Simultaneous Linear Equations

Algebra, STD1 EQ-Bank 24

Zara is looking for a new mobile phone plan. She has found two plans that suit her needs and wants to work out which plan is cheaper depending on how many minutes she uses.

Plan $\text{A}$: $20 per month fixed charge plus $0.10 per minute

Plan $\text{B}$: $0.30 per minute, no fixed charge

Let $C$ = total monthly cost in dollars, and $m$ = number of minutes used.

Plan $\text{B}$ can be modelled by the equation $C=0.30m$.
Write an equation for the monthly cost of Plan $\text{A}$. (1 mark)

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The graph of Plan $\text{B}$ is provided on the grid below. Use the equation from (a) to add the graph of Plan $\text{A}$ to the grid. (2 marks)

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For how many minutes per month do both plans cost the same amount? (1 mark)

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Zara uses an average of 140 minutes per month.
Which plan she should choose and how much does she save compared to the other plan. (1 mark)

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a. $\text{Plan A: }C=20+0.10m$

c. $100\ \text{minutes}$

d. $\text{Plan A, cheaper by } \$8$

Show Worked Solution

a. $\text{Plan A: }\ C=20+0.10m$

b. $\text{Table of values}$

$\begin{array}{|c|c|c|c|c|c|} \hline m & 0 & 50 & 100 & 150 & 200 \\ \hline \text{Plan A} & 20 & 25 & 30 & 35 & 40 \\ \hline \end{array}$

c. $\text{From the graph, the lines intersect at }\ m=100.$

$\therefore\ \text{Both plans cost the same at } 100\ \text{minutes}$

d. $\text{Plan A: }\ C=20+0.10\times140=\$34$

$\text{Plan B: }\ C=0.30\times140=\$42$

$\text{Difference} = 42-34=\$8$

$\therefore\ \text{Zara should choose Plan A, which is \$8 cheaper than Plan B.}$

Algebra, STD1 EQ-Bank 36

Two cyclists, Aiko and Ben, are riding along the same straight track in the same direction.

Aiko starts 2000 m ahead of Ben. Aiko rides at a constant speed of 250 metres/minute and Ben rides at a constant speed of 500 metres/minute.

Let $d$ = distance from the starting point in metres, and

$t$ = time in minutes.

The equation $d=2000+250t$ models Aiko's distance from the starting point.
Write an equation to model Ben's distance. (1 mark)

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Use the equations from (a) to graph Aiko and Ben's journeys on the grid below. (2 marks)

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After how many minutes does Ben catch Aiko? (1 mark)

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How far has each cyclist travelled from their own starting point when they meet? (2 marks)

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a. $d=500t$

c. $8\ \text{minutes}$

d. $\text{Aiko travelled 2000 m, Ben travelled 4000 m.}$

Show Worked Solution

a. $d=500t$

b. $\text{Table of values:}$

$\begin{array}{|c|c|c|c|c|c|c|} \hline t & 0 & 2 & 4 & 6 & 8 & 10 \\ \hline \text{Aiko} & 2000 & 2500 & 3000 & 3500 & 4000 & 4500 \\ \hline \text{Ben} & 0 & 1000 & 2000 & 3000 & 4000 & 5000 \\ \hline \end{array}$

c. $\text{From the graph, the lines intersect at }\ t=8.$

$\therefore\ \text{Ben catches Aiko after 8 minutes}$

d. $\text{When}\ \ t=8, d=4000:$

$\text{Aiko started 2000 m ahead, so the distance travelled}$

$=4000-2000=2000\ \text{m}$

$\text{Ben started at the origin, so the distance travelled =4000 m}$

Algebra, STD1 EQ-Bank 34

Two water tanks sit side by side on a farm.

Tank A has a capacity of 1000 litres and is full. It is being emptied at a constant rate of 60 litres per minute.

At the same time, Tank B is empty and is being filled at a constant rate of 40 litres per minute.

Let $V$ = volume of water in litres, and

$t$ = time in minutes.

