smc-5237-40-Sketch Linear Equations

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 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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$(2, 1)$

Show Worked Solution

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

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

$\text{Gradient }=2$

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

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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$x=4,\ y=2$

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

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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$(1,-2)$

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

\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 = $315$

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

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

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

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

\(y\)	\(=x-3\)
\(y+3x\)	\(=1\)