Solve a pair of simultaneous linear equations graphically and algebraically, and use simultaneous equations to model practical situations

What this dot point is asking

NESA wants two things. First, solve a pair of linear equations in two unknowns three ways: substitution, elimination, and reading off a graph. Second, and just as important, set up the equations yourself from a worded scenario. Most of the marks sit in the worded version: comparing two phone plans, two taxi fares, or two job offers, or finding the break-even point of a small business (the point where money coming in just equals money going out).

A pair of simultaneous equations gives you two facts about two unknowns at once. One fact on its own is not enough to fix two unknowns: $x + y = 10$ is true for $(1, 9)$, $(4, 6)$, $(7, 3)$ and endlessly many other pairs. Add a second fact about the same two unknowns and you usually narrow it down to one pair that fits both at the same time. That phrase, at the same time, is what "simultaneous" means, and it is the whole idea behind the picture below.

The answer

What "simultaneous" means, geometrically

Each linear equation in $x$ and $y$ draws a straight line: every point on the line is an $(x, y)$ pair that makes that equation true. So the answer that works for both equations has to sit on both lines, which means it is the point where the two lines cross (their intersection). Solving simultaneous equations and finding where two lines cross are the same job said two ways, in algebra and in pictures.

This picture also explains the three possible outcomes, which a question can ask you to recognise:

• One solution. The lines have different gradients, so they cross exactly once. This is the usual case and the only one that gives a single $(x, y)$ answer.
• No solution. The lines are parallel (same gradient, different $y$-intercept). They never meet, so no pair satisfies both. Algebraically, both variables cancel and you are left with a false statement such as $0 = 5$.
• Infinitely many solutions. The two equations are really the same line in disguise (one is a multiple of the other). Every point on the line works. Algebraically, both variables cancel and you are left with a true statement such as $0 = 0$.

Algebraic method 1: substitution

Use substitution when one equation already gives $y$ (or $x$) on its own, or rearranges to that in one easy step. The idea is to turn two equations in two unknowns into one equation in one unknown.

1. Rearrange one equation so a single variable is the subject, e.g. $y = \dots$.
2. Substitute that expression into the other equation, so it now contains only one variable.
3. Solve that single-variable equation.
4. Substitute the value back into the simpler equation to get the second variable.

Substitution is the natural choice for worded comparison problems, because each scenario usually arrives already in the form "cost $= \dots$".

Algebraic method 2: elimination

Use elimination when the coefficients line up (or can be lined up by multiplying), so that adding or subtracting the equations makes one variable vanish.

1. If needed, multiply one or both equations so the coefficients of one variable are equal in size.
2. Add the equations if those coefficients are opposite in sign; subtract if they are the same sign. Either way that variable is eliminated.
3. Solve the resulting single-variable equation.
4. Substitute back to find the other variable.

The single most common elimination slip is adding when you should subtract. The safe rule: opposite signs add to zero, so add; matching signs need subtracting. The "same sign, subtract" pairing is the trap most students fall into under time pressure.

Graphical method

Plot both lines on the same axes and read off the point where they cross, stating it as a pair $(x, y)$. The HSC often supplies a pre-drawn graph and asks you to read the solution, or gives you two equations and asks you to draw and solve. Reading a graph is only as accurate as the grid, so it suits "nice" intersections at whole or half values; for anything awkward, solve algebraically and use the graph as a check.

Read the intersection off a graph, stage by stage

The four panels below build a graphical solution one step at a time, using a real comparison: a Year 12 student weighing up two casual jobs. A Cafe job pays a $40 weekly retainer plus$20 an hour, so weekly pay is $P = 40 + 20h$. A Tutoring job pays a higher $80 base but only$10 an hour, so $P = 80 + 10h$. The question is how many hours make the two jobs pay the same.