The equation $V=1000-60t$ models the volume of water in Tank A.
Write an equation to model the volume of water in Tank B. (1 mark)

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Complete the table below and use the values to graph the volume of water in Tank A and Tank B on the grid below. (3 marks)

$\begin{array}{|c|c|c|c|c|c|c|c|} \hline t & 0 & 2 & 4 & 6 & 8 & 10 & 12 \\ \hline \text{Tank A} & \ \ \ \ \ \ \ \ & 880 & 760 & 640 & 520 & 400 & 280 \\ \hline \text{Tank B} & 0 & \ \ \ \ \ \ \ \ \ & 160 & 240 & 320 & 400 & \ \ \ \ \ \ \ \ \\ \hline \end{array}$

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After how many minutes do both tanks contain the same volume of water? (1 mark)

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What is the volume of water in each tank at this time? (1 mark)

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a. $V=40t$

b. $\text{Table of values:}$

$\begin{array}{|c|c|c|c|c|c|c|c|} \hline t & 0 & 2 & 4 & 6 & 8 & 10 & 12 \\ \hline \text{Tank A} & 1000 & 880 & 760 & 640 & 520 & 400 & 280 \\ \hline \text{Tank B} & 0 & 80 & 160 & 240 & 320 & 400 & 480 \\ \hline \end{array}$

c. $10\ \text{minutes}$

d. $400\ \text{litres}$

Show Worked Solution

a. $V=40t$

b. $\text{Table of values:}$

c. $\text{From the graph, the lines intersect at }\ t=10.$

$\therefore\ \text{Both tanks contain the same volume after 10 minutes}$

d. $\text{Method 1: Graphically}$

$\text{From the graph, when }\ t=10,\ \text{the volume in both tanks = 400 litres}$

$\text{Method 2: Algebraically}$

$\text{When }\ t=10:$

$\text{Tank A volume:}\ \ V=1000-60\times10=400\ \text{L}$

$\text{Tank B volume:}\ \ V=40\times10=400\ \text{L}\ \checkmark$

$\therefore\ \text{Each tank contains } 400\ \text{litres}$

Algebra, STD1 EQ-Bank 6 MC

An AFL football team scored 15 times in a game, made up of goals and behinds.

Goals are worth 6 points and behinds are worth 1 point. The team's total score was 45 points.

Let $g$ = number of goals and $b$ = number of behinds.

Which pair of equations correctly represents this situation?

$g+b=45$ and $6g+b=15$
$6g+b=15$ and $g+b=45$
$g+b=15$ and $6g+b=45$
$g+b=15$ and $g+6b=45$

Show Answers Only

$C$

Show Worked Solution

$g+b=15-\text{Eliminate A and B}$

$\text{Total points from goals} = 6g$

$\text{Total points from behinds} = b$

$\text{Total points} = 6g+b\ \ \Rightarrow\ \ 6g+b=45$

$\Rightarrow C$

Algebra, STD1 EQ-Bank 31

A cake-shop owner sells muffins for $2.50 each. It costs $1 to make each muffin and $300 for the equipment needed to make the muffins.

The owner uses a spreadsheet with formulas to model this situation.

How many muffins need to be sold to ‘break-even’? (1 mark)

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How much profit is made if 400 muffins are sold? (2 marks)

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a. $\text{200 muffins}$

b. $\text{Profit} =\$ 300$

Show Worked Solution

a. $\text{By inspection of the spreadsheet:}$

$\text{Total cost = Revenue = \$500}\ \ \Rightarrow\ \ \text{200 muffins}$

$\text{Breakeven when 200 muffins sold.}$

b. $\text{When 400 muffins are sold:}$

$\text{Revenue} =400 \times \$ 2.50=\$ 1000$

$\text{Cost} =400 \times 1+\$ 300=\$ 700$

$\text{Profit} = 1000-700=\$ 300$

Algebra, STD1 EQ-Bank 15

A local council invests in a community solar farm that sells electricity back to the grid.

The solar farm is expected to operate for 12 years. Each year the farm incurs a maintenance cost of $3000.

The council uses a spreadsheet to model the costs and revenue of the project.

