How do you solve the following system: #8x-3y=3, x+3y=3
#?
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"Find the sum of the integers between 2 and 100 which are divisible by 3 ?"
This ordered pair is #(2/3, 7/9)# or #(0.bar6, 0.bar7)#.
Use the elimination method, and then substitution to solve. Line up the two equations, one on top of the other:
#8x - 3y = 3#
#color(white)(8)##x + 3y = 3#
Now add them together, and notice that #y# will be eliminated because #-3y + 3y = 0#:
#8x color(blue)(- 3y) = 3#
#color(white)(8)##x color(blue)(+ 3y) = 3#
#color(white)(x)9x color(white)(+ 0y) = 6#
#9x = 6#
#x = 6/9 rarr 2/3#
Now substitute that value for #x# into another equation, and solve for #y#:
#x + 3y = 3#
#2/3 + 3y = 3#
#2 + 3(3)y = 3(3)#
#2 + 9y = 9#
#9y = 7#
#y = 7/9#
So this ordered pair is #(2/3, 7/9)#. As a repeating decimal, the coordinates are #(0.bar6, 0.bar7)#.
Solution: # x= 2/3 and y= 7/9#
#8 x-3 y=3 ; (1) , x +3 y =3; (2)# , adding equation (1)
and equation (2) we get, # 9 x = 6 :. x = 6/9 or x= 2/3#
Putting #x=2/3# in equation (2) we get, #2/3 +3 y =3; # or
#3 y =3- 2/3 or 3 y = 7/3 or y = 7/9 #
Solution: # x= 2/3 and y= 7/9# [Ans]
