Question 969338
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The sum of the distances traveled by car 1 and car 2 was 1425 miles, so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ +\ x_2\ =\ 1425]


Since the assignment of variable names is arbitrary we can assume without loss of generality that car 1 is the car that used 40 gallons of gas.  So if *[tex \Large x_1] is the distance traveled by that car, the fuel efficiency of that car is *[tex \Large \frac{x_1}{40}].  Likewise the efficiency of the second car is *[tex \Large \frac{x_2}{25}].  The sum of these two fuel efficencies is given as 45, hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x_1}{40}\ +\ \frac{x_2}{25}\ =\ 45]


Solve the 2X2 system of equations for *[tex \Large x_1] and *[tex \Large x_2] 


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \