How do you solve #x - 4y - 1 = 0# and #x + 5y - 4 = 0 # using substitution?

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Answer 1 · 2016-03-20T16:27:27

The solutions are:

#color(green)(x = 7/3 #
#color(green)( y = 1/ 3#

#x- 4y - 1 = 0#
#x- 4y = 1#
#color(green)(x= 1 + 4y # ..........equation #(1)#

#x + 5y - 4 = 0#
#x + 5y = 4#.............equation #(2)#

Substituting equation #(1) # in #(2): #

#color(green)(1 + 4y ) + 5y = 4#

#1 + 9y = 4#

# 9y = 4 -1 #

# 9y = 3 #

# y = 3 / 9 = color(green)(1/ 3#

Finding the value of #x# from equation #(1)#:
#color(green)(x= 1 + 4y #

#x = 1 + 4 * (1/3) #

#x = 1 + ( 4 /3) #

#x = 3/3 + ( 4 /3) #

#color(green)(x = 7/3 #