What are the total fixed costs for the solar farm project? (1 mark)

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After how many years does the solar farm break even? (1 mark)
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Calculate the profit the council makes over the full 12-year lifespan of the solar farm. (2 marks)

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a. $$35\,000$

b. $7\ \text{years}$

c. $$25\,000$

Show Worked Solution

a. $\text{Total fixed costs}=$24\,000+$8000+$3000=$35\,000$

b. $\text{From the spreadsheet, at year}\ 7:$

$\text{Total cost}=$56\,000,\ \text{Revenue}=$56\,000\ \checkmark$

$\therefore\ \text{Break-even}=7\ \text{years}$

c. $\text{Project lifespan}=12\ \text{years}$

$\text{Variable cost} =12\times \$3000= $36\,000$

$\text{Total costs} =$35\,000+$36\,000= \$71\,000$

$\text{Revenue} =12\times \$8000 = \$96\,000$

$\therefore\ \text{Profit} = \$96\,000-\$71\,000 = \$25\,000$

Algebra, STD1 EQ-Bank 23

Maya operates Sydney City Walking Tours, a city walking tour business.

The bus she hires holds up to 40 people. Each person on the tour pays $35 and receives a complimentary bottle of water.

Maya uses a spreadsheet to model the costs and revenue for each tour.

What are the total fixed costs for each tour? (1 mark)

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How many people need to attend the tour for Maya to break-even? (1 mark)

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Calculate the profit Maya makes if the tour is fully booked. (2 marks)

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a. $$600$

b. $20\ \text{people}$

c. $$600$

Show Worked Solution

a. $\text{Total fixed costs}=$350+$250=$600$

b. $\text{From the spreadsheet, Maya’s break-even is when }$

$\text{Total cost}=\text{Revenue}=$700$

$\therefore\ \text{People to break-even} = 20$

c. $\text{Fully booked}=40\ \text{people}$

$\text{Variable cost} =40\times \$5= \$200$

$\text{Total costs} =$600+$200= \$800$

$\text{Revenue} =40\times \$35 = \$1400$

$\therefore\ \text{Profit} = \$1400-\$800 = \$600$

Algebra, STD1 EQ-Bank 4 MC

A preschool has 20 bikes in total for the children to use, made up of bicycles and tricycles.

Each bicycle has 2 wheels and each tricycle has 3 wheels. Altogether, there is a total of 46 wheels.

Let $b$ = number of bicycles and $t$ = number of tricycles.

Which pair of equations correctly represents this situation?

$b+t=20$ and $2b+3t=46$
$b+t=46$ and $2b+3t=20$
$b+t=20$ and $3b+2t=46$
$b+t=46$ and $3b+2t=20$

Show Answers Only

$A$

Show Worked Solution

$b+t=20-\text{Eliminate B and D.}$

$\text{Total wheels on bicycles} = 2b$

$\text{Total wheels on tricycles} = 3t$

$\text{Total wheels} = 46\ \ \Rightarrow\ \ 2b+3t=46$

$\Rightarrow A$

Algebra, STD1 EQ-Bank 16

A cafe owner uses a spreadsheet to model the costs and revenue from selling cups of coffee.

The cafe has a fixed setup cost of $450 and a cost per cup 0f $1.50 to make each coffee. The owner sells each coffee for $4.50.

Part of the spreadsheet is shown.

What does it mean for the cafe owner to break even? (1 mark)

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Use the spreadsheet to identify the break-even point for the cafe owner. (1 mark)

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If the cafe owner sells 250 cups of coffee, what profit is made? (2 marks)

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a. $\text{Break-even is when the total cost equals the revenue}$

$\text{i.e. the owner makes neither a profit nor a loss.}$

b. $150\ \text{cups}$

c. $$300$

Show Worked Solution

a. $\text{Break-even is when the total cost equals the revenue}$

$\text{i.e. the owner makes neither a profit nor a loss.}$

b. $\text{From the spreadsheet, total cost equals revenue when }$

$\text{Cost}=\text{Revenue}=$675$

$\therefore\ \text{Cups sold to break-even} = 150$

$\text{Confirmed by cell C15:}$

$\text{Break-even (cups)}=\dfrac{450}{4.50-1.50}=\dfrac{450}{3}=150 \text{ cups}$

c. $\text{At 250 cups (row 9): Revenue} = \$1125,\ \text{Total cost} = \$825$

$\therefore\ \text{Profit} = 1125-825 = \$300$

Algebra, STD1 EQ-Bank 22

Gemstar Promotions is organising the annual Concert Under the Stars event to take place on the last weekend in January.

