Question 728761
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The short answer to the question you posed is "no and no".  It is not right, and even if it were right, it would not be "good enough".


I presume that at this point you want to know why.


The word substitution, in any mathematical context, means to take something and replace it with something else that has <i><b>exactly the same value</b></i>.  Not almost, not "sorta-kinda", but <i><b>exactly</b></i>.  How do we know that things have exactly the same value? They have an equals sign between them.  In fact, you will never be wrong to read *[tex \LARGE \varphi\ =\ w\ +\ 1] as either: "Phi is equal to w plus 1" or "Phi has exactly the same value as w plus 1".


Having said all of that, let's go back and examine your problem.  You started with:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ y\ -\ 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ =\ y\ +\ 9]


Let's first read the first of these two equations using our new interpretation of the equals sign:


"x has exactly the same value as y minus 6"


Now if we go back to the meaning of substitution in this context, we know that, within the system of equations with which we are dealing, we can, anytime we find it convenient, write *[tex \LARGE y\ -\ 6] in place of *[tex \LARGE x] or vice versa.


But what did you do?  You replaced *[tex \LARGE 2x] with *[tex \LARGE y\ -\ 6].  What's the matter with that?  Go back to the definition of substitution.  The thing substituted has to be exactly equal.  But if *[tex \LARGE y\ -\ 6] is equal to *[tex \LARGE x] as asserted by the first equation, then *[tex \LARGE y\ -\ 6] cannot, in general, be equal to *[tex \LARGE 2x], and therefore the substitution you made is incorrect.


Here it is correctly done:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(y\ -\ 6)\ =\ y\ +\ 9]


Now, the above is "right" as far as it goes, but not "good enough"; you still have to solve it for *[tex \LARGE y].  Meaning that you have to end up, by appropriate algebraic manipulations with an expression that looks like *[tex \LARGE y\ =\ \ ]{some number}.  I leave that part to you.  I also leave to you to convince yourself that putting the expression *[tex \LARGE y\ -\ 6] into parentheses was the correct thing to do.


Having done all of that, there is still a couple of more steps.  Once you know the value of *[tex \LARGE y] because you got to *[tex \LARGE y\ =\ \ ]{some number}, you have to take that {some number} and substitute it for *[tex \LARGE y] in either of your original equations and solve what remains for the value of *[tex \LARGE x].  Once you know *[tex \LARGE x] and *[tex \LARGE y], you can report your solution set as the set containing the single element comprised of the the ordered pair *[tex \LARGE (x,\,y)].  Your answer should look like:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left\{(x,\,y)\right}]


with the correct numbers in the place of *[tex \LARGE x] and *[tex \LARGE y] of course.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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