How do you solve #7x + 6y = 1# and #x = 7 - 2y# using substitution?

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Answer 1 · 2016-04-24T16:00:24

Point of intersection#" "->(x,y) = (-5,6)#

Given:

#x=7-2y# ............................(1)
#7x+6y=1# .........................(2)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute for #x# in equation (2) using equation (1)

#=> 7x+6y=1" "->" "7(7-2y)+6y=1#

#49-14y+6y=1#

#49-8y=1#

#8y=48#

#y=48/8 = 6#

#color(blue)(y=6)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute for #y# in equation (1)

#x=7-2y" "->" "x=7-2(6)#

#color(blue)(x=-5)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Tony B