\left. \begin{array} { l } { p = 2600 } \\ { R = 10 \% \text { pa } } \end{array} \right.
Solve for p, R
p=2600
R=\frac{1}{10}=0.1
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R=\frac{1}{10}
Consider the second equation. Reduce the fraction \frac{10}{100} to lowest terms by extracting and canceling out 10.
p=2600 R=\frac{1}{10}
The system is now solved.
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