How do you solve the following system using substitution?: #5x-y=3, 4y-x=1#

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Answer 1 · 2018-05-15T00:09:56

Solve one equation for one variable

Take one of the equations and solve for either x or y
#4y-x=1 => x=4y-1#

Plug into the other equation for x
#5(4y-1) - y = 3 #

Solve
#20y-5-y=3#

#19y=8#

#y=8/19#

Plug y into original equation to find #x#

#4(8/19)-x=1#

#32/19 - x =1#

#x=13/19#

Answer 2 · 2018-05-15T00:16:05

#x=13/19# and #y=8/19#

show below:

#5x-y=3#...............1

#4y-x=1#................2

from the equation 1

#y=5x-3#.............3

put 3 in 2

#4(5x-3)-x=1#

#20x-12-x=1#

#19x=13#

#x=13/19#

#y=5(13/19)-3=65/19-3=[65-57]/19=8/19#