Simultaneous Equations and Applications (Std 2-X)

v1 Algebra, STD2 A4 2010 HSC 24b

Damo hires paddle boats in summertime as part of his water sports business. To calculate the cost, $C$, in dollars, of hiring $x$ paddle boats, he uses the equation $C=40+25x$.

He hires the paddle boats for $35 per hour and determines his income, $I$, in dollars, using the equation $I=35x$.

Use the graph to solve the two equations simultaneously for $x$ and explain the significance of this solution for Damo's business. (2 marks)

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$x=4\ \ \text{See worked solution}$

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$\text{From the graph, intersection occurs at}\ x=4$

$\rightarrow\ \text{Break-even point occurs at}\ x=4$

$\text{i.e. when 4 hours of paddle board hire occurs}$

$\text{Income}$	$=35\times 4=$140\ \ \text{is equal to}$
$\text{Costs}$	$=40+(25\times 4)=$140$

$\text{If}\ <4\ \text{hours of board hire}\ \rightarrow\ \text{LOSS for business}$

$\text{If}\ >4\ \text{hours of board hire}\ \rightarrow\ \text{PROFIT}$

♦ Mean mark 36%.
MARKER’S COMMENT: The intersection on the graph is the same point at which the two simultaneous equations are solved for the given value of $x$.

v1 Algebra, STD2 A4 SM-Bank 4

Bec is a baker and makes cookies to sell every week.

The cost of making $n$ cookies, $$C$, can be calculated using the equation

$C=400+2.5n$

Bec sells the cookies for $4.50 each, and her income is calculated using the equation

$I=450n$

On the grid above, draw the graphs of $C$ and $I$. (2 marks)

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On the graph, label the breakeven point and the loss zone. (2 marks)

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(i) and (ii)

Show Worked Solution

ii. $\text{Loss zone occurs when}\ C > I,\ \text{which is shaded}$

$\text{in the diagram above.}$

v1 Algebra, STD2 A4 2005 HSC 28b

Jake and Preston are planning a fund-raising event at the local swim centre. They can have access to the giant pool float for $550 and the party room hire for $250. A sausage sizzle and drinks will cost them $9 per person.

Write a formula for the cost ($C) of running the event for $x$ people. (1 mark)

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The graph shows planned income and costs when the ticket price is $15.

Estimate the minimum number of people needed at the fund raising event to cover the costs. (1 mark)

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How much profit will be made if 200 people attend the fund raiser? (1 mark)

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Jake and Preston have 300 tickets to sell. They want to make a profit of $1510.

What should be the price of a ticket, assuming all 300 tickets will be sold? (3 marks)

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$C=800+9x$
$\text{Approximately }135$
$$400$
$$16.70$

Show Worked Solution

i.	$$C$	$=550+250+(9\times x)$
		$=800+9x$

ii. $\text{Using the graph intersection}$

$\text{Approximately 135 people are needed}$

$\text{to cover the costs.}$

iii. $\text{If 200 people attend}$

$\text{Income}$	$=200\times $15$
	$=$3000$

$\text{Costs}$	$=800+(9\times 200)$
	$=$2600$

$\therefore\ \text{Profit}$	$=3000-2600$
	$=$400$

iv. $\text{Costs when}\ x=300:$

$C$	$=800+(9\times 300)$
	$=$3500$

$\text{Income required to make }$1510\ \text{profit}$

$=3500+1510$

$=$5010$

$\therefore\ \text{Price per ticket}$	$=\dfrac{5010}{300}$
	$=$16.70$

v1 Algebra, STD2 A4 2020 HSC 24

There are two tanks at an industrial plant, Tank A and Tank B. Initially, Tank A holds 2520 litres of liquid fertiliser and Tank B is empty.

Tank A begins to empty liquid fertiliser into a transport vehicle at a constant rate of 40 litres per minute.

The volume of liquid fertiliser in Tank A is modelled by $V=1400-40t$ where $V$ is the volume in litres and $t$ is the time in minutes from when the tank begins to drain the fertiliser.

