Algebraic Relationships

Algebraic relationships is a major Year 12 focus area in Standard 1. It is about using straight lines and exponential curves to describe real situations, and reading answers off them.

• Simultaneous linear equations: graph two linear relationships (from a table of values or with graphing software), find the point of intersection, and explain what it means in context.
• Break-even analysis: model cost and revenue with linear equations, find the break-even point, and work out profit or loss (including with a spreadsheet).
• Graphs of practical situations: build a linear or exponential graph from a table of values and use it to make predictions.
• Strengths and limitations of linear and exponential models.

In Standard 1 you solve simultaneous equations mainly by graphs and tables. Solving them with algebra is a useful check, but you are not required to use an algebraic method. Quadratic and reciprocal relationships are Standard 2 only.

A linear relationship has the form $y=mx+c$, where $m$ is the gradient (the rate of change, how much $y$ goes up for each 1 unit of $x$) and $c$ is the $y$-intercept (the value of $y$ when $x=0$). In money questions, $c$ is usually a fixed cost and $m$ is a cost per item, per hour or per kilometre.

To graph a line, make a table of values: choose a few $x$-values, calculate $y$ for each, plot the points and join them with a ruler. Three points are safer than two, because a point that doesn't line up shows a calculation error.

Two linear equations are simultaneous when we want the values of $x$ and $y$ that make both true at the same time. On a graph, that is the point of intersection of the two lines.

Worked example: comparing two car-hire companies

Company P charges $60 plus 20 cents per km. Company Q charges $30 plus 35 cents per km. Let $d$ be the distance driven (km) and $C$ the cost ($):

Step 1: make a table of values for both.

Step 2: find where the costs are equal. Both are $100 when $d=200$, so the lines intersect at (200, 100).

Step 3: interpret. For trips under 200 km, Q is cheaper (lower fixed fee). For trips over 200 km, P is cheaper (lower cost per km). At exactly 200 km both cost $100.

If the table doesn't show an exact match, draw both lines on the same axes and read the intersection from the graph, or use graphing software. You can check by substituting into both equations.

What the point of intersection means

Always answer in the language of the question. A point of intersection on a cost graph means "both options cost the same for this amount of use". It is not a rate, and it is not a fixed fee. Then say which option is better on each side of that point: that is usually what a 3 or 4 mark question is really asking.

A business has:

• Cost $C$: fixed costs (rent, equipment) plus a cost for each item made.
• Revenue $R$: the money from sales, which is the selling price × the number sold.
• Profit $P=R-C$. A negative profit is a loss.

The break-even point is where $R=C$, which is where the cost and revenue lines intersect. Below the break-even point the business makes a loss; above it, a profit.

Worked example: printing T-shirts

A student prints T-shirts. Fixed costs are $1500, each shirt costs $9 to print, and shirts sell for $24.

Break-even: each shirt gives $24-9=15$ dollars towards the fixed cost.

The student breaks even after selling 100 shirts. Profit from 160 shirts:

On a graph, the profit at 160 shirts is the vertical gap between the revenue line and the cost line.

Using a spreadsheet

A spreadsheet is a quick way to model break-even problems. Put the number of items in column A, cost in column B, revenue in column C and profit in column D. In row 2 the formulas would be “=1500+9*A2” for cost, “=24*A2” for revenue and “=C2-B2” for profit. Fill down and look for where profit changes from negative to positive. You may be asked to read or complete a spreadsheet like this, or to write one of its formulas.

In a linear relationship, $y$ changes by the same amount each step. In an exponential relationship, $y$ is multiplied by the same factor each step:

• $k$ is the starting value (the value when $x=0$).
• $a$ is the growth or decay factor. If $a>1$ the quantity grows; if $0<a<1$ it decays.
• A growth of 30% per step means $a=1.3$; a decrease of 15% per step means $a=0.85$.

Worked example: growth from a table

A new streaming service has 2000 subscribers, and the number grows by 30% each month: $S=2000\times 1.3^t$.

The gaps between values (600, 780, 1014, 1318.2) keep getting bigger, so the graph curves upwards more and more steeply. To recognise an exponential pattern in a table, divide each value by the one before: here the answer is always 1.3. To write the equation from a table, $a$ is that common factor and $k$ is the value when $x=0$, so this table gives $S=2000\times 1.3^t$.

Worked example: decay

The chlorine level in a pool, in parts per million, is modelled by $C=3\times 0.85^t$, where $t$ is in days.

(Values rounded to 3 decimal places.) If chlorine should be added once the level drops below 1.5 ppm, the table shows this happens between 4 and 5 days after the pool was treated: the level is still 1.566 after 4 days but 1.331 after 5 days. A decay graph starts high and flattens out towards zero, but never reaches it.

Using the calculator

To evaluate $2000\times 1.3^4$, type “2000 × 1.3 ^ 4”. The power applies only to 1.3. A very common error is to multiply $2000\times 1.3$ first and then raise the result to the power.

Every model is only reliable over the range of values where its assumption holds. A linear model of a child's growth is fine from age 5 to 12, but people stop growing, so it fails for adults. An exponential model of a virus spreading through a town fails once most people have already caught it.

• Reading the point of intersection the wrong way round: write coordinates as (independent variable, dependent variable), for example (km, $).
• Giving only the intersection when the question asks you to advise or compare: say which option is better on each side of it.
• Forgetting the fixed cost when calculating profit.
• Treating percentage growth as adding the same amount each time (that's linear, not exponential).
• Typing the power in the wrong place on the calculator.
• Using an exponential model far beyond the data without commenting on its limitations.

• Find or calculate wants a number with working; interpret or explain wants a sentence in context, using the question's words (cost, rides, candles, kilometres).
• Show the equation you set up, for example $24x=1500+9x$, even if you then use a table or graph. Marks are given for setting up the relationship.
• For a 4-mark "which option is better" question, include: the two equations or tables, the intersection, and a conclusion for both sides of it.
• When a question gives a graph, read values carefully from the gridlines and state them, then use them.
• Check answers by substituting back into the original equations.

• Linear relationships (Year 11): gradient, intercept and direct variation are the basis of this topic.
• Investment (Year 12): simple interest gives a linear graph and compound interest gives an exponential graph.
• Depreciation and loans (Year 12): straight-line depreciation is linear decay and declining-balance depreciation is exponential decay.
• Bivariate data analysis (Year 12): a line of best fit is a linear model fitted to data.