Linear Equations and Basic Graphs (Std 2-X)

v1 Algebra, STD2 A2 2012 HSC 5 MC

The line below has intercepts $m$ and $n$, where $m$ and $n$ are positive integers.

What is the gradient of the line?

$\dfrac{m}{n}$
$\dfrac{n}{m}$
$-\dfrac{m}{n}$
$-\dfrac{n}{m}$

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$C$

Show Worked Solution

♦ Mean mark 45%

$\text{Gradient}$	$=\dfrac{\text{rise}}{\text{run}}$
	$=-\dfrac{m}{n}$

$\Rightarrow C$

v1 Algebra, STD2 A2 2012 HSC 8 MC

Dots were used to create a pattern. The first three shapes in the pattern are shown.

The number of dots used in each shape is recorded in the table.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Shape $(S)$} \rule[-1ex]{0pt}{0pt} &\;\;\; 1 \;\;\; & \;\; \;2 \;\;\; & \;\;\; 3 \;\;\; \\
\hline
\rule{0pt}{2.5ex} \text{Number of dots $(N)$} \rule[-1ex]{0pt}{0pt} &\;\;\; 8 \;\;\; & \;\; \;10 \;\;\; & \;\; \;12\; \;\; \\
\hline
\end{array}

How many dots would be required for Shape 182?

363
370
546
1092

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$B$

Show Worked Solution

$\text{Linear relationship where}$

$N=6+(2\times S)$

$\text{When}\ \ S=182$

$N$	$=6+(2\times 182)$
	$=370$

$\Rightarrow B$

v1 Algebra, STD2 A2 2009 HSC 14 MC

If $C=5x+4$, and $x$ is increased by 3, what will be the corresponding increase in $C$ ?

$3$
$15$
$3x$
$5x$

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$B$

Show Worked Solution

$C=5x+4$

$\text{If}\ x\ \text{increases by 3}$

$C\ \text{increases by}\ 5\times 3=15$

$\Rightarrow B$

♦ Mean mark 50%.
STRATEGY: Substituting real numbers into the equation can work well in these type of questions. eg. If $x=0,\ C=4$ and when $x=3,\ C=19$.

EXAMCOPY Functions, 2ADV F1 2009 HSC 1a

Sketch the graph of `y-2x = 3`, showing the intercepts on both axes. (2 marks)

--- 8 WORK AREA LINES (style=blank) ---

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Show Worked Solution

`y-2x=3\ \ =>\ \ y=2x+3`

`ytext{-intercept}\ = 3`

`text{Find}\ x\ text{when}\ y=0:`

`0-2x=3\ \ =>\ \ x=-3/2`

v1 Algebra, STD2 A2 2014 HSC 7 MC

Which of the following is the graph of $y=-3x-3$?

A.		B.

C.		D.

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$A$

Show Worked Solution

♦ Mean mark 46%

$y=-3x-3$

$\text{By elimination:}$

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

$\rightarrow\ \text{Cannot be}\ B\ \text{or}\ D$

$\text{Gradient}=-3$

$\rightarrow\ \text{Cannot be}\ C$

$\Rightarrow A$

v1 Algebra, STD2 A2 2007 HSC 18 MC

Art started to make this pattern of shapes using matchsticks.

If the pattern of shapes is continued, which shape would use exactly 416 matchsticks?

Shape 83
Shape 103
Shape 104
Shape 138

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$D$

Show Worked Solution

\begin{array} {|l|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Shape}\ \textit(S) \rule[-1ex]{0pt}{0pt}\ \ &\ \ 1\ \ &\ \ 2\ \ &\ \ 3\ \ \\
\hline
\rule{0pt}{2.5ex} \text{Matches}\ \textit(M) \rule[-1ex]{0pt}{0pt} \ \ & \ \ 5\ \ &\ \ 8\ \ &\ \ 11\ \ \\
\hline
\end{array}

$\text{Equation rule:}$

$M=3S+2$

$\text{Find}\ \ S\ \text{when}\ \ M=416:$

$416$	$=3S+2$
$3S$	$=414$
$S$	$=138$

$\therefore\ \text{The 138th shape uses 416 matchsticks.}$

$\Rightarrow D$

EXAMCOPY Algebra, STD2 A1 2009 HSC 16 MC

The time for a car to travel a certain distance varies inversely with its speed.