The outdoor venue holds up to 800 people. Each ticket holder receives a complimentary souvenir program valued at $15.

Garth from Gemstar Promotions uses a spreadsheet to model the costs and revenue for the concert.

What are the total fixed costs for the concert? (1 mark)

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The promoter will only proceed with the concert if the loss is no more than $3000.
What is the minimum number of tickets that must be sold for the concert to go ahead? (2 marks)

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Use the formula in cell C9 to calculate the number of tickets that must be sold for the promoter to break even? (1 mark)

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Calculate the profit for Gemstar Promotions if the concert is fully booked. (1 mark)

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a. $$30\,000$

b. $450\ \text{people}$

c. $500\ \text{people}$

d. $$18\,000$

Show Worked Solution

a. $\text{Total fixed costs}=$8000+$6000+$16\,000=$30\,000$

b. $\text{Maximum allowable loss}=$3000$

$\text{From the spreadsheet, at}\ 450\ \text{people}:$

$\text{Total cost}=$36\,750,\ \text{Revenue}=$33\,750$

$\text{Loss}=$36\,750-$33\,750=$3000\ \checkmark$

$\therefore\ \text{Minimum ticket sales}=450$

c. $\text{Breakeven = C6/(C7-C8)}$

$\text{Breakeven}\ = \dfrac{30\,000}{75.00-15.00}=500\ \text{people}$

d. $\text{Venue capacity}=800\ \text{people}$

$\text{Variable cost} =800\times \$15= $12\,000$

$\text{Total costs} =$30\,000+$12\,000= \$42\,000$

$\text{Revenue} =800\times \$75 = \$60\,000$

$\therefore\ \text{Profit} = \$60\,000-\$42\,000 = \$18\,000$

Algebra, STD1 A3 2025 HSC 23

It costs $465 to register a passenger car and $350 to register a motorcycle.

Let	$P$	$\ =\ \text{the number of passenger cars, and}$
	$B$	$\ =\ \text{the number of motorcycles}$

Write TWO linear equations that represent the relationship below.

There are 11 times as many passenger cars as motorcycles.
The total registration fees for passenger cars and motorcycles is $494 million. (2 marks)

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$\text{Equation 1: }\ P=11B$

$\text{Equation 2: }\ 465P+350B=494\ 000\ 000$

Show Worked Solution

$\text{Equation 1: }\ P=11B$

$\text{Equation 2: }\ 465P+350B=494\ 000\ 000$

♦♦♦ Mean mark 11%.

Algebra, STD1 A3 2025 HSC 5 MC

A baker makes and sells cakes.

The straight-line graphs represent cost $(C )$ and revenue $(R)$ in dollars, and $n$ is the number of cakes.

What profit will the baker make by selling 6 cakes?

Show Answers Only

$A$

Show Worked Solution

$\text{When }n=6$

$\text{Revenue}=10\times 6=60$

$\text{Cost}=20+5\times 6=50$

$\therefore\ \text{Profit}=$60-$50=$10$

$\Rightarrow A$

♦♦ Mean mark 53%.

Algebra, STD1 A3 2024 HSC 23

Carrie is organising a fundraiser.

The cost of hiring the venue and the band is $2500. The cost of providing meals is $50 per person.