On the grid below, draw the graph of this model and label it as Tank A. (1 mark)
Tank B remains empty until $t=10$ when liquid fertiliser is added to it at a constant rate of 60 litres per minute.
By drawing a line on the grid (above), or otherwise, find the value of $t$ when the two tanks contain the same volume of liquid fertiliser. (2 marks)

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Using the graphs drawn, or otherwise, find the value of $t$ (where $t > 0$) when the total volume of liquid fertiliser in the two tanks is 1400 litres. (1 mark)

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$\text{Tank} \ A \ \text{will pass through (0, 1400) and (35, 0)}$
$20 \ \text{minutes}$
$30 \ \text{minutes}$

Show Worked Solution

a. $\text{Tank} \ A \ \text{will pass through (0, 1400) and (35, 0)}$

b. $\text{Tank} \ B \ \text{will pass through (10, 0) and (30, 1200)}$

$\text{By inspection, the two graphs intersect at} \ \ t = 20 \ \text{minutes}$

c. $\text{Strategy 1}$

$\text{By inspection of the graph, consider} \ \ t = 30$

$\text{Tank A} = 200 \ \text{L} , \ \text{Tank B} =1200 \ \text{L}$

$\therefore\ \text{Total volume = 1400 L when t = 30}$

$\text{Strategy 2}$

$\text{Total Volume}$	$=\text{Tank A} + \text{Tank B}$
$1400$	$=1400-40t+(t-10)\times 60$
$1400$	$=1400-40t+60t-600$
$20t$	$= 600$
$t$	$= 30 \ \text{minutes}$

♦♦ Mean mark part (c) 22%.

v1 Algebra, STD2 A4 2019 HSC 36

A small business makes and dog kennels.

Technology was used to draw straight-line graphs to represent the cost of making the dog kennels $(C)$ and the revenue from selling dog kennels $(R)$. The $x$-axis displays the number of dog kennels and the $y$-axis displays the cost/revenue in dollars.

How many dog kennels need to sold to break even? (1 mark)

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By first forming equations for cost `(C)` and revenue `(R)`, determine how many dog kennels need to be sold to earn a profit of $2500. (3 marks)

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$20$
$145$

Show Worked Solution

a. $20\ \ (x\text{-value at intersection})$

b. $\text{Find equations of both lines}:$

$(0, 400)\ \text{and}\ (20, 600)\ \text{lie on}\ \ C$

$\text{gradient}_C = \dfrac{600-400}{20-0}=10$

$\rightarrow\ C=400+10x$

$(0,0)\ \text{and}\ (20, 600)\ \text{lie on}\ \ R$

$\text{gradient}_R =\dfrac{600-0}{20-0}=30$

$\rightarrow\ R=30x$

$\text{Profit} = R-C$

$\text{Find}\ \ x\ \text{when Profit }= $2500:$

$2500$	$=30x-(400+10x)$
$20x$	$=2900$
$x$	$=145$

$\therefore\ 145\ \text{dog kennels need to be sold to earn }$2500\ \text{profit}$

♦♦ Mean mark 28%.

v1 Algebra, STD2 A4 2020 VCAA 3

Noah's business manufactures car seat covers.

The monthly oncome, $I$, in dollars, from selling $n$ seat covers is given by

$I=60n$

This relationship is shown on the graph below.

The monthly cost, $C$, in dollars, of making $n$ seat covers is given by

$C=35n+5000$

On the graph above, sketch the monthly cost, $C$, of making $n$ seat covers. (1 mark)
Find the number of seat covers that need to be sold in order to break even and state the profit made at this point. (2 marks)

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b. $\text{200 seat covers and zero profit at this point}$

Show Worked Solution

a. $\text{Draw graph through points (0, 5000) and (250, 13 750)}$

b. $C=35n+5000\ \text{and }I=60n$

$\text{Break-even occurs when} \ \ I=C$

$\text{Method 2: Graphically}$

$\text{Point of intersection is }\rightarrow (200, 12\ 000)$

$\text{Method 2: Algebraically}$

$\text{Solve for} \ n:$

$60n$	$=35n+5000$
$25n$	$=5000$
$n$	$=200$

$\therefore\ \text{200 seat covers must be sold to break even}$

$\text{and the profit at this point is zero}$

v1 Algebra, STD2 A4 EQ-Bank 8

Two farmers, River and Jacqueline, sold organic honey at the the local market.

River sold his 500 gram jars of honey for $8 each and Jacqueline sold her 200 gram jars of honey for $4 each. In the first hour, their total combined sales were $40.

If River sold $x$ jars of honey and Jacqueline sold $y$ jars of honey, then the following equation can be formed:

$8x+4y=40$

In the first hour, the friends sold a total of 6 jars of honey between them.

Find the number of jars of honey each of the friends sold during this time by forming a second equation and solving the simultaneous equations graphically. (5 marks)

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$\text{River sold 4 and Jacqueline sold 2.}$

Show Worked Solution

$\text{Graphing}\ \ 8x + 4y$	$=40$
$y$	$=-2x+10$

$y\text{-intercept}\ = (0, 10)$

$x\text{-intercept}\ = (5, 0)$

$\text{Gradient}\ = -2$

$\text{Second equation:}$

$x+y=6$

$\text{From the graph,}$

$\text{River sold 4 and Jacqueline sold 2.}$

v1 Algebra, STD2 A4 SM-Bank 7

The graph of the line $x+y=3$ is shown.

