Question 218305
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Multiply the first equation by -3 giving you:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -6x - 3y = -15]


Now the coefficient on *[tex \Large y] in the new first equation is the additive inverse of the coefficient on *[tex \Large y] in the second equation.  The multiplier in the first step was selected for the purpose of obtaining this result.  Add the two equations, term-by-term:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -6x + 4x + (-3y) + 3y = (-15) + 11]


Notice that the *[tex \Large y]-terms have been <i>eliminated</i>, hence the name of the method.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2x = -4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 2]


Two ways to proceed from here:


Go back to the original set of equations and multiply the first by -2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -4x - 2y = -10]


Again, add term by term:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (-4x) + 4x + (-2y) + 3y = (-10) + 11]


This time eliminating the *[tex \Large x]-terms


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = 1]


OR:


Just take the value you got for *[tex \Large x], namely *[tex \Large x = 2] and substitute it into either of the original equations, then solve for *[tex \Large y]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(2) + y = 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = 1]


In either case, the solution you presented was correct, namely (2, 1).



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