values, or the intercepts and a check point).

(2) Draw both lines on the same axes, with the same scale, and label each with its equation.

(3) Read the coordinates of the point where the lines meet.

(4) Check the values in both equations, and write the answer as $x = \ldots$ , $y = \ldots$

Example 2: Two lines in the form y = mx + c

Solve the simultaneous equations $y = x + 1$ and $y = -x + 5$ by drawing their graphs for $x$ from $0$ to $4$ .

Solution

Tables of values:

$x$	$0$	$2$	$4$
$y = x + 1$	$1$	$3$	$5$
$y = -x + 5$	$5$	$3$	$1$

Both tables have the pair $(2, 3)$ , so the lines meet there.

Check: in $y = x + 1$ , $3 = 2 + 1$ , true. In $y = -x + 5$ , $3 = -2 + 5$ , true.

$x = 2$ , $y = 3$

Example 3: Two lines in the form ax + by = c

By drawing the graphs of $x + 2y = 2$ and $2x - 3y = 11$ for $-2 \le x \le 7$ , solve the simultaneous equations.

Solution

$x + 2y = 2$ : the intercepts are easy. $y = 0$ gives $x = 2$ ; $x = 0$ gives $y = 1$ . At the ends, $x = -2$ gives $2y = 4$ , $y = 2$ , and $x = 7$ gives $2y = -5$ , $y = -2.5$ .

$2x - 3y = 11$ : the intercepts are $5.5$ and $-3\frac{2}{3}$ , which are awkward, so make a table. $y = \frac{2x - 11}{3}$ is a whole number when $x = -2, 1, 4, 7$ :

$x$	$-2$	$1$	$4$	$7$
$y$	$-5$	$-3$	$-1$	$1$

The lines meet at $(4, -1)$ . Check: $4 + 2(-1) = 4 - 2 = 2\text{,}$ true; $2(4) - 3(-1) = 8 + 3 = 11\text{,}$ true.

$x = 4$ , $y = -1$

Example 4: Exam style, with given scales

(a) 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 graphs of $y = 2x + 2$ and $x + y = 5$ for $-2 \le x \le 4$ . (b) Use your graphs to solve the simultaneous equations $y = 2x + 2$ and $x + y = 5$ .

Solution

(a) For $x + y = 5$ , make $y$ the subject: $y = 5 - x$ .

$x$	$-2$	$0$	$2$	$4$
$y = 2x + 2$	$-2$	$2$	$6$	$10$
$y = 5 - x$	$7$	$5$	$3$	$1$

The values of $y$ run from $-2$ to $10$ , so the $y$ -axis goes from $-2$ to $10$ , one square for every $2$ units.

(b) The lines meet where $x = 1$ and $y = 4$ (on the $y$ -axis, $4$ is two squares up). Check: $2(1) + 2 = 4$ , true; $1 + 4 = 5$ , true.

(b) $x = 1$ , $y = 4$

When the Lines Do Not Meet at a Grid Point

A graph gives an estimate

A drawn line has thickness, and a meeting point between grid lines must be read by eye. So the graphical method gives the solution only as accurately as the drawing allows, usually to $1$ decimal place on 2 mm graph paper.

Algebra gives the exact solution. The next lessons show how; Example 5 already does it in one step.

Example 5: Estimate, then confirm

Draw the graphs of $y = 2x - 1$ and $x + y = 4$ for $-1 \le x \le 4$ . Estimate the solution of the simultaneous equations to $1$ decimal place, then find it exactly.

Solution

$x$	$-1$	$0$	$2$	$4$
$y = 2x - 1$	$-3$	$-1$	$3$	$7$
$y = 4 - x$	$5$	$4$	$2$	$0$

At $x = 2$ the first line is above the second ( $3 > 2$ ), and at $x = 0$ it is below ( $-1 < 4$ ), so they cross between $x = 0$ and $x = 2$ .

Estimate. Each square is $0.5$ unit. The meeting point is a little past $1.5$ across (about $1.7$ ) and a little below $2.5$ up (about $2.3$ ).

Exactly. At the meeting point the two values of $y$ are equal. Put $y = 2x - 1$ into $x + y = 4$ : $x + (2x - 1) = 4$ , so $3x - 1 = 4$ , $3x = 5$ and $x = \frac{5}{3} = 1\frac{2}{3}$ . Then $y = 2 \times \frac{5}{3} - 1 = \frac{10}{3} - \frac{3}{3} = \frac{7}{3} = 2\frac{1}{3}\text{.}$

Check: $x + y = \frac{5}{3} + \frac{7}{3} = \frac{12}{3} = 4\text{,}$ true, and $y = 2x - 1$ gave $y$ from $x$ , so it is true too. To $1$ decimal place, $1\frac{2}{3} \approx 1.7$ and $2\frac{1}{3} \approx 2.3$ , which agrees with the estimate.

