SIT190 - Module 2 - Quiz
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "\\begin{align}
\\var{h}x+\\var{k}y&=\\var{m}\\text{,}\\\\
\\var{j}x-\\var{l}y&=\\var{n}\\text{.}\\\\
\\end{align}
To find the solution to these equations, we need to cancel one of the unknowns.
\nNotice that $\\var{h}x$ in the first equation can be multiplied by $\\var{j/h}$ to match $\\var{j}x$ in the second equation. This means that we will only have to manipulate the first equation and can leave the second equation as it is.
\nWe have to multiply the entire first equation by $\\var{j/h}$, not just the $x$ term to ensure the equation still holds.
\n$\\var{h}x+\\var{k}y=\\var{m}$ multiplied by $\\var{j/h}$ gives $\\var{j}x+\\var{k*(j/h)}y=\\var{m*(j/h)}.$
\nWe now have a common $x$ term and we can cancel this by subtracting one equation from the other to find the $y$ term.
\n\\begin{align}
&&\\var{j}x+\\var{k*{j/h}}y&=\\var{m*(j/h)}\\\\
-&&\\var{j}x-\\var{l}y&=\\var{n}\\\\
&&\\overline{\\qquad} & \\overline{\\qquad}\\\\
&&0x+\\var{k*(j/h)+l}y&=\\var{m*(j/h)-n}\\\\[1em]
&&y&=\\frac{\\var{m*j/h-n}}{\\var{k*j/h+l}}\\\\
&&y&=\\var{y1}
\\end{align}
We can find the corresponding value of $x$ by substituting this value for $y$ back into either of the original equations.
\n\\begin{align}
\\var{h}x+(\\var{k}\\times\\var{y1})&=\\var{m}\\text{,}\\\\
\\var{h}x+\\var{k*y1}&=\\var{m}\\text{,}\\\\
\\var{h}x&=\\var{m-(k*y1)}\\text{,}\\\\
x&=\\var{x1}\\text{.}\\\\
\\end{align}
Therefore, $x=\\var{x1}$ and $y=\\var{y1}$.
\n\\begin{align}
\\var{a}x+\\var{b}y&=\\var{c}\\text{,}\\\\
\\var{d}x+\\var{f}y&=\\var{g}\\text{.}\\\\
\\end{align}
To be able to solve the equations, we need to cancel one of the unknowns by manipulating the two equations so that the variable we wish to cancel is of the same value in each equation.
\nAlthough we can choose to cancel either variable, $x$ or $y$, a good rule of thumb is to look at the lowest common multiples of the coefficients for each variable and cancel the variable with the lowest LCM.
\nThe LCM of the coefficients of the $x$ terms is $\\var{lcm(a,d)}$.
\nThe LCM of the coefficients of the $y$ terms is $\\var{lcm(b,f)}$.
\nTherefore, we will choose to cancel the $x$ terms.
\nWe need to multiply the equations individually to achieve the lowest common multiple identified.
\n\\begin{align}
\\simplify{ {a}x + {b}y } &= \\var{c} &\\text{multiply by } \\var{lcm(a,d)/a} \\text { to obtain } && \\simplify{ {lcm(a,d)}x + {b*lcm(a,d)/a}y} &= \\var{c*lcm(a,d)/a} \\\\
\\simplify{ {d}x + {f}y } &= \\var{g} &\\text{multiply by } \\var{lcm(a,d)/d} \\text { to obtain } && \\simplify{ {lcm(a,d)}x + {b*lcm(a,d)/d}y} &= \\var{c*lcm(a,d)/d}
\\end{align}
We now have a common $x$ term, and can cancel this by subtracting one equation from the other.
\n\\begin{align}
&& \\simplify{ {lcm(a,d)}x+{b*lcm(a,d)/a}y } = \\var{c*lcm(a,d)/a} \\\\
- && \\simplify{ {lcm(a,d)}x + {f*lcm(a,d)/d}y } = \\var{g*lcm(a,d)/d} \\\\
&& \\overline{\\simplify[]{ 0x+{b*lcm(a,d)/a-f*lcm(a,d)/d}y} = \\var{c*lcm(a,d)/a-g*lcm(a,d)/d}}
\\end{align}
\\begin{align}
\\var{(b*lcm(a,d)/a)-(f*lcm(a,d)/d)}y &= \\var{(c*lcm(a,d)/a)-(g*lcm(a,d)/d)}\\text{,}\\\\
y &= \\var{y2}\\text{.}
\\end{align}
We can find the corresponding value of $x$ by substituting thsi value of $y$ value back into either of the original equations.
\n\\begin{align}
\\simplify[]{ {a}x + {b}{y2}} &= \\var{c} \\\\
\\simplify[]{ {a}x + {b*y2}} &= \\var{c} \\\\
\\var{a}x&=\\var{c-b*y2} \\\\
x &= \\var{x2} \\text{.}
\\end{align}
Therefore, $x=\\var{x2}$ and $y=\\var{y2}$.
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\\simplify{{h}x+{k}y} &= \\var{m} \\text{,} \\\\
\\simplify{{j}x+{l}y} &= \\var{n} \\text{.}
\\end{align}
$x =$ [[0]]
\n$y =$ [[1]]
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\n\\begin{align}
\\simplify{{a}x + {b}y} &= \\var{c} \\text{,} \\\\
\\simplify{{d}x + {f}y} &= \\var{g} \\text{.}
\\end{align}
$x =$ [[0]]
\n$y =$ [[1]]
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\n\\begin{align}
\\simplify{{a}x + {b}y} &= \\var{c} \\text{,} \\\\
\\simplify{(2*{a})x + (2*{b})y} &= \\var{2c} \\text{.}
\\end{align}
$x =$ [[0]]
\n$y =$ [[1]]
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\n\\begin{align}
\\simplify{{2a}x + {2b}y} &= \\var{c+2} \\text{,} \\\\
\\simplify{{(2a+3)a}x + (({2a+3})*{b})y} &= \\var{c+10} \\text{.}
\\end{align}
$x =$ [[0]]
\n$y =$ [[1]]
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Equations of line in the form $y=mx+c$ and $ax + by = c$
", "advice": "Click 'Try another question like this one' if you need more practice.