By graphing $y=2x-3$ on the same grid, find the point of intersection of $x+y=3$ and $y=2x-3$. (3 marks)

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$(2, 1)$

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$\text{Graphing}\ y=2x-3:$

$y\text{-intercept }=-3$

$\text{Gradient }=2$

$\therefore\ \text{Point of intersection is (2, 1).}$

v1 Algebra, STD2 A4 SM-Bank 6

A student was asked to solve the following simultaneous equations.

$y=2x-5$

$x-2y+2=0$

After graphing the equations, the student found the point of intersection to be $(4,3)$?

Is the student correct? Support your answer with calculations. (2 marks)

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$\text{See Worked Solutions}$

Show Worked Solution

$\text{If the student is correct, the co-ordinates will}$

$\text{satisfy both equations.}$

$\text{Substitute (4, 3) into}\ \ y=2x-5$

$\text{LHS}$	$=3$
$\text{RHS}$	$= 2(4)-5$
	$=3$

$\therefore\ \text{LHS = RHS}$

$\text{Substitute (4, 3) into}\ \ x-2y+2=0$

$\text{LHS}$	$=4-2(3)+2$
	$=0$
	$\ =\ \text{RHS}$

$\rightarrow\ (4,3)\ \text{satisfies both equations.}$

$\therefore\ \text{Student is correct.}$

v1 Algebra, STD2 A4 2014 HSC 26d

Draw each graph on the grid below and hence solve the simultaneous equations. (3 marks)

$y=2x-6$

$y-x+2=0$

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$x=4,\ y=2$

Show Worked Solution

$\text{Solution is at the intersection:}\ \ x=4,\ y=2$

v1 Algebra, STD2 A4 SM-Bank 27

Morgan and Beau are to host a 21st birthday party for their friend's Zac and Peattie. They can hire a function room for $900 and a DJ for $450. Drinks will cost them $33 per person.

Write a formula for the cost ($$C$) of holding the birthday party for $x$ people. (1 mark)

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The graph below shows the planned income and costs if they charge $60 per person. Estimate the number of friends they need to invite to break even. (1 mark)

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How much money will Morgan and Beau have to purchase a travel voucher as a group gift if 90 people attend the party? (1 mark)

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$$C=1350+33x$
$50$
$$1080$

Show Worked Solution

i.	$\text{Fixed Costs}$	$=900+450$
		$=$1350$

$\text{Variable Costs}=$33x$

$\therefore\ $C=1350+33x$

ii.	$\text{From the graph}$
	$\text{Costs = Income when}\ x=50$
	$\text{(i.e. where lines intersect)}$

$\text{Algebraically}$
$\text{Income }$	$=\ \text{Costs}$
$60x$	$=1350+33x$
$27x$	$=1350$
$x$	$=50$

$\therefore\ \text{Breakeven when }50\ \text{people attend}$

iii. $\text{When}\ \ x=90:$

$\text{Income}$	$=90\times 60$
	$=$5400$

$$C$	$=1350+33\times 90$
	$=$4320$

$\therefore\ \text{Travel Voucher}$	$=5400-4320$
	$=$1080$

v1 Algebra, STD2 A4 2018 HSC 27b

$y$	$=x-3$
$y+3x$	$=1$

Draw these two linear graphs on the number plane below and determine their intersection. (3 marks)

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$(1,-2)$

Show Worked Solution

$\text{Table of values:}\ \ y=x-3$

\begin{array} {|c|c|c|c|c|}
\hline x & -2 & -1 & 0 & \colorbox{lightblue}{ 1 } \\
\hline \ \ y \ \ & \ \ -5 \ \ & \ \ -4 \ \ & \ \ -3 \ \ & \ \colorbox{lightblue}{ – 2} \\
\hline \end{array}

$\text{Table of values:}\ \ y+3x=1 \ \rightarrow \ y=-3x+1$

\begin{array} {|c|c|c|c|c|}
\hline x & -1 & 0 & \colorbox{lightblue}{ 1 } & 2 \\
\hline \ \ y \ \ & \ \ \ 4\ \ \ & \ \ \ 1\ \ \ & \ \colorbox{lightblue}{ – 2} & \ \ -5 \ \ \\
\hline \end{array}

$\text{From graph (and table), intersection occurs}$

$\text{at}\ \ (1, -2).$

v1 Algebra, STD2 A4 2023 HSC 21

Electricity provider $A$ charges 30 cents per kilowatt hour (kWh) for electricity, plus a fixed monthly charge of $90.

