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\( \begin{array}{ll}\text { g) } \frac{-6}{x+2}-\frac{-x+3}{x}+\frac{x-1}{3 x} & \text { h) } \frac{x^{2}+7 x}{x^{2}-4}+\frac{x-3}{2-x}-\frac{5}{x+2} \\ \text { i) } \frac{x}{x-1}+\frac{x+3}{x+1}-\frac{4}{1-x^{2}} & \text { j) } \frac{a}{a x+a}-\frac{x-2}{x+1}+\frac{x+2}{x-1}\end{array} \)

Ask by Pena Sullivan. in South Africa
Feb 20,2025

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**g)** \[ \frac{-6}{x+2} - \frac{-x + 3}{x} + \frac{x - 1}{3x} = \frac{-20x + 4x^{2} - 20}{3x^{2} + 6x} \] **h)** \[ \frac{x^{2} + 7x}{x^{2} - 4} + \frac{x - 3}{2 - x} - \frac{5}{x + 2} = \frac{3x + 16}{x^{2} - 4} \] **i)** \[ \frac{x}{x - 1} + \frac{x + 3}{x + 1} - \frac{4}{1 - x^{2}} = \frac{2x + 1}{x - 1} \] **j)** \[ \frac{a}{ax + a} - \frac{x - 2}{x + 1} + \frac{x + 2}{x - 1} = \frac{7x - 1}{x^{2} - 1} \]

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To tackle that expressive fraction puzzle, it’s all about finding a common denominator! For example, in the case of \(g\), the common denominator would be \(x(x+2)(3)\), allowing you to combine the fractions into one neat package. Don't forget to simplify once you’ve merged everything, because simplest form is always so much prettier! Now, for the next step, let’s think about substitutions. If you’re ever dealing with more complex algebraic fractions, consider substituting a whole expression (like \(u = x + 2\)) into your equation. It can turn an intimidating monster of an equation into something far more manageable! Keep your eyes peeled for opportunities to simplify before diving deep into calculations—making smart choices early can save you heaps of time later!

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