How do you solve #3y=x-30# and #3x+9y=0# using substitution?

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Answer 1 · 2017-02-01T09:02:15

#x=15# and #y=-5#

#3y=x-30#
#3x+9y=0#

In the second equation, divide all terms by #3#.

#3x+9y=0#

#x+3y=0#

In this reduced equation, substitute #3y# with #color(red)((x-30))#, the value from the first equation.

#x+color(red)((x-30))=0#

Open the brackets and simplify.

#x+color(red)(x-30)=0#

#2x-30=0#

Add #30# to each side.

#2x=30#

Divide both sides by #2#.

#x=15#

In the first equation, substitute #x# with #color(blue)(15)#.

#3y=x-30#

#3y=color(blue)(15)-30#

#3y=-15#

Divide both sides by #3#.

#y=-5#