[tex] \large \mathcal\colorbox{red}{SOLUTION:} [/tex]
[tex] \begin{array}{l} \large \bold{Simple\:Interest\:Formula:} \\ \\ \quad \huge I = Prt \\ \\ \textsf{where:} \\ \begin{aligned} \quad I &= \textsf{interest earned after }t\textsf{ years} \\ P &= \textsf{principal or initial amount} \\ r &= \textsf{annual interest rate} \\ t &= \textsf{time (in years)} \end{aligned} \\ \\ \bold{Given:} \begin{cases} \: r_1 = 4\% = 0.04 \\ \: r_2 = 5\% = 0.05 \\ \: t = 1 \\ \: I_{\textsf{Total}} = \textsf{Php }1,\!800 \end{cases} \\ \\ \bold{Required:}\: P\textsf{ or the equal amount invested at} \\ \quad \qquad \qquad \textsf{both rates} \\ \\ \textsf{Solving for }P, \\ \\ \begin{aligned} \quad I_{\textsf{Total}} &= I_1 + I_2 \\ \\ 1800 &= P(0.04)(1) + P(0.05)(1) \\ \\ 1800 &= 0.04P + 0.05P \\ \\ 0.09P &= 1800 \\ \\ \dfrac{\cancel{0.09}P}{\cancel{0.09}} & = \dfrac{1800}{0.09} \\ \\ \implies P &= \textsf{Php }20,\!000 \end{aligned} \\ \\ \therefore \: \boxed{P_{\textsf{at }4\%} = P_{\textsf{at }5\%} = \textsf{Php }20,\!000}\:\:\textit{Answer}\end{array} [/tex]
[tex] \mathfrak\colorbox{blue}{\#CarryOnLearning} [/tex]