From the graph, $x \approx 1.7$ , $y \approx 2.3$ . Exactly, $x = 1\frac{2}{3}$ , $y = 2\frac{1}{3}$ .

More Uses of the Graphical Method

Example 6: Reading solutions from a drawn graph

The graph shows the lines $x + y = 6$ , $y = 2x$ and $y = x - 2$ . Use it to solve each pair of simultaneous equations. (a) $x + y = 6$ and $y = 2x$ (b) $x + y = 6$ and $y = x - 2$ (c) $y = 2x$ and $y = x - 2$

Solution

(a) $x + y = 6$ and $y = 2x$ meet at $(2, 4)$ . Check: $2 + 4 = 6$ ; $4 = 2 \times 2$ .

(b) $x + y = 6$ and $y = x - 2$ meet at $(4, 2)$ . Check: $4 + 2 = 6$ ; $2 = 4 - 2$ .

(a) $x = 2$ , $y = 4$ (b) $x = 4$ , $y = 2$ (c) $x = -2$ , $y = -4$

Example 7: With a vertical or horizontal line

Solve graphically (a) $x = 3$ and $x + y = 5$ , (b) $y = -2$ and $2x - y = 6$ .

Solution

(a) $x = 3$ is vertical. $x + y = 5$ passes through $(0, 5)$ and $(5, 0)$ . They meet at $(3, 2)$ .

Check: $3 + 2 = 5$ .

(b) $y = -2$ is horizontal. $2x - y = 6$ passes through $(3, 0)$ and $(0, -6)$ , and through $(4, 2)$ . They meet at $(2, -2)$ .

Check: $2(2) - (-2) = 4 + 2 = 6\text{.}$

(a) $x = 3$ , $y = 2$ (b) $x = 2$ , $y = -2$

Two graphs can also solve one equation in $x$ . At the point where $y = 2x + 1$ meets $y = 7 - x$ , the two values of $y$ are equal, so $2x + 1 = 7 - x$ there. The $x$ -coordinate of the meeting point is the solution of that equation.

Example 8: Solving one equation with two graphs

By drawing the graphs of $y = 2x + 1$ and $y = 7 - x$ for $0 \le x \le 4$ , solve the equation $2x + 1 = 7 - x$ .

Solution

$x$	$0$	$2$	$4$
$y = 2x + 1$	$1$	$5$	$9$
$y = 7 - x$	$7$	$5$	$3$

The lines meet at $(2, 5)$ , so the solution is the $x$ -coordinate, $x = 2$ . Check: $2(2) + 1 = 5$ and $7 - 2 = 5$ .

$x = 2$

A Graph from a Real Setting

Example 9: Two bicycle-hire shops

At East Coast Park, shop $A$ hires out bicycles for $ $6$ plus $ $3$ an hour, and shop $B$ for $ $12$ plus $ $1.50$ an hour. Let the cost for $t$ hours be $ $C$ . (a) Write an equation for each shop and draw both graphs for $0 \le t \le 8$ . (b) For how many hours do the two shops charge the same, and how much? (c) Which shop is cheaper for a short ride, and which for a long one?

Solution

(a) Shop $A$ : $C = 6 + 3t$ . Shop $B$ : $C = 12 + 1.5t$ .

$t$	$0$	$4$	$8$
Shop $A$ , $C = 6 + 3t$	$6$	$18$	$30$
Shop $B$ , $C = 12 + 1.5t$	$12$	$18$	$24$

(b) The lines meet at $(4, 18)$ . Check: $6 + 3(4) = 18$ and $12 + 1.5(4) = 12 + 6 = 18\text{.}$

(b) For 4 hours both shops charge $ $18$ . (c) Shop $A$ is cheaper for less than 4 hours; shop $B$ is cheaper for more than 4 hours.

Common Mistakes

Checking in one equation only. $(4, 1)$ satisfies $x + y = 5$ but not $x - y = 1$ . A solution must satisfy both.
Drawing the two lines on different axes or scales. Both lines go on the same pair of axes.
Reading the coordinates the wrong way round. Read $x$ on the horizontal axis first.
Reading squares instead of numbers when a square stands for $2$ units or half a unit.
Giving an estimate as if it were exact. Between grid lines, a graph gives about $1.7$ , not exactly $1\frac{2}{3}$ .
Giving both coordinates when one equation in $x$ was asked. For $2x + 1 = 7 - x$ the answer is $x = 2$ only.

