Question 1145102
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If *[tex \Large P] was invested in two accounts, *[tex \Large p_1] at *[tex \Large i_1%] and *[tex \Large p_2] at *[tex \Large i_2%] and the total interest earned at the end of one year is *[tex \Large Y], then we know two things:


*[tex \LARGE \[1\]\ \ \ \ \ \ \ \ \ \  p_1\ +\ p_2\ =\ P]


and


*[tex \LARGE \[2\]\ \ \ \ \ \ \ \ \ \  \frac{p_1i_1}{100}\ +\ \frac{p_2i_2}{100}\ =\ Y]


But from [1] we can deduce:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p_2\ =\ P\ -\ p_1]


Then substituting into [2] we have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{p_1i_1}{100}\ +\ \frac{\(P\,-\,p_1\)i_2}{100}\ =\ Y]


Since you are given *[tex \Large P] and *[tex \Large Y], you now have a single equation in a single variable, namely:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.045p_1\ +\ 0.05(6000\,-\,p_1\)\ =\ 290]


Solve for *[tex \Large p_1] and then calculate *[tex \Large 6000\ -\ p_1]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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