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

Question

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

Ethan Adkins · Accepted Answer

#x=5, y=2# Considering x+2y=9# and #x=3y-1 Change #x=3y-1# to #x+2y=9#. #5y-1=9# #(3y-1)+2y=9# #5y=10# #y=2# In the first equation, replace y=2 with #x=3(2)-1# and #x=5#.

Eli Assouline · Answer

#(5,2)#

You can enter the value of the variable #x#, which you know, into the equation as follows: #overbrace((3y - 1))^(x) + 2y = 9#

Subtract 1 from 2 and solve #3y - 1 = 9#.

#=> 1 - 5y = 9#

#=> 10# x 5y

#=> y is equal to 2#

To find #x#, enter #y# into the appropriate equation: #x = 3overbrace((2))^(y) - 1#.

#=> x is equal to six minus one.

#=> x is equal to 5#

#(x,y) => (5,2)#

Abigail Allen · Answer

#x = 5#
#y = 2# Should #x = 3y -1# subsequently apply that formula to the second equation. This implies that #(3y - 1) + 2y = 9# #5y - 1 = #9# 9 + 1# - 1 + 1 = #5y #5y equals #10# #(5y)/5 = 10/5# #y = 2# Having said that, to obtain the #x#, simply swap out the #y# in the first equation. #x = 3(2) -1# #x equals 6 -1# #x equals 5# Simply verify that the values make sense after that: #x = 3y - 1# #5 = 3(2) -1# #5 is equal to 6 - 1. #5 is equivalent to 5#. Regarding the second: #x + 2y = #9# #5 plus 2(2) equals 9# #5 plus #4 equals #9. 9# = #9 Since both solutions satisfy both equations, they are both accurate.

Jameson Allen · Answer

undefined To solve the system of equations using substitution, you would first solve one of the equations for one variable and then substitute that expression into the other equation.

For example, from the equation x = 3y - 1, you can isolate x to get x = 3y - 1. Then, substitute this expression for x into the second equation x + 2y = 9.

After substitution, you'll get (3y - 1) + 2y = 9.

Solve this equation for y, then substitute the value of y back into one of the original equations to solve for x.