Solving Problems Involving Linear Equations

The concept of linearity, whether we are aware of it or not, occurs in all sort of places in real life. The key understanding is that the general mathematical relationship $y=mx+c$y=mx+c comprising of a constant term $c$c (which may or may not be zero) and a variable term $mx$mx (rising or falling by an an amount $m$m as the count of $x$x itself increases unit by unit) that together determine outcome $y$y. 

The letters $y,m,x$y,m,x and $c$c  are purely mathematical, and in real applications they are often replaced by more meaningful letters. Nevertheless the idea is the same.

Example 1

Stock price trends

A certain double-glazing company, listed on the Australian Stock Exchange, shows its opening weekly stock price for the first five weeks of the 2015/2016 financial year:

Week	1	2	3	4	5
Price	$1.87	$2.06	$2.25	$2.44	$2.63

A graph of the rise in stock value shows an unmistakeable linear trend, and we thus proceed to find an equation of the form $p=mw+c$p=mw+c where $p$p is the price at week $w$w, and the values of $m$m and $c$c need to be determined.

We could choose any two points, say $\left(2,2.06\right)$(2,2.06) and $\left(5,2.63\right)$(5,2.63) and find the slope $m$m given as $m=\frac{2.63-2.06}{5-2}=0.19$m=2.63−2.065−2​=0.19. Hence our equation becomes $p=0.19w+c$p=0.19w+c.

Then by substituting any of the five points, say, $\left(2,2.06\right)$(2,2.06) , we have that $2.06=0.19\left(2\right)+c$2.06=0.19(2)+c and so this means that $c=1.68$c=1.68. 

The linear trend over these five weeks is thus given by $p=0.19w+1.68$p=0.19w+1.68 and if this were to continue, we could use the equation to predict future price values. Predicting future price values using an equation formed from previous values is a process known as extrapolation.

Example 2

Electricity tariffs 

It is often the case that electricity is charged to the consumer at a certain amount per kilowatt hour (say $15$15 cents) plus a monthly connection fee (say $\$20$$20). 

This arrangement can be modelled by a linear function given by $C=0.15x+20$C=0.15x+20 where $C$C is the total cost levied for using $x$x kilowatt hours. Note that using no electricity still means incurring a monthly connection charge of $\$20$$20. Our model, expressed mathematically, has a slope of $0.15$0.15 and a $C$C-intercept of $20$20.

If for example I use $528$528 kilowatt hours, my cost will be given by $C=0.15\left(528\right)+20=99.20$C=0.15(528)+20=99.20. 

Example 3

Income tax 

The following individual income tax rates for the financial year 2015–16 apply in Australia.  

Income Range	Tax payable if your income is in this range
0 – $18,200	Nil
$18,201 – $37,000	19c for each $1 over $18,200
$37,001 – $80,000	$3,572 plus 32.5c for each $1 over $37,000
$80,001 – $180,000	$17,547 plus 37c for each $1 over $80,000
$180,001 and over	$54,547 plus 45c for each $1 over $180,000

We could interpret this table as five different linear constructions, so that a second, more mathematical table, can be constructed with the same information might look like this:

Income Range	Tax payable if your income is in this range
0 – $18,200	$y_1=0$y1​=0
$18,201 – $37,000	$y_2=0.19x-3458$y2​=0.19x−3458
$37,001 – $80,000	$y_3=0.325x-8453$y3​=0.325x−8453
$80,001 – $180,000	$y_4=0.37x-12053$y4​=0.37x−12053
$180,001 and over	$y_5=0.45x-26453$y5​=0.45x−26453

To work these equations out is straight-forward. For example, the first table states that if my income was between $\$37001$$37001 and $\$80000$$80000, I would need to pay $\$3572$$3572 plus $32.5$32.5 cents for each dollar over $\$37,000$$37,000.

As an equation this means that $y_2=3572+0.325\left(x-37000\right)$y2​=3572+0.325(x−37000). By expanding then simplifying, this becomes $y_2=0.19x-3458$y2​=0.19x−3458. 

Note that our tax system is designed to tax incomes proportionately more as the income rises. The slope of each line becomes higher on the additional dollars that are earned above each threshold. 

Example 4

Break even points

As a graphical example, imagine a sole-trader that manufactures a famous teddy bear. The fixed costs, which include the factory rent, the machines and his wages amounts to $\$3600$$3600 per week. The variable costs, including the material costs and packaging amount to $\$52$$52 per bear. 

From this information we can write a cost equation given by $C=52n+3600$C=52n+3600 where $C$C is the total cost of producing $n$n bears.

The trader intends on selling each bear for $\$100$$100 each, including postage and handling. This means that we can write a revenue equation as $R=100n$R=100n. 

Here is a sketch both the cost and revenue lines:

The sole trade sees that the two lines intersect when sales reach $75$75 bears. We can verify this by setting $C=R$C=R so that $52n+3600=100n$52n+3600=100n. This is not hard to solve. Here, $48n=3600$48n=3600 and so $n=75$n=75 bears. This is known as the break even point. At that point, the revenue and costs are equal. f the trader sells less than $75$75 bears, then she makes a loss. If the trader sells more than $75$75 bears,then the trader begins to make a profit.

For example, if the trader sells $100$100 bears, the revenue will be given by $R=100\times100=10000$R=100×100=10000 dollars and the costs will be given by $C=52\times100+3600=8800$C=52×100+3600=8800, and so the profit will be the difference $1200$1200 dollars. 

 

Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3

QUESTION 4