7.05 Applications of linear systems

The magic of systems of equations comes to life when we see how useful it is in real life applications. It is usually used when we have at least two unknown quantities and at least two pieces of information involving both of these quantities. The first part is to use some variables to represent these quantities and then figuring out how to write the bits of information down as equations. Then it's just a matter of using either the substitution method, elimination method or a graphical method to solve the equations simultaneously.

Let's explore further:

Worked examples

question 1

The sum of two numbers, $a$a and $b$b, is $7$7, while their product is $10$10. What are the values of $a$a and $b$b if $a$a is larger?

Think: What conditions do $a$a and $b$b need to satisfy and what possible solutions are there for each condition?

Do:

Let's take a look at $a$a & $b$b having a sum of $7$7 first. If we choose a random number like $0$0 to be $a$a, then the only possible value for $b$b is $7$7, as $0+7=7$0+7=7. Similarly if $a=1$a=1, then $b$b has to equal $6$6 as $1+6=7$1+6=7. Therefore possible number pairs include (but are not limited to):

$a$a	$b$b
$0$0	$7$7
$1$1	$6$6
$2$2	$5$5
$3$3	$4$4
$4$4	$3$3
$5$5	$2$2
$6$6	$1$1
$7$7	$0$0

Now if we look at the second condition, we can choose say a value of $1$1 for $a$a, then $b$b has to be $10$10, as $1\times10=10$1×10=10. Similarly if $a$a was $2$2, then $b$b has to be $5$5 as $2\times5=10$2×5=10. Possible pairs that give us a product of $10$10 include (but are not limited to):

$a$a	$b$b
$1$1	$10$10
$2$2	$5$5
$5$5	$2$2
$10$10	$1$1

We can now see that the only two pairs satisfying both conditions are $a=2$a=2 & $b=5$b=5, and $a=5$a=5 & $b=2$b=2. We know $a$a has to be larger so the answer must be $a=5$a=5 & $b=2$b=2.

 

Question 2

Leah got a quote from two photographers for an event. Photographer A charges $\$48$$48 for a booking fee plus $\$17$$17 per hour. Photographer B charges $\$28$$28 for a booking fee plus $\$21$$21 per hour.

a) Fill out a table comparing how much it would cost in total to hire each photographer for the following hours: $2,3,4,5,6$2,3,4,5,6

Think about how much it would cost to hire Photographer A for $x$x hours, make a formula and use it to see the total cost for different hours. Do the same for Photographer B.

Do

$\text{cost of hiring}=\text{booking fee}+\text{hourly rate}\times x$cost of hiring=booking fee+hourly rate×x where $x$x is the number of hours

That means the formula for

Photographer A is $\text{cost }=48+17x$cost =48+17x.

Photographer B is $\text{cost }=28+21x$cost =28+21x.

We can use these formulas to write up a table.

eg. for two hours with Photographer A, it costs $48+17\times2=\$82$48+17×2=$82.

Hours	Hiring cost for Photographer A (S)	Hiring cost for Photographer B ($)
$2$2	$82$82	$70$70
$3$3	$99$99	$91$91
$4$4	$116$116	$112$112
$5$5	$133$133	$133$133
$6$6	$150$150	$154$154

 

b) How many hours would've cost the same for Photographer A or Photographer B?

Think about looking at the table, seeing where the costs are the same and finding the appropriate time hired

Do

Comparing the columns of the table, we can see that the two photographers charge different amounts for all the hours except for one, where they both cost $\$133$$133 in total. Looking across along the same row we see this occurs when the hiring time is $5$5 hours.

Reflect: How can we check the reasonableness of our answers?

 

question 3

Rachel is buying drinks for her friends, and some of them want coffee while others want tea. In the cafe she went to a cup of coffee costs $\$4$$4 and a cup of tea costs $\$3$$3 If she bought $7$7 drinks and spent a total of $\$26$$26, how many teas and coffees did she get of each?

Think: What are the two amounts we need to find and the two bits of information relating the two amounts?

Do: We need to find the number of teas bought and the number of coffees bought. Let's let these be $x$x and $y$y.

The first bit of information tells us that the total number of drinks is $7$7, therefore:

$x+y=7$x+y=7     - $(1)$(1)

The next bit of information tell us that the total charge was $\$26$$26, where coffees cost $\$4$$4 and teas $\$3$$3. Therefore:

$3x+4y=26$3x+4y=26     - $(2)$(2)

So now that we have our two equations, substitution looks better if we rearrange $(1)$(1):

$y=7-x$y=7−x     - $(3)$(3)

Substituting this equation into the $y$y of $(2)$(2):

$(3)$(3)→$(2)$(2):
$3x+4y$3x+4y	$=$=	$26$26
$3x+4\left(7-x\right)$3x+4(7−x)	$=$=	$26$26
$3x+28-4x$3x+28−4x	$=$=	$26$26
$-x+28$−x+28	$=$=	$26$26
$-x$−x	$=$=	$-2$−2
$x$x	$=$=	$2$2

Now substituting in this $x$x value to $(3)$(3): 

$x$x→$(3)$(3):
$y$y	$=$=	$7-x$7−x
	$=$=	$7-2$7−2
	$=$=	$5$5

Therefore Rachel bought $2$2 cups of tea and $5$5 cups of coffee. 

Reflect: Does the answer make sense in the context of the situation?

 

Practice questions

QUESTION 4

question 5