Complete the table showing Provider $A$'s monthly charges for different levels of electricity usage. (1 mark)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ 0 \ \ & \ \ 400 \ \ & \ \ 1000 \ \ \\
\hline
\rule{0pt}{2.5ex} \textit{Monthly Charge (\$)} \rule[-1ex]{0pt}{0pt} & \ \ 90 \ \ & \ \ 210 \ \ & \ \ 390 \ \ \\
\hline
\end{array}

Provider $B$ charges 52.5 cents per kWh, with no fixed monthly charge. The graph shows how Provider $B$'s charges vary with the amount of electricity used in a month.

On the grid above, graph Provider $A$'s charges from the table in part (a). (1 mark)
Use the two graphs to determine the number of kilowatt hours per month for which Provider $A$ and Provider $B$ charge the same amount. (1 mark)

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A customer uses an average of 600 kWh per month.
Which provider, $A$ or $B$, would be the cheaper option and by how much? (2 marks)

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a. $\text{When kWh} =400$

$\text{Monthly charge}\ =$90+0.30\times 400=$210$

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ 0 \ \ & \ \ 400 \ \ & \ 1000 \ \\
\hline
\rule{0pt}{2.5ex} \textit{Monthly Charge (\$)} \rule[-1ex]{0pt}{0pt} & \ \ 90 \ \ & \ \ 210 \ \ & \ \ 390 \ \ \\
\hline
\end{array}

c. $\text{400 kWh}$

d. $\text{Provider}\ A\ \text{is cheaper by \$45.}$

Show Worked Solution

a. $\text{When kWh} =400$

$\text{Monthly charge}\ =$90+0.30\times 400=$210$

c. $A_{\text{charge}} = B_{\text{charge}}\ \text{at intersection.}$

$\therefore\ \text{Same charge at 400 kWh}$

d. $\text{Cost at 600 kWh:}$

$\text{Method 1: Using graph}\rightarrow\ $315-270=$45$

$\text{Method 2: Algebraically}:$

$\text{Provider}\ A: \ 90 + 0.30 \times 600 = $270$

$\text{Provider}\ B: \ 0.525 \times 600 = $315$

$\therefore \text{Provider}\ A\ \text{is cheaper by \$45.}$

v1 Algebra, STD2 A4 2018 HSC 27d

The graph displays the cost ($$c$) charged by two companies for the hire of a jetski for $x$ hours.

Both companies charge $450 for the hire of a jetski for 5 hours.

What is the hourly rate charged by Company A? (1 mark)

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Company B charges an initial booking fee of $80.

Write a formula, in the of $c=b+mx$, for the cost of hiring a jetski from Company B for $x$ hours. (2 marks)

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A jetski is hired for 7 hours from Company B.

Calculate how much cheaper this is than hiring from Company A. (2 marks)

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$$90$
$c=80+74x$
$$32$

Show Worked Solution

i.	$\text{Hourly rate}\ (A)$	$=\dfrac{450}{5}$
		$=$90$

ii. $m=\text{hourly rate}$

$\text{Find}\ m,\ \text{given}\ c = 450,\ \text{when}\ \ x = 5\ \text{and}\ \ b = 80$

$450$	$=80+m\times 5$
$5m$	$=370$
$m$	$=\dfrac{370}{5}=74$

$\therefore\ c=80+74x$

iii.	$\text{Cost}\ (A)$	$=90\times 7=$630$
	$\text{Cost}\ (B)$	$=80+74\times 7=$598$

$\therefore\ \text{Company}\ B’\text{s hiring cost is }$32\ \text{cheaper.}$

v1 Algebra, STD2 A4 2004 HSC 16 MC

Uri drew a correct diagram that gave the solution to the simultaneous equations

$y=2x+3$ and $y=x+4$.

Which diagram did he draw?

