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Add math (2010/2A)

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Question One The first question asks the student to solve the following simultaneous equations: $$ \left\{ \begin{array}{l} x - 2y = 7\\xy - x = 9y \end{array} \right. $$ The first thing we notice is that the second equation can be factorized and rewritten as $$x(y-1) = 9y$$ Next, we can introduce this expression to the first equation. But before introducing this term, we need to multiply each of the term in the first equation by \(y-1\) $$\color{brown}{x(y-1)}- 2y(y-1) = 7(y-1) $$ $$\color{brown}{9y}- 2y(y-1) = 7(y-1) $$ $$11y- 2y^2 = 7y - 7 $$ $$2y^2-4y = 7 $$ $$2y^2-4y + 2 = 9 $$ $$2(y-1)^2 = 9 $$ $$ y-1 = \pm \tfrac{3}{\sqrt{2}} $$ $$ y = 1\pm \tfrac{3}{\sqrt{2}} $$ $$ x = \left(2\pm \tfrac{6}{...

Add math (2010/2A)

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Question One The first question asks the student to solve the following simultaneous equations: $$ \left\{ \begin{array}{l} x - 2y = 7\\xy - x = 9y \end{array} \right. $$ The first thing we notice is that the second equation can be factorized and rewritten as $$x(y-1) = 9y$$ Next, we can introduce this expression to the first equation. But before introducing this term, we need to multiply each of the term in the first equation by \(y-1\) $$\color{brown}{x(y-1)}- 2y(y-1) = 7(y-1) $$ $$\color{brown}{9y}- 2y(y-1) = 7(y-1) $$ $$11y- 2y^2 = 7y - 7 $$ $$2y^2-4y = 7 $$ $$2y^2-4y + 2 = 9 $$ $$2(y-1)^2 = 9 $$ $$ y-1 = \pm \tfrac{3}{\sqrt{2}} $$ $$ y = 1\pm \tfrac{3}{\sqrt{2}} $$ $$ x = \left(2\pm \tfrac{6}{...