Which of the following graphs shows this relationship?

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`A`

Show Worked Solution

`T`	`prop 1/S`
`T`	`= k/S`

`text{By elimination:}`

`text(As S) uarr text(, T) darr => text(cannot be B or D)`

♦ Mean mark 38%

`text(C is incorrect because it graphs a linear relationship)`

`=> A`

v1 Algebra, STD2 A2 2011 HSC 23b

Sticks were used to create the following pattern.

The number of sticks used is recorded in the table.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Shape $(S)$} \rule[-1ex]{0pt}{0pt} & \;\;\; 1 \;\;\; & \;\;\; 2 \;\;\; & \;\;\; 3 \;\;\; \\
\hline
\rule{0pt}{2.5ex} \text{Number of sticks $(N)$}\; \rule[-1ex]{0pt}{0pt} & \;\;\; 6 \;\;\; & \;\;\; 10 \;\;\; & \;\;\; 14 \;\;\; \\
\hline
\end{array}

Draw Shape 4 of this pattern. (1 mark)

--- 3 WORK AREA LINES (style=blank) ---
How many sticks would be required for Shape 128? (1 mark)

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Is it possible to create a shape in this pattern using exactly 609 sticks?

Show suitable calculations to support your answer. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

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$\text{See Worked Solutions.}$
$514$
$\text{No (See worked solution)}$

Show Worked Solution

i. $\text{Shape 4 is shown below:}$

ii. $\text{Since}\ \ N=2+4S$

♦ Mean mark 48%.
MARKER’S COMMENT: Students should attempt to find a “rule” in such questions, and use this formula to solve the question, as per the Worked Solution.

$\text{If }S$	$=128$
$N$	$=2+(4\times 128)$
	$=514$

iii.	$609$	$=2+4S$
	$4S$	$=607$
	$S$	$=151.75$

$\text{Since}\ S\ \text{is not a whole number, 609 sticks}$

$\text{will not create a shape in this pattern.}$

v1 Algebra, STD2 A2 2004 HSC 2 MC

Michael drew a graph of the height of a plant.

What is the gradient of the line?

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$B$

Show Worked Solution

$\text{2 points on graph}\ \ (0, 6),\ (1, 9)$

$\text{Gradient}$	$=\dfrac{y_2-y_1}{x_2-x_1}$
	$=\dfrac{9-6}{1-0}$
	$=3$

$\Rightarrow B$

v1 Algebra, STD2 A2 2006 HSC 7 MC

Which equation represents the relationship between $x$ and $y$ in this table?

\begin{array} {|c|c|c|}
\hline \ \ x\ \ & \ \ 0\ \ &\ \ 2\ \ & \ \ 4\ \ & \ \ 6\ \ & \ \ 8\ \ \\
\hline y & 3 & 4 & 5 & 6 & 7 \\
\hline \end{array}

$y=2x+3$
$y=\dfrac{1}{2}x+3$
$y=\dfrac{1}{2}x-3$
$y=3x-2$

Show Answers Only

$B$

Show Worked Solution

$\text{By elimination (using the table)}$

$(0, 3)\ \text{must satisfy}$

$\therefore\ \text{NOT}\ C\ \text{or}\ D$

$(2, 4)\ \text{must satisfy}$

$\therefore\ \text{NOT}\ A\ \text{as}\ 2\times 2+3\neq\ 5$

$\Rightarrow B$

v1 Algebra, STD2 A2 2005 HSC 17 MC

The total cost, $$C$, of a school excursion is given by $C=4n+9$, where $n$ is the number of students.

If five extra students go on the excursion, by how much does the total cost increase?

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$B$

Show Worked Solution

$C=4n+9$

$\text{If}\ n\ \text{increases to}\ n+5$

$C$	$=4(n+5)+9$
	$=4n+20+9$
	$=4n+29$

$\therefore \text{Total cost increases by }$20$

$\Rightarrow B$

EXAMCOPY Functions, 2ADV F1 2015 HSC 2 MC

What is the slope of the line with equation `2x - 4y + 3 = 0`?

`-2`
`-1/2`
`1/2`
`2`

Show Answers Only

`C`

Show Worked Solution

`2x – 4y + 3`	`= 0`
`4y`	`= 2x + 3`
`y`	`= 1/2 x + 3/4`

`:.\ text(Slope)\ = 1/2`

`=> C`

v1 Algebra, STD2 A2 2015 HSC 13 MC

What is the equation of the line $l$?

