How do you solve #x= 3y-1# and #x+2y=9# using substitution?
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You know the value of the variable #x#, so you can substitute that into the equation.
#overbrace((3y - 1))^(x) + 2y = 9#
Remove the parentheses and solve.
#3y - 1 + 2y = 9#
#=> 5y - 1 = 9#
#=> 5y = 10#
#=> y = 2#
Plug #y# into either equation to find #x#.
#x = 3overbrace((2))^(y) - 1#
#=> x = 6 - 1#
#=> x = 5#
#(x,y) => (5,2)#
Given #x=3y-1 and x+2y=9#
Substitute #x=3y-1# into #x+2y=9#,
#(3y-1)+2y=9#
#5y-1=9#
#5y=10#
#y=2#
Substitute y=2 into the first equation,
#x=3(2)-1#
#x=5#
If
#x = 3y -1#
then use that equation in the second equation. This means that
#(3y - 1) + 2y = 9#
#5y - 1 = 9#
#5y - 1 + 1 = 9 + 1#
#5y = 10#
#(5y)/5 = 10/5#
#y = 2#
Having said this, just replace the #y# in the first equation in order to get the #x#.
#x = 3(2) -1#
#x = 6 -1#
#x = 5#
After that, just check that the values make sense:
#x = 3y - 1#
#5 = 3(2) -1#
#5 = 6 - 1#
#5 = 5#
And for the second one:
#x + 2y = 9#
#5 + 2(2) = 9#
#5 + 4 = 9#
#9 = 9#
Both answers satisfy both equations, which makes them correct.
