How do you solve #-6y - 2x = 0# and #11y + 3x = 4# using substitution?

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Answer 1 · 2017-01-05T21:46:04

#x=-6 and y=2#

#-6y -2x =0 to -6y = 2x#
Therefore #-3y = x#

We can substitute #color(blue)(-3y = x)# into the second equation;

#11y + 3 color(red)((x)) = 4#
#11y + 3 color(red)((-3y)) = 4#

This is simplified to;

#11y -9y =4#
#2y=4#
#y=2#

As we know that #color(blue)(-3y = x)#, we can substitute the value of #y# into that to get #x#

#-3y = x to -3(2) =x#

#-6=x#