Year 10 Maths - 10 Linear Relations

Solving Simultaneous Equations Graphically

Q: How do you solve one equation like 2x - 3 = 5 graphically?

Graph y = 2x - 3 and the horizontal line y = 5. They cross at (4, 5), so the solution is the x-coordinate, x = 4.

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Theory

Every point on a graph satisfies its equation, so the point where two graphs cross satisfies both equations: it is the solution of the pair. This Year 10 maths page covers drawing the lines accurately, reading or estimating the point of intersection, checking with algebra, parallel lines and the same line, break-even problems, and solving one equation by graphing each side. Curriculum: VC2M10A09 / AC9M10A02.

The graph of an equation is the set of all points whose coordinates make it true. The point of intersection of two graphs lies on both, so its coordinates satisfy both equations at once.

To solve graphically, draw both graphs on the same axes with the same scale and read the coordinates of each crossing point.

Two straight lines meet at one point, are parallel (same gradient, different $y$-intercepts: no solution), or are the same line (infinitely many solutions).

One equation in $x$, such as $3 - x = 2x - 3$, can be solved the same way: graph each side as a line. The $x$-coordinate of the crossing is the solution.

The solution of a pair of equations is the point on both graphs. Read the $x$-value straight down and the $y$-value straight across.

To solve $x-1=5-\dfrac{x}{2}$, graph each side as a line. They cross at $(4,3)$, so the solution is $x=4$: the $x$-coordinate only.

Two quick ways to draw a straight line:

\[y = mx + c:\ \text{start at } (0, c),\ \text{then rise } m\]\[\text{for each } 1 \text{ across}\]\[ax + by = d:\ \text{put } x = 0,\ \text{then } y = 0\]

y = m x + c

Gradient from two points on a graph, then the equation:

\[m = \dfrac{y_2 - y_1}{x_2 - x_1}\]\[y = mx + c\]

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

Break-even. Cost \(C = \text{fixed} + \text{cost per item} \times n\); revenue \(R = \text{price} \times n\). Break-even is where \(C = R\). Profit \(= R - C\).

One equation, two lines. To solve one equation such as \(3 - x = 2x - 3\) graphically, graph each side as a line: \(y = 3 - x\) and \(y = 2x - 3\). They cross at \((2,\ 1)\), so the solution is the \(x\)-coordinate only: \(x = 2\). If one side is a number, such as \(4x - 1 = 7\), that side is the horizontal line \(y = 7\).

Solving a pair graphically

Prepare each line: from $y = mx + c$ use the intercept and gradient; from $ax + by = d$ find the two intercepts. Plot at least two points (three is safer).
Draw both graphs on one set of axes, with the same scale, using a ruler.
Read the point of intersection: $x$ straight down, $y$ straight across.
Check by substituting into both equations. If the crossing is between grid lines, estimate it, then solve algebraically for the exact value.

Using the intersection

Lines from a graph: find each equation from two marked points, then solve them together.
Unknown constant: if $y = 3x + c$ meets $y = -x + 8$ on the $x$-axis, the meeting point is $(8, 0)$, so $0 = 24 + c$ and $c = -24$.
Area between lines: $y = x + 3$ and $y = -x + 5$ meet at $(1, 4)$ and cut the $x$-axis at $-3$ and $5$. The triangle has base $8$, height $4$, area $\dfrac{1}{2} \times 8 \times 4 = 16$ units$^2$.
Number of solutions: two lines cross once, never (parallel: same gradient, different $y$-intercepts) or everywhere (the same line).
One equation: graph $y = \text{left side}$ and $y = \text{right side}$. The solution is the $x$-coordinate of the crossing (Example 4).

Example 1 — Draw both lines, read the crossing

Solve $3x + 2y = 12$ and $y = x + 1$ graphically, then check algebraically.

Solution

For $3x + 2y = 12$ use the intercepts. Put $x = 0$: $2y = 12$, point $(0, 6)$. Put $y = 0$: $3x = 12$, point $(4, 0)$.

For $y = x + 1$ start at the $y$-intercept $(0, 1)$ and go up $1$ for every $1$ across, e.g. $(3, 4)$.

The lines cross at $(2, 3)$. Check in both equations:

$3(2) + 2(3)$	$=$	$6 + 6 = 12\ \checkmark$
$2 + 1$	$=$	$3\ \checkmark$
$\therefore\ x$	$=$	$\boldsymbol{2},\quad y = \boldsymbol{3}$

(2, 3)

Example 2 — Estimate, then find the exact point

The lines $y = 3x - 1$ and $x + 2y = 6$ are graphed below.

(i) Estimate the point of intersection to one decimal place.
(ii) Find the exact point algebraically.

Solution

(i) The dot is a little to the right of $x = 1$, and a little less than halfway from $y = 2$ to $y = 3$. Estimate $\boldsymbol{(1.1,\ 2.4)}$.

(ii) Substitute $y = 3x - 1$ into $x + 2y = 6$:

$x + 2(3x - 1)$	$=$	$6$
$7x - 2$	$=$	$6$
$x$	$=$	$\dfrac{8}{7}$
$y$	$=$	$3 \times \dfrac{8}{7} - 1$
	$=$	$\dfrac{17}{7}$

The exact point is $\boldsymbol{\left(\dfrac{8}{7},\ \dfrac{17}{7}\right)}$. As decimals, $(1.14\ldots,\ 2.43\ldots)$, which agrees with the estimate.

