<p>Solve the following equations.</p>
<ol class="i-list"><li><div class="list-item-paragraph">\(\frac{4(-2x+3)}{9}-3=-\frac{(-7)(x+6)}{12}-\frac{13}{6}\)<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">\(\frac{\frac{4}{3}\left(9x+\frac{2}{5}\right)}{8}=-\frac{\left(-\frac{3}{4}\right)\left(\frac{1}{2}x-3\right)}{\frac{3}{2}}\)<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">\(\frac{1}{5-2x}+\frac{3}{2x+5}=\frac{1}{4x^2-25}\)<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">\(\frac{1}{(3x-8)}+\frac{2}{(3x-8)^2}=\frac{1}{9x^2-48x+64}\)<span class="inline-marks">[2]</span></div></li></ol>

\[\frac{4(-2x+3)}{9}-3=-\frac{(-7)(x+6)}{12}-\frac{13}{6}\]

\(\frac{4(-2x+3)}{9}-3=-\frac{(-7)(x+6)}{12}-\frac{13}{6}\)

\[\frac{\frac{4}{3}\left(9x+\frac{2}{5}\right)}{8}=-\frac{\left(-\frac{3}{4}\right)\left(\frac{1}{2}x-3\right)}{\frac{3}{2}}\]

\(\frac{\frac{4}{3}\left(9x+\frac{2}{5}\right)}{8}=-\frac{\left(-\frac{3}{4}\right)\left(\frac{1}{2}x-3\right)}{\frac{3}{2}}\)

\[\frac{1}{5-2x}+\frac{3}{2x+5}=\frac{1}{4x^2-25}\]

\(\frac{1}{5-2x}+\frac{3}{2x+5}=\frac{1}{4x^2-25}\)

\[\frac{1}{(3x-8)}+\frac{2}{(3x-8)^2}=\frac{1}{9x^2-48x+64}\]

\(\frac{1}{(3x-8)}+\frac{2}{(3x-8)^2}=\frac{1}{9x^2-48x+64}\)

IP1 Class Test (May 2025) Q11

IP1 Class Test (May 2025) Q11

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