Complete the table of values to show the total cost of the fundraiser. (1 mark)

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\begin{array} {|l|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Number of people} \rule[-1ex]{0pt}{0pt} & \ \ \ \ 0\ \ \ \ & 25 & 50 & 75 & 100 & 125 & 150 \\
\hline
\rule{0pt}{2.5ex} \text{Cost} \rule[-1ex]{0pt}{0pt} & & 3750 & 5000 & 6250 & 7500 & 8750 & 10\,000 \\
\hline
\end{array}

Carrie decides that tickets should be sold at $70 per person. The graph shows the expected revenue at this ticket price. Using the information in part (a), plot the line that shows the cost of the fundraiser. (2 marks)
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How many tickets need to be sold for the fundraiser to break even? (1 mark)

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Carrie sold 300 tickets. How much profit did the fundraiser make? (3 marks)

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a. $\text{Cost when 0 people attend}= $2500$

c. $100\ \text{tickets}$

d. $\text{Profit }=$3500$

Show Worked Solution

a. $\text{Cost when 0 people }= $2500$

c. $\text{Point of intersection}\ \ \Rightarrow\ \ \text{break-even}$

$\therefore\ \text{Break-even when 100 tickets sold.}$

Mean mark (c) 53%.

d. $\text{Revenue}\ (R)=70n\ \ (n=\ \text{number of people)}$

$\text{Cost}\ (C)=2500 + \Big(\dfrac{1250}{25}\Big)n=2500+50n$

$\text{Find profit}\ (P)\ \text{when}\ \ n=300:$

$P$	$=R-C$
	$=70\times 300-(2500+50\times300)$
	$=21\,000-17\,500=$3500$

♦ Mean mark (d) 46%.

Algebra, STD1 A3 2023 HSC 26

Electricity provider $A$ charges 25 cents per kilowatt hour (kWh) for electricity, plus a fixed monthly charge of $40.

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} \text{Electricity used in a month (kWh)} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ 0\ \ \ \ \ & \ \ \ 400\ \ \ & \ \ 1000\ \ \\
\hline
\rule{0pt}{2.5ex} \text{Monthly charge (\$)} \rule[-1ex]{0pt}{0pt} & 40 & & 290 \\
\hline
\end{array}

Provider $B$ charges 35 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 on 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 800 kWh per month.
Which provider, $A$ or $B$, would be the cheaper option and by how much? (2 marks)

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c. $400\text{ kWh}$

d. $A\text{ is cheaper by }$40.$

Show Worked Solution

♦♦ Mean mark (b) 40%.

c. $\text{Same charge when Provider }A = \text{Provider } B \text{ i.e. where the lines intersect}$

$=400\text{ kWh (see graph above)}$

d. $\text{When kWh}= 800, \ \ A=$240 \text{ and } B=$280$

$\therefore A\text{ is cheaper by }$40.$

Algebra, STD1 A3 2022 HSC 25

Sam is making cupcakes to sell at a market. It costs Sam $60 to hire a stall, and each cupcake costs $1.50 to make. Sam intends to sell each cupcake for $4.00.

The equations representing Sam's cost `($ C)` and revenue `($ R)`, are

`C=1.5 x+60` and `R=4 x`, where `x` is the number of cupcakes sold.

The graphs of `C` and `R` are shown below.

How many cupcakes must Sam sell in order to break even? (1 mark)

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If Sam sells 60 cupcakes, what profit is made?
You may assume that Profit = Revenue – Cost. (2 marks)

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a. `24`

b. `$90`

Show Worked Solution

a. `\text(Break even occurs where the two graphs intersect.)`

`\text(→ 24 cupcakes)`

b. `\text(If cupcakes sold) (x) = 60:`

`C = 1.5 xx 60 + 60= $150`

`R = 4 xx 60=$240`

`:.\ text{Profit}\ = 240-150 = $90`

Algebra, STD1 A3 2021 HSC 29

In a park the only animals are goannas and emus. Let `x` be the number of goannas and let `y` be the number of emus.

The number of goannas plus the number of emus in the park is 31. Hence `x + y = 31`.

Each goanna has four legs and each emu has two legs. In total the emus and goannas have 76 legs.

By writing another relevant equation and graphing both equations on the grid, find the number of goannas and the number of emus in the park. (4 marks)

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Number of goannas = _______________

Number of emus = _________________

Show Answers Only

`text{7 goannas, 24 emus}`

Show Worked Solution

`text{Total goanna legs} = 4x`

♦♦♦ Mean mark 8%.

