Year 10 Maths Advanced Simultaneous Equations

Graphically

20 practice questions 1 video lesson Theory + worked examples

Solving simultaneous equations graphically means drawing both graphs on the same axes and reading the point where they meet. That point satisfies both equations. In NSW Year 10 Advanced maths the method is used for pairs of lines, for a line and a curve, and to show when there are no solutions or infinitely many (NESA outcome MA5‑EQU‑P‑02).

You will graph pairs of lines using intercepts or gradients, read the point of intersection, and check it algebraically. You will explain that parallel lines give no solution, estimate an intersection on a grid of half units, and find a break-even point for a market stall. You will also read where a line meets a parabola, a circle and $y=2^x$, find the area of a shaded triangle, and use horizontal lines to count solutions of a quadratic. It is part of Year 10 Advanced Simultaneous Equations.

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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 Advanced maths page covers drawing the lines accurately, reading or estimating the intersection, checking with algebra, break-even problems, and how many times a line can meet a parabola, circle or exponential curve.

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).

A line can meet a parabola or a circle at $2$, $1$ or $0$ points. Each crossing is one 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.

A line and a curve can meet more than once. Each meeting point is a solution, so count the points to count the solutions.

Two quick ways to draw a straight line:

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

y = m x + c

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

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

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 \(= R - C\).

Curves. The \(x\)-values where \(y = f(x)\) meets \(y = g(x)\) solve \(f(x) = g(x)\). For example, where \(y = 3^x\) meets \(y = 2x + 1\) gives the solutions of \(3^x = 2x + 1\). Check a point read from a graph by substituting it into both equations, e.g. \((5, 12)\) is on the circle \(x^2 + y^2 = 169\) because \(25 + 144 = 169\).

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: count the crossings. For $f(x) = k$, slide the horizontal line $y = k$ up and down and compare it with the turning point.

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 $y = 2x - 1$ and $x + y = 4$ are graphed on a grid of half-unit squares.

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

Solution

(i) The dot is about a third of a small square past $x = 1.5$ and about two-thirds of a square past $y = 2$. Estimate $\boldsymbol{(1.7,\ 2.3)}$.

(ii) Substitute $y = 2x - 1$ into $x + y = 4$:

$x + 2x - 1$	$=$	$4$
$3x$	$=$	$5$
$x$	$=$	$\dfrac{5}{3}$
$y$	$=$	$2 \times \dfrac{5}{3} - 1 = \dfrac{7}{3}$

The exact point is $\boldsymbol{\left(\dfrac{5}{3},\ \dfrac{7}{3}\right)}$. As decimals, $(1.66\ldots,\ 2.33\ldots)$, which agrees with the estimate.

(\frac{5}{3}, \frac{7}{3})

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:

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

150 + 6 n = 16 n

Example 4 — A line and a parabola

The graph shows $y = x^2 - 4x + 5$, with turning point $(2, 1)$, and the line $y = x + 1$.

(i) Read off the points of intersection and verify them. (ii) Show that their $x$-values solve $x^2 - 5x + 4 = 0$. (iii) For which values of $k$ does $x^2 - 4x + 5 = k$ have two solutions?

Solution

(i) The graphs cross at $(1, 2)$ and $(4, 5)$.

$(1, 2):\quad 1 - 4 + 5$	$=$	$2,\quad 1 + 1 = 2\ \checkmark$
$(4, 5):\quad 16 - 16 + 5$	$=$	$5,\quad 4 + 1 = 5\ \checkmark$

(ii) At a meeting point the two $y$-values are equal:

$x^2 - 4x + 5$	$=$	$x + 1$
$x^2 - 5x + 4$	$=$	$0$
$(x - 1)(x - 4)$	$=$	$0$

So $x = 1$ or $x = 4$, the same $x$-values as the graph.

(iii) The solutions are where the horizontal line $y = k$ meets the parabola. It meets twice when it is above the turning point: $\boldsymbol{k > 1}$. (At $k = 1$ it touches once; below, it misses.)

x^{2} - 5 x + 4 = 0

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.

Find every crossing. A line can cut a parabola or circle twice. Extend the graphs far enough to see both points.

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 many times can a line meet a parabola?

Twice, once or not at all. Each meeting point gives one solution of the equation formed by setting the two expressions equal.

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

\(x + 2x - 1\)	\(=\)	\(4\)
\(3x\)	\(=\)	\(5\)
\(x\)	\(=\)	\(\dfrac{5}{3}\)
\(y\)	\(=\)	\(2 \times \dfrac{5}{3} - 1 = \dfrac{7}{3}\)

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

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

\((1, 2):\quad 1 - 4 + 5\)	\(=\)	\(2,\quad 1 + 1 = 2\ \checkmark\)
\((4, 5):\quad 16 - 16 + 5\)	\(=\)	\(5,\quad 4 + 1 = 5\ \checkmark\)

\(x^2 - 4x + 5\)	\(=\)	\(x + 1\)
\(x^2 - 5x + 4\)	\(=\)	\(0\)
\((x - 1)(x - 4)\)	\(=\)	\(0\)

Graphically

📖 Theory

Solving a pair graphically

Using the intersection

Common pitfalls

Frequently asked questions

Theory