Question 181614
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*[tex \LARGE \text {(1)          }\math 9x + 3y = 10.02]
*[tex \LARGE \text {(2)          }\math 6x + 4y =  8.26]


Multiply (1) by 4:


*[tex \LARGE \text {(3)          }\math  36x + 12y = 40.08]


Multiply (2) by -3:


*[tex \LARGE \text {(4)          }\math -18x - 12y = -24.78]


Now that the coefficients on <i>y</i> are additive inverses, add (3) to (4):


*[tex \LARGE \text {(5)          }\math  18x + 0y = 15.3]


Multiply by *[tex \LARGE \frac {1}{18}]


*[tex \LARGE \text {(6)          }\math  x = \frac {15.3}{18} = 0.85]


Substitute the value for <i>x</i> given in (6) into either (1) or (2) (I choose (1)):


*[tex \LARGE \text {                 }\math 9(0.85) + 3y = 10.02]
*[tex \LARGE \text {             \Rightarrow}\math 7.65 + 3y = 10.02]
*[tex \LARGE \text {             \Rightarrow}\math 3y = 2.37]
*[tex \LARGE \text {             \Rightarrow}\math y = 0.79]


The solution set is the ordered pair (0.85, 0.79)


{{{drawing(
400, 400, -1, 2, -1, 2,
grid(1),
circle(.85,.79,.02),
graph(
400, 400, -1, 2, -1, 2,
y = -3x + 3.34,
y = (-3x)/2 + 2.065
))}}}


Looks pretty close to me.


You could also have multiplied (1) by 2 and (2) by -3 (or by -2 and 3) so as to eliminate the <i>x</i> instead of the <i>y</i>, but ultimately the result would have been the same.  Verification is left as an exercise for the student.


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