How do you solve #5x+y=1# and #3x+2y=2# using substitution?

2 Answers
sankarankalyanam
Mar 5, 2018

#color(green)(x = 0, y = 1#

Explanation:

#5x + y = 1# Eqn (1)

#y = 1 - 5x#

#3x + 2y = 2# Eqn (2)

Substituting value of y in second equation in terms of x,

#3x + (2* (1-5x) = 2#

#3x + 2 - 10x = 2#

#-7x = 2 - 2 = 0# or #x = 0#

Substituting value of x in Equation (1),

#(5*0) + y = 1# or #y = 1#

Solution #color(green)(x = 0, y = 1)#

Nallasivam V
Mar 5, 2018

#x=0#
#y=1#

Explanation:

Given -

#5x+y=1# ---------------(1)
#3x+2y=2# ---------------(2)

#5x+y=1# ---------------(1)#xx2#
#10x+2y=2# ------(3)
#3x+2y=2# -------(2) #(3) - (2)#
#7x=0#
#x=0/7=0#

Substitute #x=0# in equation (1)

#5(0)+y=1#
#y=1#

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