How do you solve this system of equations: #y+ 3= x and 3x + 4y = 16#?

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2 Answers
Sonnhard
Jun 9, 2018

#x=4,y=1#

Explanation:

Writing #y+3# for #x# in the second equation

#3(y+3)+4y=16#
expanding
#3y+9+4y=16#
so #7y=7#
#y=1#

so
#x=4#

Jonny
Aug 9, 2018

#y=1#
#x=4#

Explanation:

The easiest way here is by substitution. We know what #x# is, because of the first equation so simply take what #x# is and substitute it for #x# in the second equation like so:

#3(y+3) +4y = 16#

Distribute the 3

#3y + 9 +4y = 16#

Combine like terms

#7y+ 9= 16#

solve for #y# by subtracting 9 from both sides then dividing both sides by 7:

#y=1#

plug 1 in for #y# into either of your original equations and solve for #x#

#x=4#

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