More about equations Examples


  • Examples
    It is given that the simultaneous equations \(\left\{ \begin{array}{1}y = {x^2} - kx + 34\\y = 6x - 3k\end{array} \right.\) have only one set of real solution and k is a positive constant.
    Solve the simultaneous equations.
  • Solutions

    \(\left\{ \begin{array}{l}y = {x^2} - kx + 34\;.....................\;(1)\\y = 6x - 3k\;............................\;(2)\end{array} \right.\)
    Substitute (2) into (1),
    \(\begin{array}{1}6x - 3k = {x^2} - kx + 34\\{x^2} - (k + 6)x + 3k + 34 = 0\;........\;(3)\end{array}\)
    ∵ The simultaneous equations have only one set of real solution.
    i.e. Equation (3) has only one double real root.
    ∴ \(\Delta = 0\)
    i.e. \(\begin{array}{1}{[ - (k + 6)]^2} - 4(1)(3k + 34) = 0\\{k^2} + 12k + 36 - 12k - 136 = 0\\{k^2} = 100\end{array}\)
    \(k = 10\) or \(k = - 10\) (rejected)
    Substitute k = 10 into (3),
    \(\begin{array}{1}{x^2} - (10 + 6)x + 3(10) + 34 = 0\\{x^2} - 16x + 64 = 0\\{(x - 8)^2} = 0\\x = 8\end{array}\)
    Substitute k = 10 and x = 8 into (2),
    \(\begin{array}{1}y = 6(8) - 3(10)\\ = 18\end{array}\)
    ∴ The solution is \(x = 8\), \(y = 18\).


  • Examples
    (a) Solve the simultaneous equations \(\left\{ \begin{array}{l}y = {x^2} + 4x + 2\\y - 2x - 2 = 0\end{array} \right.\).
    (b) Using the result of (a), solve the simultaneous equations \(\left\{ \begin{array}{l}a + b = {(a - b)^2} + 4(a - b) + 2\\3b - a - 2 = 0\end{array} \right.\).
  • Solutions

    (a) \(\left\{ \begin{array}{l}y = {x^2} + 4x + 2\;.........\;(1)\\y - 2x - 2 = 0\;...........\;(2)\end{array} \right.\)
    Substitute (1) into (2),
    \(\begin{array}{1}{x^2} + 4x + 2 - 2x - 2 = 0\\{x^2} + 2x = 0\\x(x + 2) = {0^{}}\end{array}\)
    \(x = 0\) or \(x = - 2\)
    Substitute x = 0 into (1), \(\begin{array}{1}y = {(0)^2} + 4(0) + 2\\ = 2\end{array}\)
    Substitute x = -2 into (1), \(\begin{array}{1}y = {( - 2)^2} + 4( - 2) + 2\\ = - 2\end{array}\)
    ∴ The solutions are \(x = 0\), \(y = 2\) and\(x = - 2\), \(y = - 2\).

    (b) Rewrite \(\left\{ \begin{array}{l}a + b = {(a - b)^2} + 4(a - b) + 2\\3b - a - 2 = 0\end{array} \right.\) as \(\left\{ \begin{array}{l}a + b = {(a - b)^2} + 4(a - b) + 2\\(a + b) - 2(a - b) - 2 = 0\end{array} \right.\).
    Let \(x = a - b\) and\(y = a + b\), then the simultaneous equations become \(\left\{ \begin{array}{l}y = {x^2} + 4x + 2\\y - 2x - 2 = 0\end{array} \right.\).
    Using the result of (a), we have
    \(\left\{ \begin{array}{l}x = 0\\y = 2\end{array} \right.\) or \(\left\{ \begin{array}{l}x = - 2\\y = - 2\end{array} \right.\)
    i.e. \(\left\{ \begin{array}{l}a - b = 0\;........\;(3)\\a + b = 2\;........\;(4)\end{array} \right.\) or \(\left\{ \begin{array}{l}a - b = - 2\;........\;(5)\\a + b = - 2\;........\;(6)\end{array} \right.\)
    (3) + (4), \(\begin{array}{1}2a = 2\\a = 1\end{array}\) (5) + (6), \(\begin{array}{1}2a = - {\rm{ }}4\\a = - 2\end{array}\)
    (4) - (3), \(\begin{array}{1}2b = 2\\b = 1\end{array}\) (6) - (5), \(\begin{array}{1}2b = 0\\b = 0\end{array}\)
    ∴ The solutions are \(a = 1\), \(b = 1\) and \(a = - 2\), \(b = 0\).


  • Examples
    The figure shows the graphs of \(y = 2{m^2}\) and \(y = {x^2} - 5mx + 6{m^2}\), where m is a positive constant.
    [GraphMissing Latex_More about equations Q29]
    (a) Express the coordinates of C and D in terms of m.
    (b) Express the coordinates of A and B in terms of m.
    (c) If the area of trapezium ABCD is 32, find the value of m
  • Solutions

    (a) For \(y = {x^2} - 5mx + 6{m^2}\), when \(y = 0\),
    \(\begin{array}{1}{x^2} - 5mx + 6{m^2} = 0\\(x - 2m)(x - 3m) = 0\end{array}\)
    \(x = 2m\) or \(x = 3m\)
    ∴ The coordinates of C and D are \((2m{\rm{ }},{\rm{ }}0)\) and \((3m{\rm{ }},{\rm{ }}0)\) respectively.

    (b) \(\left\{ \begin{array}{l}y = {x^2} - 5mx + 6{m^2}\;........\;(1)\\y = 2{m^2}\;..........................\;(2)\end{array} \right.\)
    Substitute (2) into (1),
    \(\begin{array}{1}2{m^2} = {x^2} - 5mx + 6{m^2}\\{x^2} - 5mx + 4{m^2} = 0\\(x - m)(x - 4m) = {0^{}}\end{array}\)
    \(x = m\) or \(x = 4m\)
    ∴ The coordinates of A and B are \((m{\rm{ }},{\rm{ }}2{m^2})\) and \((4m{\rm{ }},{\rm{ }}2{m^2})\) respectively.

    (c) Area of trapezium ABCD = 32
    \(\begin{array}{1}\frac{1}{2} \times [(4m - m) + (3m - 2m)] \times 2{m^2} = 32\\4{m^3} = 32\\{m^3} = 8\\{m^3} = {2^3}\\m = \underline{\underline 2} \end{array}\)