$y=-\dfrac{1}{5}x+5$
$y=\dfrac{1}{5}x+5$
$y=-5x+5$
$y=5x+5$

Show Answers Only

$C$

Show Worked Solution

$l\ \ \text{passes through (0, 5) and (1, 0)}$

$\text{Gradient}$	$=\dfrac{y_2-y_1}{x_2-x_1}$
	$=\dfrac{5-0}{0-1}$
	$=-5$

$y\ \text{-intercept}= 5$

$\therefore\ y=-5x+5$

$\Rightarrow C$

♦ Mean mark 48%.

v1 Algebra, STD2 A2 2016 HSC 14 MC

The graph shows a line which has an equation in the form $y=mx+c$.

Which of the following statements is true?

$m$ is positive and $c$ is negative
$m$ is negative and $c$ is positive
$m$ and $c$ are both positive
$m$ and $c$ are both negative

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$B$

Show Worked Solution

$y\text{-intercept}\ (c)\ \text{is positive}$

$\rightarrow\ \text{eliminate A and D}$

$\text{gradient}\ (m)\ \text{is negative}$

$\rightarrow\ \text{eliminate C}$

$\Rightarrow B$

v1 Algebra, STD2 A2 2017 HSC 20 MC

A pentagon is created using matches.

By adding more matches, a row of two pentagons is formed.

Continuing to add matches, a row of three pentagons can be formed.

Continuing this pattern, what is the maximum number of complete pentagons that can be formed if 230 matches in total are available?

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$C$

Show Worked Solution

$\text{1 pentagon:}\ 5+4\times 0=5$

$\text{2 pentagons:}\ 5+4\times 1=9$

$\text{3 pentagons:}\ 5+4\times 2 = 13$

$\vdots$

$n\ \text{pentagons:}\ 5 + 4(n – 1)$

$5+4(n – 1)$	$=230$
$4n-4$	$=225$
$4n$	$=229$
$n$	$=57.25$

$\text{Complete pentagons possible}\ =57$

$\Rightarrow C$

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

$\text{Method 1: Graphically}$

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

$\text{Method 2: Algebraically}$

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

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

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

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

$\Rightarrow C$

EXAMCOPY Functions, 2ADV F1 2017 HSC 1 MC

What is the gradient of the line `2x + 3y + 4 = 0`?

`-2/3`
`2/3`
`-3/2`
`3/2`

Show Answers Only

`A`

Show Worked Solution

`2x + 3y + 4`	`= 0`
`3y`	`= -2x-4`
`y`	`= -2/3 x-4/3`
`:.\ text(Gradient)`	`= -2/3`

`=> A`

v1 Algebra, STD2 A2 SM-Bank 1

Sketch the graph of $y-3x=5$, showing the intercepts on both axes. (2 marks)

--- 8 WORK AREA LINES (style=blank) ---

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Show Worked Solution

$y-3x=5\ \ \rightarrow\ y=3x+5$

$y\text{-intercept}=(0 , 3)$

$x\text{-intercept}=\bigg(-\dfrac{5}{3} , 0\bigg)$

v1 Algebra, STD2 A2 SM-Bank 1 MC

What is the slope of the line with equation $3x-9y+5=0$?

$3$
$\dfrac{1}{3}$
$-\dfrac{1}{3}$
$-3$

Show Answers Only

$B$

Show Worked Solution

$3x-9y+5$	$=0$
$9y$	$=3x+5$
$y$	$=\dfrac{1}{3}x+\dfrac{5}{9}$

$\therefore\ \text{Slope}\ =\dfrac{1}{3}$

$\Rightarrow B$

v1 Algebra, STD2 A2 SM-Bank 2 MC

What is the gradient of the line $5x+7y+3=0$?

