Question 270305
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 2x\ +\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x\ -\ 3y\ =\ 1]


This is already set up for substitution because you have one of your equations as *[tex \Large y] in terms of everything else.  So just substitute the expression in *[tex \Large x] into the second equation in the place of *[tex \Large y], like this:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x\ -\ 3\left(2x\ +\ 3\right)\ =\ 1]


Simplify and collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x\ -\ 6x\ -\ 9\ =\ 1]


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


Finally, solve for *[tex \Large x].  Once you have a value for *[tex \Large x], substitute that value into either of the original equations and then solve for *[tex \Large y].  The calculated *[tex \Large x] and *[tex \Large y] form an ordered pair *[tex \Large \left(x\,,\,y\right)] that is the solution set of your system.



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