`text{Total emu legs} = 2y`

`text{Total legs} = 76`

`4x + 2y`	`= 76`
`2x + y`	`= 38\ \ …\ (1)`
`x + y`	`= 31\ \ …\ (2)`

`text{Number of goannas}\ (x) = 7`

`text{Number of emus}\ (y) = 24`

Algebra, STD1 A3 2020 HSC 29

There are two tanks on a property, Tank A and Tank B. Initially, Tank A holds 1000 litres of water and Tank B is empty.

Tank A begins to lose water at a constant rate of 20 litres per minute.

The volume of water in Tank A is modelled by `V = 1000-20t` where `V` is the volume in litres and `t` is the time in minutes from when the tank begins to lose water.

On the grid below, draw the graph of this model and label it as Tank A. (1 mark)
Tank B remains empty until `t=15` when water is added to it at a constant rate of 30 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 water. (2 marks)
Using the graphs drawn, or otherwise, find the value of `t` (where `t > 0`) when the total volume of water in the two tanks is 1000 litres. (1 mark)

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a. `text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}`

b. `29 \ text{minutes}`

c. `45 \ text{minutes}`

Show Worked Solution

a. `text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}`

♦ Mean mark part (a) 45%.

b. `text{T} text{ank} \ B \ text{will pass through (15, 0) and (45, 900)}`

♦♦ Mean mark part (b) 24%.

`text{By inspection, the two graphs intersect at} \ \ t = 29 \ text{minutes}`

c. `text{Strategy 1}`

♦♦♦ Mean mark part (c) 5%.

`text{By inspection of the graph, consider} \ \ t = 45`

`text{T} text{ank A} = 100 \ text{L} , \ text{T} text{ank B} =900 \ text{L} `

`:.\ text(Total volume = 1000 L when t = 45)`

`text{Strategy 2}`

`text{Total Volume}`	`=text{T} text{ank A} + text{T} text{ank B}`
`1000`	`= 1000-20t + (t-15) xx 30`
`1000`	`= 1000-20t + 30t-450 `
`10t`	`= 450`
`t`	`= 45 \ text{minutes}`

Algebra, STD1 A3 2019 HSC 30

A small business makes and sells bird houses.

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

How many bird houses 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 bird houses need to be sold to earn a profit of $1900. (3 marks)

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a. `20`

b. `96`

Show Worked Solution

a. `20\ \ (xtext(-value at intersection))`

b. `text(Find equations of both lines):`

♦ Mean mark part (b) 47%.

`(0, 500)\ text(and)\ (20, 800)\ text(lie on)\ \ C`

`m_C = (800-500)/(20-1) = 15`

`=> C = 500 + 15x`

`(0,0)\ text(and)\ (20, 800)\ text(lie on)\ \ R`

`m_R = (800-0)/(20-0) = 40`

`=> R = 40x`

`R-C =text(Profit)`

`text(Find)\ \ x\ \ text(when Profit = $1900:)`

`40x-(500 + 15x)`	`=1900`
`40x-500-15x`	`=1900`
`25x`	`= 2400`
`x`	`= 96`

Algebra, STD2 A4 EQ-Bank 5 MC

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

`x + y = 4` and `x-y = 4`

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?

Show Answers Only

`A`

Show Worked Solution

`text(Line 1:) \ \ x + y = 4`

`text(Line 2:) \ \ x-y = 4`

`text(Intersection at) \ (4, 0).`

`=> A`

Algebra, STD2 A4 EQ-Bank 2 MC

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

`x-y = 4` and `x + y = 4`

Part of the screen is shown.

What is the solution when the equations are solved simultaneously?

`x = 4, y = 4`
`x = 4, y = 0`
`x = 0, y = 4`
`x = 0, y =-4`

Show Answers Only

`B`

Show Worked Solution

`text(Solution occurs at the intersection of the two lines.)`

`=> B`

Algebra, STD2 A4 2014 HSC 26d

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

`y = 2x + 1`

`x-2y-4 = 0`

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`x=-2,\ y=-3`

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`text(Solution is at the intersection:)\ \ x=-2,\ y=-3`

Algebra, STD2 A4 EQ-Bank 19

The graph of the line `y = 4-x` is shown.