Practice Questions

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

Question 1

(a) Is $(2, 5)$ the solution of $x + 2y = 12$ and $3x - y = 1$ ? (b) Is $(1, 3)$ the solution of $2x + y = 5$ and $x - y = 2$ ?

ShowHide solution

(a) $2 + 2(5) = 12$ , true; $3(2) - 5 = 1$ , true. Yes.

(b) $2(1) + 3 = 5$ , true; $1 - 3 = -2$ , not $2$ . No: it satisfies the first equation only.

(a) Yes (b) No

Question 2

Draw the graphs of $y = 2x + 1$ and $y = -x + 7$ for $-1 \le x \le 4$ , and use them to solve the simultaneous equations.

ShowHide solution

$x$	$-1$	$1$	$4$
$y = 2x + 1$	$-1$	$3$	$9$
$y = -x + 7$	$8$	$6$	$3$

The lines meet at $(2, 5)$ . Check: $2(2) + 1 = 5$ ; $-2 + 7 = 5$ .

$x = 2$ , $y = 5$

Question 3

Solve $2x + y = 1$ and $x - y = 5$ by drawing their graphs for $-1 \le x \le 4$ .

ShowHide solution

$2x + y = 1$ gives $y = 1 - 2x$ ; $x - y = 5$ gives $y = x - 5$ .

$x$	$-1$	$0$	$4$
$y = 1 - 2x$	$3$	$1$	$-7$
$y = x - 5$	$-6$	$-5$	$-1$

The lines meet at $(2, -3)$ . Check: $4 + (-3) = 1$ ; $2 - (-3) = 5$ .

$x = 2$ , $y = -3$

Question 4

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 graphs of $x + y = 3$ and $y = 3x - 5$ for $0 \le x \le 3$ . Use them to solve the simultaneous equations.

ShowHide solution

$x$	$0$	$1$	$2$	$3$
$y = 3 - x$	$3$	$2$	$1$	$0$
$y = 3x - 5$	$-5$	$-2$	$1$	$4$

Both tables have $(2, 1)$ , and the lines meet there. Check: $2 + 1 = 3$ ; $3(2) - 5 = 1$ .

$x = 2$ , $y = 1$

Question 5

Draw the graphs of $y = x + 2$ and $3x + 2y = 6$ for $-2 \le x \le 2$ . Estimate the solution to $1$ decimal place, then find it exactly as in Example 5.

ShowHide solution

$3x + 2y = 6$ has intercepts $2$ and $3$ ; with $x = -2$ , $2y = 12$ and $y = 6$ .

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

Estimate: a little under one square ( $0.5$ ) to the right of the $y$ -axis, about $0.4$ , and a little under $2.5$ up, about $2.4$ .

Exactly: put $y = x + 2$ into $3x + 2y = 6$ : $3x + 2(x + 2) = 6$ , so $5x + 4 = 6$ , $5x = 2$ , $x = 0.4$ ; then $y = 0.4 + 2 = 2.4$ .

Check: $3(0.4) + 2(2.4) = 1.2 + 4.8 = 6\text{.}$

$x = 0.4$ , $y = 2.4$ (here the estimate happens to be exact)

Question 6

Use the graph to solve each pair of simultaneous equations. (a) $x - y = -2$ and $2x + y = 8$ (b) $2x + y = 8$ and $y = 1$ (c) $x - y = -2$ and $y = 1$

ShowHide solution

(a) The first two lines meet at $(2, 4)$ . Check: $2 - 4 = -2$ ; $4 + 4 = 8$ .

(b) $2x + y = 8$ meets $y = 1$ halfway between $3$ and $4$ : $(3.5, 1)$ . Check: $7 + 1 = 8$ .

(a) $x = 2$ , $y = 4$ (b) $x = 3.5$ , $y = 1$ (c) $x = -1$ , $y = 1$

Question 7

Solve graphically (a) $x = -2$ and $3x + y = 1$ , (b) $y = 4$ and $x - 2y = -5$ .

ShowHide solution

(a) $3x + y = 1$ is $y = 1 - 3x$ : through $(-2, 7)$ , $(0, 1)$ and $(1, -2)$ . The vertical line $x = -2$ meets it at $(-2, 7)$ . Check: $-6 + 7 = 1$ .

(b) $x - 2y = -5$ has intercepts $-5$ and $2.5$ , and passes through $(3, 4)$ and $(1, 3)$ . The horizontal line $y = 4$ meets it at $(3, 4)$ . Check: $3 - 8 = -5$ .

(a) $x = -2$ , $y = 7$ (b) $x = 3$ , $y = 4$

Question 8

Solve the equation $x + 4 = 10 - x$ by drawing two graphs for $0 \le x \le 6$ .