$-\dfrac{7}{5}$
$-\dfrac{5}{7}$
$\dfrac{7}{5}$
$\dfrac{5}{7}$

Show Answers Only

$B$

Show Worked Solution

$5x+7y+3$	$=0$
$7y$	$=-5x-3$
$y$	$=-\dfrac{5}{7}x-\dfrac{3}{7}$

$\therefore\ \text{Gradient}=-\dfrac{5}{7}$

$\Rightarrow B$

v1 Algebra, STD2 A2 SM-Bank 5 MC

The equation of the line drawn in the diagram below is:

$5y=8x$
$y=-\dfrac{8}{5}x+8$
$y=\dfrac{8}{5}x+8$
$y=-\dfrac{5}{8}x+8$

Show Answers Only

$B$

Show Worked Solution

$y\text{-intercept}=+8$

$\text{Gradient}$	$=\dfrac{\text{rise}}{\text{run}}$
	$=-\dfrac{8}{5}$

$\therefore\ \text{Equation is}:\ y=-\dfrac{8}{5}x+8$

$\Rightarrow B$

v1 Algebra, 2ADV F1 2018 HSC 3 MC

What is the $x$-intercept of the line $x-4y+8=0$?

$(-2, 0)$
$(-8, 0)$
$(0, -8)$
$(0, -2)$

Show Answers Only

$B$

Show Worked Solution

$x\text{-intercept occurs when}\ y = 0:$

$x-4y+8$	$=0$
$x$	$=-8$

$\therefore\ x\text{-intercept is}\ (-8, 0)$

$\Rightarrow B$

v1 Algebra, STD2 A2 SM-Bank 5

The diagram shows the graph of a line.

What is the equation of this line? (2 marks)

Show Answers Only

$y=\dfrac{1}{6}x+1$

Show Worked Solution

$y\text{-intercept}=1$

$\text{Gradient}$	$=\dfrac{\text{rise}}{\text{run}}$
	$=\dfrac{1}{6}$

$\therefore\ \text{Equation:}\ \ y=\dfrac{1}{6}x+1$

v1 Algebra, STD2 A2 SM-Bank 6 MC

Hannah went paddle boarding on a holiday.

The hiring charges are listed in the table below:

\begin{array} {|l|c|c|}
\hline \text{Hours hired} \ (h) & 1 & 2 & 3 & 4 & 5 \\
\hline \text{Cost} \ (C) & 15 & 23 & 31 & 39 & 47 \\
\hline \end{array}

Which linear equation shows the relationship between $C$ and $h$?

$C=8+7h$
$C=7+8h$
$C=15+8h$
$C=8+15h$

Show Answers Only

$B$

Show Worked Solution

$\text{Consider Option 2:}$

$7+8\times 1=7+8=15$

$7+8\times 2=7+16=23$

$7+8\times 3=7+24=31\ \ \ \ \text{etc …}$

$\text{The linear equation is:}\ \ C=7+8h$

$\Rightarrow B$

v1 Algebra, STD2 A2 SM-Bank 20

Brett uses matchsticks to make a pattern of shapes, as shown in the table below.

How many sticks ($S$) will be needed to make Shape Number 24? (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

$121$

Show Worked Solution

$\text{Shape 1:}\ \ S=5\times 1+1=6$

$\text{Shape 2:}\ \ S=5\times 2+1=11$

$\vdots$

$\text{Shape}\ N:\ \ S=5N+1$

$\therefore\ \text{Sticks required when}\ \ N=24$

$=24\times 5+1$

$=121$

v1 Algebra, STD2 A2 SM-Bank 10 MC

Petra drew a straight line through the points $(0, 4)$ and $(5, -3)$ as shown in the diagram below.

What is the gradient of the line that Petra drew?

$-\dfrac{7}{5}$
$-\dfrac{5}{7}$
$\dfrac{7}{5}$
$\dfrac{5}{7}$

Show Answers Only

$A$

Show Worked Solution

$\text{Line passes through } (0, 4)\ \text{and } (5, -3)$

$\text{Gradient}$	$=\dfrac{y_2-y_1}{x_2-x_1}$
	$=\dfrac{4- -3}{0-5}$
	$=-\dfrac{7}{5}$

$\Rightarrow A$

v1 Algebra, STD2 A2 SM-Bank 11 MC

What is the equation for the line shown?

$y=-\dfrac{4}{3}x+1$
$y=-\dfrac{3}{4}x+1$
$y=\dfrac{4}{3}x+1$
$y=\dfrac{3}{4}x+1$

Show Answers Only

$D$

Show Worked Solution

$\text{Line cuts}\ \ y\text{-axis at 1}$

$\text{Gradient}=\dfrac{\text{rise}}{\text{run}}=\dfrac{3}{4}$

$\therefore\ y=\dfrac{3}{4}x+1$

$\Rightarrow D$

v1 Algebra, STD2 A2 SM-Bank 12 MC

Which of these equations represents the line in the graph?

