Linear Modelling and Basic Graphs (Y11-X)

Algebra, STD2 A2 2012 HSC 5 MC (Adapted)

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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$C$

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♦ Mean mark 45%

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

$\Rightarrow C$

Algebra, STD2 A2 2009 HSC 14 MC (Adapted)

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 $x=3,\ C=19$.

Algebra, STD2 A2 2014 HSC 7 MC (Adapted)

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%

$y=-3x-3$

$\text{By elimination:}$

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

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

$\text{Gradient}=-3$

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

$\Rightarrow A$

Algebra, STD2 A2 2004 HSC 2 MC (Adapted)

Michael drew a graph of the height of a plant.

What is the gradient of the line?

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$B$

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

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

$\Rightarrow B$

Algebra, STD2 A2 2006 HSC 7 MC (Adapted)

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}

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

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$B$

Show Worked Solution

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

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

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

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

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

$\Rightarrow B$

Algebra, STD2 A2 2005 HSC 17 MC (Adapted)

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 }$20$

$\Rightarrow B$

Algebra, STD2 A2 2015 HSC 13 MC (Adapted)

What is the equation of the line $l$?

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

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$C$

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%.

Algebra, STD2 A2 2016 HSC 14 MC (Adapted)

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

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

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

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

$\rightarrow\ \text{eliminate C}$

$\Rightarrow B$

Algebra, STD2 A4 2017 HSC 17 MC (Adapted)

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?

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

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$C$

Show Worked Solution

$\text{Method 1: Graphically}$

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

$\text{Method 2: Algebraically}$

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

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

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

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

$\Rightarrow C$

Algebra, STD2 EO-Bank 20

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

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

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

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

Algebra, STD2 EO-Bank 1 MC

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

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

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$B$

Show Worked Solution

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

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

$\Rightarrow B$

Algebra, STD2 EO-Bank 2 MC

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

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

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$B$

Show Worked Solution

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

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

$\Rightarrow B$

Algebra, STD2 EO-Bank 9 MC

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

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

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$B$

Show Worked Solution

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

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

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

$\Rightarrow B$

Algebra, 2ADV F1 2018 HSC 3 MC (Adapted)

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

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

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$B$

Show Worked Solution

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

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

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

$\Rightarrow B$

Algebra, STD2 EO-Bank 28

The diagram shows the graph of a line.

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

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

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$y\text{-intercept}=1$

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

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

Algebra, STD2 EO-Bank 10 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 $h$?

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

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$B$

Show Worked Solution

$\text{Consider Option 2:}$

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

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

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

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

$\Rightarrow B$

Algebra, STD2 EO-Bank 11 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?

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

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$A$

Show Worked Solution

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

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

$\Rightarrow A$

Algebra, STD2 EO-Bank 3 MC

What is the equation for the line shown?

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

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$D$

Show Worked Solution

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

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

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

$\Rightarrow D$

Algebra, STD2 EO-Bank 12 MC

Which of these equations represents the line in the graph?

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

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$C$

Show Worked Solution

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

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

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

$\Rightarrow C$

Algebra, STD2 A2 2020 HSC 6 MC (Adapted)

Suppose $y=-2-3x$.

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

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

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$C$

Show Worked Solution

$\text{Strategy 1}$

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

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

$\text{Strategy 2}$

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

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

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

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

$\Rightarrow C$

Algebra, STD2 A2 2021 HSC 9 MC (Adapted)

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 $n$?

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

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$D$

Show Worked Solution

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

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

$\Rightarrow D$

Algebra, STD2 A2 2022 HSC 2 MC (Adapted)

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

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$B$

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%.

Algebra, STD2 A1 2009 HSC 16 MC (Adapted)

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

Which of the following graphs shows this relationship?

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$C$

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%.

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

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

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

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

\(2x-1\)	\(=5-x\)
\(3x\)	\(=6\)
\(x\)	\(=2\)

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

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

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

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