Question 977582
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Eq 1: *[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ +\ 2y\ =\ 27]


Eq 2: *[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x\ -\ 3y\ = 7]


Multiply equation 1 by *[tex \Large \frac{5}{3}]


Eq 3: *[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x\ +\ \frac{10}{3}y\ =\ 45]


Eq 4: *[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x\ =\ -\frac{10}{3}y\ +\ 45]


Substitute 4 into 2:


Eq 5: *[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{10}{3}y\ +\ 45\ -\ 3y\ = 7]


Now just solve for *[tex \Large y].  Once you know *[tex \Large y], substitute that value into either of the original equations and solve for *[tex \Large x]


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

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