Show Answers Only

$D$

Show Worked Solution

$\text{By elimination:}$

$y=2x+3\ \text{cuts the }y \text{-axis at}\ 3$

$\rightarrow\ \text{Eliminate be A and B}$

$y=x+4\ \text{cuts the }y\text{-axis at}\ 4$

$\text{AND has a positive gradient}$

$\rightarrow\ \text{Eliminate C}$

$\Rightarrow D$

v1 Algebra, STD2 A4 2011 HSC 20 MC

A function centre hosts events for up to 500 people. The cost $C$, in dollars, for the centre
to host an event, where $x$ people attend, is given by:

$C=20\ 000+40x$

The centre charges $120 per person. Its income $I$, in dollars, is given by:

$I=120x$

How much greater is the income of the function centre when 500 people attend an event, than its income at the breakeven point?

$$10\ 000$
$$20\ 000$
$$30\ 000$
$$40\ 000$

Show Answers Only

$C$

Show Worked Solution

$\text{When}\ x=500,\ I=120\times 500=$60\ 000$

$\text{Breakeven when}\ \ x=250\ \ \text{(from graph)}$

$\text{When}\ \ x=250,\ I=120\times 250=$30\ 000$

$\text{Difference}$	$=60\ 000-30\ 000$
	$=$30\ 000$

$\Rightarrow C$

♦ Mean mark 50%
COMMENT: Students can read the income levels directly off the graph to save time and then check with the equations given.

v1 Algebra, STD2 A4 SM-Bank 7 MC

A computer application was used to draw the graphs of the equations

$x+y=-6$ and $x-y=-6$

Part of the screen is shown.

Which row of the table correctly matches the equations with the lines drawn and identifies the solution when the equations are solved simultaneously?

\begin{align*}
\begin{array}{c|c}
\text{ } \\
\textbf{ A. } \\
\textbf{ B. } \\
\textbf{ C. } \\
\textbf{ D. }
\end{array}
\begin{array}{|c|c|c|}
\hline
\ x+y=-6 & x-y=-6 & \text{Solution} \\
\hline
\text{Line 1} & \text{Line 2} & x=-6,\ y=0 \\
\hline
\text{Line 1} & \text{Line 2} & x=-6, y=-6 \\
\hline
\text{Line 2} & \text{Line 1} & x=-6,\ y=0 \\
\hline
\text{Line 2} & \text{Line 1} & x=-6, y=-6 \\
\hline
\end{array}
\end{align*}

Show Answers Only

$C$

Show Worked Solution

$\text{Line 1:} \ \ x-y=-6$

$\text{Line 2:} \ \ x+y=-6$

$\text{Intersection at} \ (-6, 0).$

$\Rightarrow C$

v1 Algebra, STD2 A4 SM-Bank 6 MC

A computer application was used to draw the graphs of the equations

$x-y=-5$ and $x+y=5$

Part of the screen is shown.

What is the solution when the equations are solved simultaneously?

$x=5,\ y=5$
$x=5,\ y=0$
$x=-5,\ y=0$
$x=0,\ y=5$

Show Answers Only

$D$

Show Worked Solution

$\text{Solution occurs at the intersection of the two lines.}$

$\Rightarrow D$

v1 Algebra, STD2 A4 2017 HSC 17 MC

The graph of the line with equation $y=5-x$ is shown.

When the graph of the line with equation $y=2x-1$ is also drawn on this number plane, what will be the point of intersection of the two lines?

$(0, 5)$
$(1, 2)$
$(2, 3)$
$(5, 0)$

Show Answers Only

$C$

Show Worked Solution

$\text{Method 1: Graphically}$

$\text{From graph, intersection is at} (2,3)$

$\text{Method 2: Algebraically}$

$y$	$=5-x$	$…\ (1)$
$y$	$=2x-1$	$…\ (2)$

$\text{Substitute (2) into (1)}$

$2x-1$	$=5-x$
$3x$	$=6$
$x$	$=2$

$\text{When}\ \ x=2,\ y=5-2=3$

$\Rightarrow C$

\(\text{Income}\)	\(=35\times 4=$140\ \ \text{is equal to}\)
\(\text{Costs}\)	\(=40+(25\times 4)=$140\)

i.	\($C\)	\(=550+250+(9\times x)\)
		\(=800+9x\)

\(\text{Total Volume}\)	\(=\text{Tank A} + \text{Tank B}\)
\(1400\)	\(=1400-40t+(t-10)\times 60\)
\(1400\)	\(=1400-40t+60t-600\)
\(20t\)	\(= 600\)
\(t\)	\(= 30 \ \text{minutes}\)

\(2500\)	\(=30x-(400+10x)\)
\(20x\)	\(=2900\)
\(x\)	\(=145\)

\(60n\)	\(=35n+5000\)
\(25n\)	\(=5000\)
\(n\)	\(=200\)

\(\text{Income}\)	\(=200\times $15\)
	\(=$3000\)

\(\text{Costs}\)	\(=800+(9\times 200)\)
	\(=$2600\)

\(\therefore\ \text{Profit}\)	\(=3000-2600\)
	\(=$400\)

\(C\)	\(=800+(9\times 300)\)
	\(=$3500\)