How do you solve the following system: #x + y = 16, 2x + 3y = 12#?

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Answer 1 · 2018-03-02T00:04:31

#x=36#, #y=-20#

We have, #x+y=16#.............#[1]#

And also #2x+3y=12#................#[2]#

From #[1]# subtracting #x # from both sides, #y=16-x# and substituting this value for # y# into #[2]# we obtain.......

#2x+3[16-x]=12#, and so after expanding the bracket we get
#2x+48-3x=12#, after arranging and collecting like terms,

#x=36#............#[3]# Substitute this value of #x# into ..#[1]#

#36+y=16# and so #y=-20#. You can check these answers by

substituting these values of #x and y# into...... #[1]# or#[2]# and they should satisfy the equations.