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Functions, MET1 2024 VCAA 2

Consider the simultaneous linear equations

\(\begin{aligned} 3 k x-2 y & =k+4 \\ (k-4) x+k y & =-k\end{aligned}\)

where  \(x, y \in R\) and \(k\) is a real constant.

Determine the value of \(k\) for which the system of equations has no real solution.   (3 marks)

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\(k=\dfrac{4}{3}\)

Show Worked Solution

\(\text{For no solutions lines must be parallel }(m_1=m_2)\)

\(\text{and y-intercepts not equal}\ (c_1\neq c_2).\)

\(3kx-2y=k+4\ \ \Rightarrow \ \ y=\dfrac{3k}{2}x-\dfrac{(k+4)}{2}\)

\(m_1=\dfrac{3k}{2},\ c_1=-\dfrac{(k+4)}{2}\)

\((k-4)x+ky-k\ \ \Rightarrow\ \ y=-\dfrac{(k-4)}{k}x-1\)

\(m_2=-\dfrac{(k-4)}{k},\ c_2=-1\)

Mean mark 56%.

\(\text{Equating gradients:}\)

\(\dfrac{3k}{2}\) \(=-\dfrac{(k-4)}{k}\)
\(3k^2\) \(=-2(k-4)\)
\(3k^2\) \(=-2k+8\)
\(3k^2+2k-8\) \(=0\)
\((3k-4)(k+2)\) \(=0\)

\(k=\dfrac{4}{3},\ k=-2\)
 

\(\text{Substituting to test }y-\text{intercepts:}\)

\(\text{Case 1:}\ \ k=-2\quad\)

\(c_1=-\dfrac{(k+4)}{2}=-\dfrac{(-2+4)}{2}=-1 = c_2\ \ \text{(no solution)}\)

\(\text{Case 2:}\ \ k= \dfrac{4}{3}\)

\(c_1=-\dfrac{(k+4)}{2}==-\dfrac{\Big(\dfrac{4}{3}+4\Big)}{2}=-\dfrac{8}{3} \neq c_2\)

\(\therefore \text{There is no real solution when}\ \ k=\dfrac{4}{3}\)

Functions, MET1 EQ-Bank 4

Consider the simultaneous equations below, where \(a\) and \(b\) are real constants.

\begin{aligned}
& (a+3) x+9 y=3 b\\
& 2 x+a y=5
\end{aligned}

Find the values of \(a\) and \(b\) for which the simultaneous equations have no solutions.   (4 marks)

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\(a=-6\ \text{and}\ b\neq -\dfrac{5}{2}\ \ \text{OR}\ \ a=3\ \text{and}\ b\neq 5\)

Show Worked Solution

(\(a+3)x+9y\) \(=3b\)  
\(y\) \(=-\dfrac{a+3}{9}x+\dfrac{b}{3}\) \(\dots (1)\)
\(\rightarrow \ m_1\) \(=-\dfrac{a+3}{9}, \ c_1=\dfrac{b}{3}\)  

 

\(2x+ay\) \(=5\)  
\(y\) \(=-\dfrac{2}{a}x+\dfrac{5}{a}\) \(\dots (2)\)
\(\rightarrow \ m_2\) \(=-\dfrac{2}{a}, \ c_2=\dfrac{5}{a}\)  

  
\(\text{No solution if  }m_1=m_2,\ \text{and}\ c_1\neq c_2.\)

\(\rightarrow\ -\dfrac{a+3}{9}=-\dfrac{2}{a}\ \text{and }\dfrac{b}{3}\neq \dfrac{5}{a}\rightarrow\ b\neq \dfrac{15}{a}\)

   

\(-\dfrac{a+3}{9}=-\dfrac{2}{a}\)

\(a^2+3a-18\) \(=0\)  
\((a+6)(a-3)\) \(=0\)  
\(\therefore a=-6\ \) \(\ \text{or}\ \) \(\ a=3\)

  

\(\text{When }\ a=-6,\  b\neq \dfrac{15}{-6}\rightarrow\ b\neq -\dfrac{5}{2}\)

\(\text{OR}\)

\(\text{When }\ a=3,\  b\neq \dfrac{15}{3}\rightarrow\ b\neq 5\)

Algebra, MET2 2015 VCAA 21 MC

The graphs of  `y = mx + c`  and  `y = ax^2`  will have no points of intersection for all values of `m, c` and `a` such that

  1. `a > 0 and c > 0`
  2. `a > 0 and c < 0`
  3. `a > 0 and c > -m^2/(4a)`
  4. `a < 0 and c > -m^2/(4a)`
  5. `m > 0 and c > 0`
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`D`

Show Worked Solution

`text(Intersect when:)`

`mx + c` `= ax^2`
`ax^2 – mx – c` `= 0`

 

`text(S)text(ince no points of intersection:)`

♦ Mean mark 37%.
`Delta` `< 0`
`m^2 – 4a(−c)` `< 0`
`m^2 + 4ac` `< 0`

 

`text(Solve for)\ c:`

`:.\ c > (−m^2)/(4a),quada < 0`

`text(or)`

`c < (−m^2)/(4a),quada > 0`

`=>   D`

Algebra, MET2 2014 VCAA 17 MC

The simultaneous linear equations  `ax - 3y = 5`  and  `3x - ay = 8 - a`  have no solution for

  1. `a = 3`
  2. `a = -3`
  3. both  `a = 3` and  ` a = -3`
  4. `a in R text(\{3})`
  5. `a in R text(\[−3, 3])`
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`B`

Show Worked Solution

`text(No solution occurs when:)`

♦ Mean mark 50%.

`=> m_1 = m_2quadtext(and)quadc_1 != c_2`

`a/3` `= 3/a` `text(and)` `-5/3` `!= -((8-a)/a)`
`a` `= ±3`   `5a` `!= 24-3a`
      `a` `!=3`

`:. a = −3`

`=>   B`

Algebra, MET1 SM-Bank 28

Consider the simultaneous linear equations below.

 `4x-2y = 18`

`3x + ky = 10`

where  `k`  is a real constant.

  1. What are the values of  `k`  where no solutions exist?   (3 marks)

  2. What values of  `k`  do the simultaneous equations have a unique solution?   (1 mark)

Show Answers Only
  1. `k=-3/2`
  2. `k\ in R text(\) {-3/2}` 
Show Worked Solution
a.    `4x-2y` `=18`
  `y` `=2x-9\ \ …\ (1)`

`=> m_1 = 2,\ \ c_1=-9`

 

`3x +ky` `=10`
`y` `=-3/k x +10/k\ \ …\ (2)`

`=> m_2 =-3/k,\ \ c_2=10/k`

 

`text(No solution if)\ \ m_1=m_2,  and  c_1!=c_2.`

`-3/k` `=2`
`k` `=- 3/2`

`text(When)\ \ k=-3/2,  c_1!=c_2.`

`:.\ text(No solution when)\ \ k=-3/2.`

 

b.   `text(A unique solution exists when)\ \ m_1 != m_2,`

`k in R\ text(\) {-3/2}`