Given:
#7x-2y=-67#.................................(1)
#x+3y=3#............................................(2)
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Rewrite equation (2) to give:
#y=-1/3x+1" ".......................(2_a)#
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Substitute #(2_a)# into (1) giving:
#7x-2(-1/3x+1)=-67" "....(1_a)#
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#color(blue)("Solving for "x)#
#7x+2/3x-2=-67#
#19/3 x=-65#
#color(blue)(x=-195/19" "->10 5/19)#
#color(purple)("Notice how I keep fractional values instead of decimals.")##color(purple)("It is more precise that way!")#
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I will let you solve for #y#. Just substitute the value for #x" in "(2_a)#
#(2_a)# looks as though it is the simplest one to use!
