Question 528174
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If two things are equal to each other, then you can substitute one for the other any time you like.


Your second equation sets *[tex \Large y] equal to *[tex \Large 8x\ +\ 5]


So you can substitute the expression *[tex \Large 8x\ +\ 5] for *[tex \Large y] whereever it is convenient to do so -- at least within the context of this problem.


That means that you can re-write your first equation thusly:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ -\ 4(8x\ +\ 5)\ =\ 20]


resulting in an equation in a single variable that can be solved by ordinary means.  Once you have determined the value of *[tex \Large x], that value can be substituted back into either equation so as to be able to calculate the value of *[tex \Large y].


Now that you have a value for *[tex \Large x] and a value for *[tex \Large y], you can create an ordered pair *[tex \Large \left(x,\,y\right)] which is the single element of the solution set of the 2X2 <i><b>system</b></i> of  equations.


If you have already done all of that and are bothered by the fact that the values of *[tex \Large x] and *[tex \Large y] are rather ugly fractions, remember that no one has ever given you a written guarantee that the answers to math problems would always turn out to be nice neat round numbers or fractions with single digit denominators.


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