\begin{tabular}{l} \( \frac{3 x^{-1}+y}{x^{-1} y} \times \frac{x}{x}=\frac{3 x^{-1} \cdot x+y \cdot x}{x^{-1} y \cdot x}=\frac{3+x y}{y} \) \\ \hline \end{tabular} These equations have more - Rearrange the equations EXERCISE 2 \( 3 \frac{9^{-1} x^{-2} y^{2}}{\left(3 x y^{-1}\right)^{-3}} \) \( 6 \frac{9^{2 x} \cdot 3^{-2}}{81^{x-1}} \) \( 1 \frac{12 a^{0} b^{3} c^{-1}}{15 a^{-2} b^{5} c^{-3}} \) (4) \( \frac{6\left(x^{2} y\right)^{3}\left(x^{-2} y\right)^{-1}}{3 x^{-3} y^{5}} \) \( 7\left(\frac{x^{2} y^{3}}{z^{-2}}\right)^{-1} \) \( 10 \frac{3^{-2}+3^{0}}{3^{-1}} \) \( 13 \frac{x y}{x^{-2}+y^{-2}} \) \( 16 \frac{\left(3 x^{4}\right)^{2}+\left(2 x^{2}\right)^{4}}{5 \cdot 2^{5} \cdot 2^{-5} \cdot x^{10}} \) \( 5 \frac{\left(2 a^{-1}\right)^{2}}{2\left(a^{2}\right)^{-1}} \) \( 8 \frac{\left(2 a^{-1}\right)^{0}}{2\left(a^{0}\right)^{-1}} \) \( 11(2.3)^{-2}+2.3^{-2} \) \( 14\left(a^{3}+3\right)^{-2} \) (17) \( \frac{\left(8.2^{x+1}\right)\left(4.2^{x-1}\right)}{2^{x-3}} \) (9) \( \frac{a^{-2}+a^{-1}+a^{0}}{a} \) \( 12\left(2^{-1}-5^{-1}\right)^{2} \) (15) \( \frac{3.3^{x}}{3^{x}+3^{x-1}} \) \( 18 \frac{12^{x+1} \cdot 27^{x-2}}{18^{2 x-1}} \) - Factorise by taking out - Solve the remaining ex WORKED EXAMIPLE Solve for \( x: 2^{x}+2^{x+1}= \) SOLUTION \[ \begin{aligned} 2^{x}+2^{x+1} & =24 \\ 2^{x}+2^{x} \cdot 2^{1} & =24 \\ 2^{x}(1+2) & =24 \\ 2^{x}(3) & =24 \\ 2^{x} & =8 \\ 2^{x} & =2^{3} \\ x & =3 \end{aligned} \]
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Bonus Knowledge
Get ready to flex those math muscles! When rearranging equations like the one given, it's all about applying some algebraic finesse. For instance, merging fractions using a common denominator or simplifying terms is key. Once organized, expressions often reveal factors that can be pulled out, or even lead you directly to solutions more smoothly—like finding hidden treasure in a math maze! Once you've shaken things up with factorization, continue diving into basic algebraic principles to solve for the unknowns. A common mistake is forgetting to apply appropriate operations consistently, especially with negative exponents. Keep those rules in check, like \( a^{-n} = \frac{1}{a^n} \), and you'll sail smoothly through the equations, just like a math superhero!