1. Add or Subtract by: \( \begin{array}{ll}\text { a) } & \begin{array}{l}\text { i) stating the LCD } \\ \text { ii) writing the equivalent fractions } \\ \text { iii) performing the operation, and } \\ \text { iv) simplifying, if necessary }\end{array} \\ \text { is }-\frac{9}{15} & \text { b) }-\frac{3}{8}-\frac{7}{12}\end{array} \) \( \begin{array}{ll}4-\frac{4}{3} & \text { d) } \frac{9}{5}+\frac{3}{y}\end{array} \)
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To start with \(-\frac{9}{15}\), the least common denominator (LCD) is 15 itself. The equivalent fraction is just \(-\frac{9}{15}\). Since this is already simplified, there's no need to perform further operations here. For the operation \(-\frac{3}{8} - \frac{7}{12}\), the LCD is 24. The equivalent fractions are \(-\frac{9}{24}\) and \(-\frac{14}{24}\). Performing the operation gives: \(-\frac{9}{24} - \frac{14}{24} = -\frac{23}{24}\), which is already simplified. When calculating \(4 - \frac{4}{3}\), the LCD is 3. We convert \(4\) to \(\frac{12}{3}\), so \(4 - \frac{4}{3} = \frac{12}{3} - \frac{4}{3} = \frac{8}{3}\). For \(\frac{9}{5} + \frac{3}{y}\), the LCD is \(5y\). We convert \(\frac{9}{5}\) to \(\frac{9y}{5y}\) and \(\frac{3}{y}\) to \(\frac{15}{5y}\). Performing the operation gives \(\frac{9y + 15}{5y}\), which may need to be simplified based on the value of \(y\).