Algebraic Fractions - SS2 Mathematics Past Questions and Answers - page 1

\(\frac{3}{2y + 4} - \frac{5}{3y + 6} = \ \frac{3}{2(y + 2)} - \frac{5}{3(y + 2)} = \ \frac{9 - 10}{6(y + 2)} = \ \frac{- 1}{6(y + 2)}\)

\(Given\) \(\frac{\frac{1}{2x} - \frac{4}{y}}{\frac{1}{x} + \frac{2}{3y}}\)

\(Numerator\): \(\frac{1}{2x} - \frac{4}{y} = \ \frac{y - 8x}{2xy}\)

\(Denominator\): \(\frac{1}{x} + \frac{2}{3y} = \ \frac{3y + 2x}{3xy}\)

\(\frac{\frac{1}{2x} - \frac{4}{y}}{\frac{1}{x} + \frac{2}{3y}} = \ \frac{\frac{y - 8x}{2xy}}{\frac{3y + 2x}{3xy}} = \ \frac{y - 8x}{2xy} \div \frac{3y + 2x}{3xy} = \frac{y - 8x}{2xy} \times \frac{3xy}{3y + 2x} = \ \frac{y - 8x}{2} \times \frac{3}{3y + 2x} = \frac{3(y - 8x)}{2(3y + 2x)}\)

\[\frac{a^{2} - 100}{8} \div \frac{2a + 20}{20} = \ \frac{a^{2} - 100}{8} \times \frac{20}{2a + 20} = \frac{a^{2}{- 10}^{2}}{8} \times \frac{20}{2(a + 10)} = \ \frac{(a + 10)(a - 10)}{2} \times \frac{5}{2(a + 10)}\]

\(= \ \frac{a - 10}{2} \times \frac{5}{2} = \frac{5(a - 10)}{4}\)

\[0.15x + 7 = 0.5x\]

\[\frac{15}{100}x + 7 = \frac{5}{10}x\]

\[15x + 700 = 50x\]

\[15x - 50x = - 700\]

\[- 35x = - 700\]

\(x = \frac{- 700}{- 35} = \ \frac{100}{5} = 20\)

\[\frac{3x + 2}{5} = \frac{8}{x}\]
\(x(3x + 2) = 40\)
\(3x^{2} + 2x = 40\)
\(3x^{2} + 2x - 40 = 0\)
\((3x - 10)(x + 4) = 0\)
\(3x - 10 = 0 or x + 4 = 0\)
\(x = \frac{10}{3} or x = - 4\)

\[\frac{3p + p^{2}}{12 + 3p} \div \frac{9p - p^{3}}{12 - p - p^{2}}\]

\[\frac{3p + p^{2}}{12 + 3p} \times \frac{12 - p - p^{2}}{9p - p^{3}}\]

\[\frac{p(3 + p)}{3(4 + p)} \times \frac{(3 - p)(p + 4)}{p(3^{2} - p^{2})}\]

\[\frac{p(3 + p)}{3(4 + p)} \times \frac{(3 - p)(p + 4)}{p(3 - p)(3 + p)}\]

\[\frac{1}{3} \times \frac{1}{1} = \frac{1}{3}\]

\[3.1x = - 31\]

\[\frac{31}{10}x = - 31\]

\[31x = - 310\]

\[x = \frac{- 310}{31} = \ - 10\]