How do you solve #x= 3y-1# and #x+2y=9# using substitution?

3 Answers
Mar 6, 2018

Answer:

#(5,2)#

Explanation:

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)#

Mar 6, 2018

Answer:

#x=5, y=2#

Explanation:

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#

Mar 6, 2018

Answer:

#x = 5#
#y = 2#

Explanation:

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.