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y &= \\simplify {{m2}x+{c2}} \\text{.}
\\end{align}
Identify the $x$ and $y$ intercepts for \\begin{align}
y &= \\simplify {{m}x+{c}} \\text{.}
\\end{align}
$x$ intercept =[[0]]
\n$y$ intercept =[[1]]
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\\simplify {{ak}{b}x+{b}y = {ck}{b}} \\text{.}
\\end{align}
a) Convert it to $y = mx+c.$
\n[[0]]
\nb) Identify the gradient
\n[[1]]
\nc) Identify the $x$ and $y$ intercepts
\n$x$ intercept = [[2]]
\n$y$ intercept = [[3]]
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y &= \\simplify {{c2}} \\text{.}
\\end{align}
Find the gradient of the line defined by \\begin{align}
x &= \\simplify {{m2}} \\text{.}
\\end{align}
[[0]]
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "\\begin{align}
\\mathbf{AB} &= \\var{A}\\var{B} \\\\
&= \\begin{pmatrix} \\simplify[]{ {a[0][0]}*{b[0][0]}+{a[0][1]}*{b[1][0]} } & \\simplify[]{ {a[0][0]}*{b[0][1]} + {a[0][1]}*{b[1][1]} } \\\\ \\simplify[]{ {a[1][0]}*{b[0][0]} + {a[1][1]}*{b[1][0]} } & \\simplify[]{ {a[1][0]}*{b[0][1]} + {a[1][1]}*{b[1][1]} } \\end{pmatrix} \\\\
&= \\var{a*b}
\\end{align}
\\begin{align}
\\mathbf{BA} &= \\var{B}\\var{A} \\\\
&= \\begin{pmatrix} \\simplify[]{ {b[0][0]}*{a[0][0]}+{b[0][1]}*{a[1][0]} } & \\simplify[]{ {b[0][0]}*{a[0][1]} + {b[0][1]}*{a[1][1]} } \\\\ \\simplify[]{ {b[1][0]}*{a[0][0]} + {b[1][1]}*{a[1][0]} } & \\simplify[]{ {b[1][0]}*{a[0][1]} + {b[1][1]}*{a[1][1]} } \\end{pmatrix} \\\\
&= \\var{b*a}
\\end{align}
\\begin{align}
\\mathbf{CB} &= \\var{C}\\var{B} \\\\
&= \\begin{pmatrix} \\simplify[]{ {c[0][0]}*{b[0][0]}+{c[0][1]}*{b[1][0]} } & \\simplify[]{ {c[0][0]}*{b[0][1]} + {c[0][1]}*{b[1][1]} } \\\\ \\simplify[]{ {c[1][0]}*{b[0][0]} + {c[1][1]}*{b[1][0]} } & \\simplify[]{ {c[1][0]}*{b[0][1]} + {c[1][1]}*{b[1][1]} } \\end{pmatrix} \\\\
&= \\var{c*b}
\\end{align}
\\begin{align}
\\mathbf{AC} &= \\var{A}\\var{C} \\\\
&= \\begin{pmatrix} \\simplify[]{ {a[0][0]}*{c[0][0]}+{a[0][1]}*{c[1][0]} } & \\simplify[]{ {a[0][0]}*{c[0][1]} + {a[0][1]}*{c[1][1]} } \\\\ \\simplify[]{ {a[1][0]}*{c[0][0]} + {a[1][1]}*{c[1][0]} } & \\simplify[]{ {a[1][0]}*{c[0][1]} + {a[1][1]}*{c[1][1]} } \\end{pmatrix} \\\\
&= \\var{a*c}
\\end{align}
Given matrices $A = \\var{A1}, B=\\var{B1}, C=\\var{C1}$:
\n1) Identify pairs that can be added:[[0]]
\n2) Identify pairs that cannot be multiplied:[[1]]
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\n1) $\\mathbf{A+B} = \\var{A1}+\\var{C1} = $ [[0]]
\n2) $\\mathbf{A \\times B} = \\var{A1}\\var{B1} = $ [[1]]
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\n| 1. | \nComplete the questions within 90 minutes and achieve 80% or higher. | \n
| 2. | \nYou can take this self-assessment as many times as you need, until you receive a satisfactory grade (you don't need to achieve 80% when you 'give it a go' the first time) | \n
| 3. | \nUse the \"Print this results summary\" and save as a pdf after you complete your attempt. You will need the printout showing all questions for your module submission. | \n
Congratulations! You have achieved the minimum threshold for this module's self-assessment.
\nUse the \"Print this results summary\" and save your attempt as a pdf. You will need the printout showing all questions for your module submission.
", "threshold": "80"}, {"message": "Unfortunately you have not achieved the minimum score.
\nIf this is your first 'Give it a go' attempt - don't despair! This is exactly why we take our first attempt - to see how we're going and whether we need more practice in order to complete the quest.
\nYou should still use the \"Print this results summary\" option to save a copy of your results as a pdf, which will help with your learning and can also be shared with your tutors so they can help with certain questions.
\n