(\frac{8}{7}, \frac{17}{7})

A graph gives a good estimate; algebra gives the exact answer.

Example 3 — Break-even point

A student prints phone cases. Set-up costs $\$150$ and each case costs $\$6$ to make, so the cost of $n$ cases is $C = 150 + 6n$. Each case sells for $\$16$, so the revenue is $R = 16n$.
(i) Graph both and find the break-even point.
(ii) Confirm it algebraically.
(iii) Find the profit on $30$ cases.

Solution

(i) The break-even point is where cost equals revenue: where the lines cross.

They cross at $n = 15$, at $\$240$.

(ii) Set $C = R$:

$150 + 6n$	$=$	$16n$
$150$	$=$	$10n$
$n$	$=$	$15$
$R$	$=$	$16 \times 15 = 240$

Break-even at $\boldsymbol{15}$ cases and $\boldsymbol{\$240}$. Fewer cases make a loss (the $C$ line is higher); more make a profit.

(iii) Profit $=$ revenue $-$ cost:

$R$	$=$	$16 \times 30 = 480$
$C$	$=$	$150 + 6 \times 30 = 330$
$\text{Profit}$	$=$	$480 - 330$
	$=$	$\boldsymbol{\$150}$

150 + 6 n = 16 n

Example 4 — One equation, two lines

Solve $5-2x=x-4$ graphically.

Solution

Graph each side of the equation as a line: $y=5-2x$ and $y=x-4$.

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

The lines cross at $(3,-1)$. Where they meet, the two sides are equal, so the solution is the $x$-coordinate: $\boldsymbol{x=3}$.

$\text{Check: LHS}$	$=$	$5-2(3) = -1$
$\text{RHS}$	$=$	$3-4 = -1\ \checkmark$

Common pitfalls

Use the same scale on both lines. Two graphs drawn on different axes cannot be compared. Draw both on one grid.

Coordinates are (x, y). Read across the horizontal axis first. $(3, 1)$ and $(1, 3)$ are different points.

A graph gives an estimate. If the crossing is between grid lines, say 'about', then solve algebraically for the exact answer.

Parallel lines are not an error. If the lines never meet there is no solution. Confirm it: equal gradients and different $y$-intercepts.

One equation has an x answer only. When you solve $5 - 2x = x - 4$ graphically, the lines cross at $(3,\ -1)$, but the solution is $x = 3$, not the point.

Frequently asked questions

How do you solve simultaneous equations graphically?

Draw both graphs on the same axes. The point where they cross lies on both graphs, so its coordinates are the solution. Check by substituting into both equations.

What if the lines do not cross at a grid point?

Estimate the coordinates from the graph, then solve the equations algebraically to get the exact answer, often a fraction.

What does it mean if the lines are parallel?

Parallel lines never meet, so the simultaneous equations have no solution. Parallel lines have the same gradient and different y-intercepts.

What is a break-even point?

It is where the cost line and the revenue line cross, so cost equals revenue and the profit is zero. Selling more gives a profit; selling fewer gives a loss.

How do you solve one equation like 2x - 3 = 5 graphically?

Graph $y = 2x - 3$ and the horizontal line $y = 5$. They cross at $(4,\ 5)$, so the solution is the $x$-coordinate, $x = 4$.

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\(3(2) + 2(3)\)	\(=\)	\(6 + 6 = 12\ \checkmark\)
\(2 + 1\)	\(=\)	\(3\ \checkmark\)
\(\therefore\ x\)	\(=\)	\(\boldsymbol{2},\quad y = \boldsymbol{3}\)

\(x + 2(3x - 1)\)	\(=\)	\(6\)
\(7x - 2\)	\(=\)	\(6\)
\(x\)	\(=\)	\(\dfrac{8}{7}\)
\(y\)	\(=\)	\(3 \times \dfrac{8}{7} - 1\)
\(\)	\(=\)	\(\dfrac{17}{7}\)

\(150 + 6n\)	\(=\)	\(16n\)
\(150\)	\(=\)	\(10n\)
\(n\)	\(=\)	\(15\)
\(R\)	\(=\)	\(16 \times 15 = 240\)

\(R\)	\(=\)	\(16 \times 30 = 480\)
\(C\)	\(=\)	\(150 + 6 \times 30 = 330\)
\(\text{Profit}\)	\(=\)	\(480 - 330\)
\(\)	\(=\)	\(\boldsymbol{\$150}\)

\(x\)	\(0\)	\(2\)	\(4\)
\(5-2x\)	\(5\)	\(1\)	\(-3\)
\(x-4\)	\(-4\)	\(-2\)	\(0\)

\(\text{Check: LHS}\)	\(=\)	\(5-2(3) = -1\)
\(\text{RHS}\)	\(=\)	\(3-4 = -1\ \checkmark\)

Solving Simultaneous Equations Graphically

📖 Theory

Solving a pair graphically

Using the intersection

Common pitfalls

Frequently asked questions

Theory