How do you solve the system of equations #5x - 3y = 1# and #9x - 4y = 6#?

1 Answer
Apr 24, 2018

Answer:

#x = 16/7#
#y = 73/21#

Explanation:

Firstly, you have to multiply each equation so there is the same number of a letter in each equation. In this case it is easier to make both #y#s equal #12#.
#5x - 3y = 1# should be multiplied by 4
#20x - 12y = 4

#9x - 4y = 6# should be multiplied by 3
#27x -12y = 18#

You can then take one equation away from the other. I will take the first away from the second:
#27x - 12y = 18#
#20x - 12y = 4# #-#
#(27x - 20x) + (-12y - -12y) = 18 - 4#
#7x = 16#
#x = 16/7#

You can then put this value of #x# into one of the original equations to get a value for #y#:
#(5 xx 16/7) - 3y = 1#
#3y = 80/7 - 1#
#3y = 73/7#
#y = 73/21#