ShowHide solution

Draw $y = x + 4$ (through $(0, 4)$ and $(6, 10)$ ) and $y = 10 - x$ (through $(0, 10)$ and $(6, 4)$ ).

They meet at $(3, 7)$ , so $x = 3$ . Check: $3 + 4 = 7$ and $10 - 3 = 7$ .

$x = 3$

Question 9

The lines $ax + y = 7$ and $x - y = 1$ meet at the point $(2, b)$ . Find $b$ and $a$ .

ShowHide solution

The point $(2, b)$ is on $x - y = 1$ : $2 - b = 1$ , so $b = 1$ .

It is also on $ax + y = 7$ : $a(2) + 1 = 7$ , so $2a = 6$ and $a = 3$ .

Check: $(2, 1)$ in $3x + y = 7$ : $6 + 1 = 7$ ; in $x - y = 1$ : $2 - 1 = 1$ .

$b = 1$ , $a = 3$

Question 10

Taxi company $P$ charges $ $3$ plus $ $0.50$ per km; company $Q$ charges $ $5$ plus $ $0.30$ per km. Draw the graphs of the fare, $ $F$ , against the distance, $d$ km, for $0 \le d \le 15$ . Find the distance for which the fares are equal, and say which company is cheaper for a 15 km ride.

ShowHide solution

$P$ : $F = 3 + 0.5d$ ; at $d = 0$ , $F = 3$ ; at $d = 15$ , $F = 10.5$ . $Q$ : $F = 5 + 0.3d$ ; at $d = 0$ , $F = 5$ ; at $d = 15$ , $F = 9.5$ .

The lines meet at $(10, 8)$ . Check: $3 + 0.5(10) = 8$ and $5 + 0.3(10) = 8$ .

For $15$ km, $Q$ costs $ $9.50$ and $P$ costs $ $10.50$ .

The fares are equal ($ $8$ ) for $10$ km. For a 15 km ride, company $Q$ is cheaper.

Question 11

Candle $X$ is 20 cm tall and burns down 2 cm every hour. Candle $Y$ is 14 cm tall and burns down 1 cm every hour. Both are lit at the same time. By drawing graphs of the height, $H$ cm, against the time, $t$ hours, find when the candles are the same height, and what that height is.

ShowHide solution

$X$ : $H = 20 - 2t$ , from $(0, 20)$ to $(10, 0)$ . $Y$ : $H = 14 - t$ , from $(0, 14)$ to $(10, 4)$ .

The lines meet at $(6, 8)$ . Check: $20 - 2(6) = 8$ and $14 - 6 = 8$ .

After 6 hours both candles are 8 cm tall.

Question 12

The graph shows $x + 3y = 6$ and $2x - y = 2$ . (a) Estimate the solution of the simultaneous equations to $1$ decimal place. (b) Find the exact solution as in Example 5, and explain why the graph could not give it.

ShowHide solution

(a) The lines meet a little less than halfway between $1.5$ and $2$ across, about $1.7$ , and a little below $1.5$ up, about $1.4$ .

(b) From $2x - y = 2$ , $y = 2x - 2$ . Put it into $x + 3y = 6$ : $x + 3(2x - 2) = 6$ , so $7x - 6 = 6$ , $7x = 12$ and $x = \frac{12}{7} = 1\frac{5}{7}$ . Then $y = 2 \times \frac{12}{7} - 2 = \frac{24}{7} - \frac{14}{7} = \frac{10}{7} = 1\frac{3}{7}\text{.}$

Check: $\frac{12}{7} + 3 \times \frac{10}{7} = \frac{12 + 30}{7} = 6\text{;}$ $2 \times \frac{12}{7} - \frac{10}{7} = \frac{14}{7} = 2\text{.}$

To $1$ decimal place, $1\frac{5}{7} \approx 1.7$ and $1\frac{3}{7} \approx 1.4$ . Sevenths cannot be read exactly by eye from a drawing.

(a) $x \approx 1.7$ , $y \approx 1.4$ (b) $x = 1\frac{5}{7}$ , $y = 1\frac{3}{7}$

Key Points

Simultaneous equations must both be true at the same time; their solution is the pair $(x, y)$ that satisfies both.
The point where the two lines meet lies on both, so its coordinates are the solution.
Draw both lines on the same axes with at least three points each, read the meeting point, and check it in both equations.
Use a table for awkward intercepts; follow the given scale exactly, and read numbers, not squares.
A graph gives an estimate when the lines meet between grid lines; algebra gives the exact answer.
The $x$ -coordinate of the point where $y = 2x + 1$ and $y = 7 - x$ meet is the solution of the one equation $2x + 1 = 7 - x$ .
In a real setting, the meeting point is where two quantities are equal; on each side of it, the lower line is the cheaper or smaller one.