$y=9-4x$
$y=9+4x$
$y=9-\dfrac{9}{4}x$
$y=9+\dfrac{9}{4}x$

Show Answers Only

$C$

Show Worked Solution

$\text{Graph passes through }(0, 9)\text{ and} (4, 0)$

$\text{Gradient}$	$=\dfrac{y_2-y_1}{x_2-x_1}$
	$=\dfrac{9-0}{0-4}$
	$=-\dfrac{9}{4}$

$\therefore\ \text{Equation is:}\ \ y=9-\dfrac{9}{4}x$

$\Rightarrow C$

v1 Algebra, STD2 A2 2020 HSC 6 MC

Suppose $y=-2-3x$.

When the value of $x$ increases by 4, the value of $y$ decreases by

$1$.
$4$.
$12$.
$14$.

Show Answers Only

$C$

Show Worked Solution

$\text{Strategy 1}$

$\text{If}\ \ x\ \ \text{increases by} \ 4$

$\rightarrow y\ \text{decreases by} \ \ 3x=3\times 4 = 12$

$\text{Strategy 2}$

$\text{Test}\ 2\ \text{values:}$

$\text{If} \ \ x=0 , \ y=-2$

$\text{If}\ \ x=4 , \ y =-2-3\times 4=-14$

$\therefore\ \ y \ \text{decreases by} \ 12.$

$\Rightarrow C$

v1 Algebra, STD2 A2 2021 HSC 9 MC

Marty is thinking of a number. Let the number be $n$.

When Marty subtracts 4 from this number and multiplies the result by 7, the answer is 8 more than $n$.

Which equation can be used to find $n$?

$7n-4=8n$
$7(n-4)=8n$
$7n-4=n+8$
$7(n-4)=n+8$

Show Answers Only

$D$

Show Worked Solution

$\text{The description defines the following equation:}$

$(n-4)\times 7$	$=n+8$
$7(n-4)$	$=n+8$

$\Rightarrow D$

v1 Algebra, STD2 A2 2022 HSC 2 MC

Which of the following could be the graph of $y=-2-2x$?

Show Answers Only

$B$

Show Worked Solution

$\text{By elimination:}$

$y\text{-intercept} =-2\ \rightarrow\ \text{Eliminate}\ A \text{ and}\ D$

$\text{Gradient is negative}\ \rightarrow\ \text{Eliminate}\ C$

$\Rightarrow B$

♦ Mean mark 48%.

v1 Algebra, STD2 A1 2009 HSC 16 MC

The time for a train to travel a certain distance varies inversely with its speed.

Which of the following graphs shows this relationship?

Show Answers Only

$C$

Show Worked Solution

$T$	$\propto \dfrac{1}{S}$
$T$	$=\dfrac{k}{S}$

$\text{By elimination:}$

$\text{As Speed} \uparrow \ \text{, Time}\downarrow\ \Rightarrow\ \text{cannot be A or B}$

$\text{D is incorrect because it graphs a linear relationship}$

$\Rightarrow C$

♦♦ Mean mark 38%.

\(416\)	\(=3S+2\)
\(3S\)	\(=414\)
\(S\)	\(=138\)

\(\text{If }S\)	\(=128\)
\(N\)	\(=2+(4\times 128)\)
	\(=514\)

iii.	\(609\)	\(=2+4S\)
	\(4S\)	\(=607\)
	\(S\)	\(=151.75\)

\(\text{Gradient}\)	\(=\dfrac{y_2-y_1}{x_2-x_1}\)
	\(=\dfrac{9-6}{1-0}\)
	\(=3\)

\(C\)	\(=4(n+5)+9\)
	\(=4n+20+9\)
	\(=4n+29\)

\(\text{Gradient}\)	\(=\dfrac{\text{rise}}{\text{run}}\)
	\(=-\dfrac{m}{n}\)

\(N\)	\(=6+(2\times 182)\)
	\(=370\)

\(\text{Gradient}\)	\(=\dfrac{y_2-y_1}{x_2-x_1}\)
	\(=\dfrac{5-0}{0-1}\)
	\(=-5\)

\(5+4(n – 1)\)	\(=230\)
\(4n-4\)	\(=225\)
\(4n\)	\(=229\)
\(n\)	\(=57.25\)