Question 354768
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Typo or error?  Your problem statement says the first loan is at 0.2% simple interest, which really looks like a typo, but if 0.2% is actually correct, then your second equation has an error and should be:  *[tex \Large 0.002x\ +\ 0.05y\ =\ 2817.23]


Either way, there are three basic ways to go about solving your system:


1.  Substitution:


Solve *[tex \Large x\ +\ y\ =\ 78825] for either of the variables in terms of the other.  Let's solve for *[tex \Large y]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 78825\ -\ x]


Now substitute into the 2nd equation.  (I'm going to go with my gut feeling and say you had a typo in the problem statement, but you can make the adjustment if necessary)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.02x\ +\ 0.05(78825\ -\ x)\ =\ 2817.23]


And then just solve for *[tex \Large x].


2.  Elimination:


Multiply your second equation by -50:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -50(0.02x\ +\ 0.05y)\ =\ -50(2817.23)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -x\ -\ 2.5y\ =\ -148861.5]


Then add the two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ (-x)\ +\ y\ +\ (-2.5)y\ =\ 78825\ -\ 148861.5]


Notice that *[tex \Large x] goes away leaving you with an equation in *[tex \Large y]


3. Cramer's Rule:


Create the coefficient matrix of your system:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left\[\ \ \ \ 1\ \ \ \ \ 1\cr 0.02\ 0.05\right\]]


And then calculate the determinant, *[tex \Large D], of that matrix.


The determinant


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left|a_1\ b_1\cr a_2\ b_2\right|\ =\ a_1b_2\ -\ a_2b_1]


So you will have


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left|\ \ \ \ 1\ \ \ \ \ 1\cr 0.02\ 0.05\right|\ =\ 0.05\ -\ 0.02\ =\ 0.03]


Then replace the first column of your matrix with the constant values from your equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left\[\ \ 78825\ \ \ \ \ 1\cr 2817.23\ 0.05\right\]]


and then using the same process, calculate the determinant, *[tex \Large D_x]


Next, replace the second column:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left\[\ \ \ \ 1\ \ \,78825\cr 0.02\ 2817.23\right\]]


From which you calculate *[tex \Large D_y]


Finally,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{D_x}{D}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{D_y}{D}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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