Question 565278
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Let *[tex \Large x_1] represent the first number and let *[tex \Large x_2] represent the second number.


The sum of the two numbers is:


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


and twice the first number less 15 is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x_1\ -\ 15]


and since "is" means "equals" we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ +\ x_2\ =\ 2x_1\ -\ 15]


Similarly, the difference is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ -\ x_2]


And 5 less than twice the second number is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x_2\ -\ 5]


Giving us:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ -\ x_2\ =\ 2x_2\ -\ 5]


Collect terms in each:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -x_1\ +\ \ x_2\ =\ -15]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \,x_1\ -\ 3x_2\ =\ -\ 5]


Leaving you with a nice little 2X2 system that is all set up to solve by the elimination method.


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