Electricity provider \(A\) charges 25 cents per kilowatt hour (kWh) for electricity, plus a fixed monthly charge of $40. --- 1 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- a. b. c. \(\text{400 kWh}\) d. \(\text{Provider}\ A\ \text{is cheaper by \$40.}\) a. b. c. \(A_{\text{charge}} = B_{\text{charge}}\ \text{at intersection.}\) \(\therefore\ \text{Same charge at 400 kWh}\) d. \(\text{Cost at 800 kWh:}\) \(\text{Provider}\ A: \ 40 + 0.25 \times 800 = $240\) \(\text{Provider}\ B: \ 0.35 \times 800 = $280\) \(\therefore \text{Provider}\ A\ \text{is cheaper by \$40.}\)
\(\textit{Electricity used in a
month (kWh)}\)\(\ \ 0\ \ \)
\(400\)
\(1000\)
\(\textit{Monthly charge (\$)}\)
\(40\)
\(290\)
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.
\(\textit{Electricity used in a
month (kWh)}\)\(\ \ 0\ \ \)
\(400\)
\(1000\)
\(\textit{Monthly charge (\$)}\)
\(40\)
\(140\)
\(290\)
\(\textit{Electricity used in a
month (kWh)}\)\(\ \ 0\ \ \)
\(400\)
\(1000\)
\(\textit{Monthly charge (\$)}\)
\(40\)
\(140\)
\(290\)
Algebra, STD2 A4 2021 HSC 34
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 on the following page, find the number of goannas and the number of emus in the park. (4 marks)
--- 1 WORK AREA LINES (style=lined) ---
Number of goannas = _______________
Number of emus = _________________
--- 0 WORK AREA LINES (style=lined) ---
Show Worked Solution
`text{Total goanna legs} = 4x`
♦♦♦ Mean mark 29%.
`text{Total emu legs} = 2y`
| `4x + 2y` | `= 76` | |
| `2x + y` | `= 38 \ …\ (1)` | |
| `x + y` | `= 31 \ …\ (2)` |
`text{Number of goannas} = 7`
`text{Number of emus} = 24`
Algebra, STD2 A4 2020 HSC 24
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)
--- 2 WORK AREA LINES (style=lined) ---
- 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)
--- 1 WORK AREA LINES (style=lined) ---
Show Answers Only
- `text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}`
- `29 \ text{minutes}`
- `45 \ text{minutes}`
Show Worked Solution
a. `text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}`
b. `text{T} text{ank} \ B \ text{will pass through (15, 0) and (45, 900)}`
`text{By inspection, the two graphs intersect at} \ \ t = 29 \ text{minutes}`
c. `text{Strategy 1}`
♦♦ Mean mark part (c) 22%.
`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, 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` (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
Show Answers Only
`x=-2,\ y=-3`
Show Worked Solution
`text(Solution is at the intersection:)\ \ x=-2,\ y=-3`
Algebra, STD2 A4 EQ-Bank 8
Two friends, Sequoia and Raven, sold organic chapsticks at the the local market.
Sequoia sold her chapsticks for $4 and Raven sold hers for $3 each. In the first hour, their total combined sales were $20.
If Sequoia sold `x` chapsticks and Raven sold `y` chapsticks, then the following equation can be formed:
`4x + 3y = 20`
In the first hour, the friends sold a total of 6 chapsticks between them.
Find the number of chapsticks each of the friends sold during this time by forming a second equation and solving the simultaneous equations graphically. (5 marks)
Show Worked Solution
| `text(Graphing)\ \ 4x + 3y` | `= 20` |
| `y` | `= −4/3x + 20/3` |
`ytext(-intercept) = (0, 20/3)`
`xtext(-intercept)= (5, 0)`
`text(Gradient)\ = -4/3`
`text(S)text(econd equation:)`
`x + y = 6`
`text(From the graph,)`
`text(Sequoia sold 2 and Raven sold 4.)`
Algebra, STD2 A4 SM-Bank 7
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)
--- 1 WORK AREA LINES (style=lined) ---
Show Worked Solution
`text(Graphing)\ y = 2x + 1 :`
`ytext(-intercept = 1)`
`text(Gradient = 2)`
`:.\ text{Point of intersection is (1, 3).}`
Algebra, STD2 A4 SM-Bank 5
| `y` | `= x + 5` |
| `y + 2x` | `= 2` |
Draw these two linear graphs on the number plane below and determine their intersection. (3 marks)
--- 2 WORK AREA LINES (style=lined) ---
Show Worked Solution
`text(Table of coordinates:)\ \ y = x + 5`
`text(Table of coordinates:)\ \ y + 2x = 2 \ => \ y = −2x + 2`
`text{From graph (and table), intersection occurs}`
`text(at)\ \ (−1,4).`
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?
A. `(0, 6)`
B. `(1, 4)`
C. `(2, 2)`
D. `(3, 0)`
Show Worked Solution
`text(Solution 1)`
`text(From graph, intersection at (1,4))`
`=>B`
`text(Solution 2)`
| `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`