By graphing `y = 2x + 1` on the grid provided, find the point of intersection of `y = 4-x` and `y = 2x + 1`. (3 marks)

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`(1, 3)`

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`text(Graphing)\ y = 2x + 1 :`

`ytext(-intercept = 1)`

`text(Gradient = 2)`

`:.\ text{Point of intersection is (1, 3).}`

Algebra, STD2 A4 EQ-Bank 18

A student was asked to solve the following simultaneous equations.

`y = 2x-8`

`x-4y + 3 = 0`

After graphing the equations, the student found the point of intersection to be `(5,2)`?

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

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`text(See Worked Solutions)`

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`text{Substitute (5, 2) into}\ \ y = 2x-8`

`text(LHS) = y = 2`

`text(RHS) = 2(5)-8 = 2=\ text{LHS}`

`text{Substitute (5, 2) into}\ \ x-4y + 3 = 0`

`text(LHS)= 5-4(2) + 3= 0=\ text(RHS)`

`=> (5,2)\ text(satisfies both equations.)`

`:.\ text(Student is correct.)`

Algebra, STD2 A4 EQ-Bank 17

`y`	`= x + 5`
`y + 2x`	`= 2`

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

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`(-1,4)`

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`text(Table of coordinates:)\ \ y = x + 5`

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

`text(Table of coordinates:)\ \ y + 2x = 2 \ => \ y =-2x + 2`

\begin{array} {|c|c|c|}
\hline \quad x \quad & \ \ -2 \ \ & \ \ -1 \ \ & \quad 0 \quad & \quad 1 \quad\\
\hline y & 6 & 4 & 2 & 0 \\
\hline \end{array}

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

`text(at)\ \ (-1,4).`

Algebra, STD2 A4 2018 HSC 27d

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

Both companies charge $360 for the hire of a minibus for 3 hours.

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

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Company B charges an initial booking fee of $75.

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

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A minibus is hired for 5 hours from Company B.

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

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a. `$120`

b. `c = 95x + 75`

c. `$50`

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a. `text(Hourly rate)\ (A)= 360 ÷ 3= $120`

b. `m = text(hourly rate)`

`text(Find)\ m,\ text(given)\ c = 360,\ text(when)\ \ x = 3\ \ text(and)\ \ b = 75`

`360`	`= m xx 3 + 75`
`3m`	`= 285`
`m`	`= 95`

`:. c = 95x + 75`

c.	`text(C)text(ost)\ (A)`	`= 120 xx 5 = $600`
	`text(C)text(ost)\ (B)`	`= 95 xx 5 + 75 = $550`

`:.\ text(The hiring cost for Company)\ B\ text(is $50 cheaper.)`

Algebra, STD2 A4 EQ-Bank 27

Penny is a baker and makes meat pies every day.

The cost of making `p` pies, `$C`, can be calculated using the equation

`C = 675 + 3.5 p`

Penny sells the pies for $5.75 each, and her income is calculated using the equation

`I = 5.75 p`

On the graph, 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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a. & b.

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b. `text(Loss zone occurs when)\ C > I,\ text(which is shaded in the diagram above.)`

Algebra, STD2 A4 EQ-Bank 24

Temperature can be measured in degrees Celsius (`C`) or degrees Fahrenheit (`F`).

The two temperature scales are related by the equation `F = (9C)/5 + 32`.

Calculate the temperature in degrees Fahrenheit when it is `-20` degrees Celsius. (1 mark)

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The following two graphs are drawn on the axes below:

`F = (9C)/5 + 32` and `F = C`

Explain what happens at the point where the two graphs intersect. (1 mark)

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a. `-4^@F`

b. `text(The two graphs intersect at a temperature where)`

`text(Celsius and Fahrenheit are the same.)`

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a. `F= (9xx(-20))/5 + 32=-4^@F`

b. `text(The two graphs intersect at a temperature where)`

`text(Celsius and Fahrenheit are the same.)`

Algebra, STD2 A4 2017 HSC 17 MC

The graph of the line with equation `y = 6-2x` is shown.

