Question 181157
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*[tex \Large \text {           }\math 2(14-4y)-y=1]


Works just fine:


*[tex \Large \text {           }\math 2(14-4y)-y=1 \ \ \rightarrow \ \ 28 - 8y - y = 1 \ \ \rightarrow \ \ -9y = -27 \ \ \rightarrow \ \ y = 3 ]


Then


*[tex \Large \text {           }\math x + 4(3) = 14 \ \ \rightarrow \ \ x + 12 = 14 \ \ \rightarrow \ \ x = 2]


Your other way, *[tex \Large y = 2x + 1] isn't going to work.  You have a sign error.



*[tex \Large \text {           }\math 2x - y = 1 \ \ \rightarrow \ \ -y = -2x + 1  \ \ \rightarrow \ \ y = 2x - 1].


Using this expression to substitute:



*[tex \Large \text {           }\math x + 4(2x - 1) = 14  \ \ \rightarrow \ \ x + 8x - 4 = 14 \ \ \rightarrow \ \ 9x = 18 \ \ \rightarrow \ \ x = 2]


And then substituting this value:


*[tex \Large \text {           }\math 2(2) - y = 1 \ \ \rightarrow \ \ 4 - y = 1 \ \ \rightarrow \ \ -y = -3 \ \ \rightarrow \ \ y = 3]


Same answer.  Your solution set consists of the single ordered pair (2,3).



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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