Stage 1, set up the axes and plot the first line. Put hours $h$ across the bottom and weekly pay $P$ up the side. Plot the Cafe line $P = 40 + 20h$: it starts at its $y$-intercept of $40$ (the pay for zero hours, the retainer) and climbs $20$ for every extra hour. Two points are enough to draw a straight line.

Stage 2, plot the second line on the same axes. Add the Tutoring line $P = 80 + 10h$ to the same grid. It starts higher (intercept $80$) but rises more gently ($10$ an hour), so it is the flatter of the two. Plotting both on one set of axes is what lets the crossing appear.

Stage 3, find the crossing point. The two lines meet at exactly one point. To read its coordinates accurately, drop a dashed line straight down from the crossing to the $h$-axis and straight across to the $P$-axis. The vertical drop lands on $h = 4$; the horizontal lands on $P = 120$.

Stage 4, state the solution and read it in context. Write the answer as a coordinate pair, $(h, P) = (4, 120)$, that is $120 at $4$ hours. In words: at $4$ hours a week both jobs pay $120. The graph also shows which job wins on each side of the crossing. Below $4$ hours the Tutoring line is higher, because its $80 base is doing most of the work, so Tutoring pays more. Above $4$ hours the steeper Cafe line is higher, so Cafe pays more. Saying who wins on each side is worth a mark in comparison questions.

You can confirm the crossing with algebra in one line: set the two pay expressions equal, $40 + 20h = 80 + 10h$, giving $10h = 40$ and $h = 4$, then $P = 40 + 20(4) = 120$. The algebra and the graph agree. Always run this quick check when a graph gives you a tidy crossing.

Graphical versus algebraic: which to use

Both methods find the same point, but they are good at different things.

• Algebra is exact. Substitution and elimination give the precise answer even when it is a messy value such as $h = \frac{20}{3}$. You cannot read a value like that off a grid. So for "solve" and "show your working" questions, algebra is the safe choice, and it is the only way to get an exact fraction or an answer right to the cent.
• A graph shows the whole story. It makes the three cases (one, none, or infinitely many solutions) easy to see at a glance. It also shows which option is better on each side of the crossing, which two numbers on their own cannot. So for "use the graph to..." or "for what range is plan A cheaper" questions, the graph is the natural tool.
• Use each to check the other. If a question gives a graph with a tidy crossing, check it with algebra. If you solve with algebra and the numbers are clean, a quick sketch confirms the lines really do cross where you say. If the two disagree, you have made a slip somewhere.

Modelling a worded problem

Most marks live in the worded version, which almost always compares two situations. The reliable recipe:

1. Define your variables explicitly. Write "let $h$ be the number of hours" and "let $P$ be the pay in dollars". Markers look for this line.
2. Write one equation per scenario, each of the form total $=$ fixed part $+$ rate $\times$ variable.
3. Solve simultaneously (substitution is usually quickest here) to find the crossover.
4. Answer in context, in the units of the question, and add the comparison: which option is better below the crossover and which above.

How exam questions ask about simultaneous equations

The wording changes but the task does not. Translate it:

• "Solve the simultaneous equations ..." A pure algebra question. Use elimination or substitution, give both coordinates, and show a check or back-substitution.
• "Use the graph to solve ..." or "Read off the point of intersection." Read the crossing and state it as $(x, y)$. Choose values that fit the grid, and remember the HSC usually picks tidy crossings like $(3, 4)$.
• "After how many [km / months / calls] do the two cost the same?" A modelling question. Define variables, write a cost equation for each option, set them equal, solve. The "same cost" condition is the simultaneous condition.
• "Which plan is cheaper for ...?" or "For what values is A better than B?" Find the crossover first, then read off which line is lower on the side the question asks about. The crossover splits the range into "A cheaper" and "B cheaper".
• "Sketch both lines and find where they meet." Plot using intercept and gradient (or two points each), then read the intersection, then confirm by algebra.
• "Explain why these equations have no solution / infinitely many solutions." A parallel-versus-same-line question. Compare gradients: equal gradients with different intercepts give no solution; identical equations give infinitely many.