When the graph of the line with equation `y = x + 3` is also drawn on this number plane, what will be the point of intersection of the two lines?

`(0, 6)`
`(1, 4)`
`(2, 2)`
`(3, 0)`

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`B`

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`text(Method 1 – Graphically)`

`text(From graph, intersection at)\ (1,4)`

`=>B`

`text(Method 2 – Algebraically)`

`y`	`= 6-2x`	`…\ (1)`
`y`	`= x + 3`	`…\ (2)`

`text{Substitute (2) into (1)}`

`x + 3`	`= 6-2x`
`3x`	`= 3`
`x`	`= 1`

`text(When)\ \ x = 1,\ y = 4`

`=>B`

Algebra, STD2 A4 2005 HSC 28b

Sue and Mikey are planning a fund-raising dance. They can hire a hall for $400 and a band for $300. Refreshments will cost them $12 per person.

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

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The graph shows planned income and costs when the ticket price is $20.

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

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How much profit will be made if 150 people attend the dance? (1 mark)

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Sue and Mikey plan to sell 200 tickets. They want to make a profit of $1500.

What should be the price of a ticket, assuming all 200 tickets will be sold? (3 marks)

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a. `700 + 12x`

b. `text(Approximately 90)`

c. `$500`

d. `$23`

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a. `$C= 400 + 300 + (12 xx x)= 700 + 12x`

b. `text(Using the graph intersection:)`

`text(Approximately 90 people are needed to cover the costs.)`

c. `text(If 150 people attend)`

`text(Income)= 150 xx $20= $3000`

`text(C)text(osts)= 700 + (12 xx 150)= $2500`

`:.\ text(Profit)= 3000-2500= $500`

d. `text(C)text(osts when)\ x = 200:`

`C=700 + (12 xx 200)= $3100`

`text(Income required to make $1500 profit)`

`= 3100 + 1500= $4600`

`:.\ text(Price per ticket)= 4600/200= $23`

Algebra, STD2 A4 2004 HSC 16 MC

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

`y = 2x-5` and `y = x + 6`.

Which diagram did he draw?

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`D`

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`text(By elimination)`

`y = 2x-5\ \ text(cuts the y-axis at)\ -5`

`:.\ text(Cannot be A or B)`

`y = x + 6\ \ text(cuts the y-axis at)\ 6`

`text(AND has a positive gradient)`

`:.\ text(Cannot be C)`

`=> D`

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 = 10\ 000 + 50x`

The centre charges $100 per person. Its income `I`, in dollars, is given by:

`I = 100x`

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

`$15\ 000`
`$20\ 000`
`$30\ 000`
`$40\ 000`

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`C`

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

`text(When)\ x=500,\ I=100xx500=$50\ 000`

`text(Breakeven when)\ \ x=200\ \ \ text{(from graph)}`

`text(When)\ \ x=200,\ I=100xx200=$20\ 000`

`text(Difference)=50\ 000-20\ 000=$30\ 000`

`=> C`

Algebra, STD2 A4 2010 HSC 24b

Ashley makes picture frames as part of her business. To calculate the cost, `C`, in dollars, of making `x` frames, she uses the equation `C=40+10x`.

She sells the frames for $20 each and determines her income, `I`, in dollars, using the equation `I=20x`.

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

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`x=4`

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`text(From the graph, intersection occurs at)\ x=4`

♦ 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`.

`=>\ text(Break-even point occurs at)\ x=4`

`text(i.e. when 4 frames sold)`

`text(Income)`	`=20xx4=$80\ \ text(is equal to)`
`text(C)text(osts)`	`=40+(10xx4)=$80`

`text(If)\ <4\ text(frames sold)=>\ text(LOSS for business)`

`text(If)\ >4\ text(frames sold)=>\ text(PROFIT)`

Let	\(P\)	\(\ =\ \text{the number of passenger cars, and}\)
	\(B\)	\(\ =\ \text{the number of motorcycles}\)

\(P\)	\(=R-C\)
	\(=70\times 300-(2500+50\times300)\)
	\(=21\,000-17\,500=$3500\)