How do you solve #3x+y=16# & #2x-3y=-4# using substitution?

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Answer 1 · 2016-09-05T18:53:04

#x = 4, y = 4#

We have: #3 x + y = 16# and #2 x - 3 y = - 4#

Let's solve the first equation for #y#:

#=> y = 16 - 3 x#

Then, let's substitute this value of #y# into the second equation:

#=> 2 x - 3 (16 - 3 x) = - 4#

#=> 2 x - 48 + 9 x = - 4#

#=> 11 x = 44#

#=> x = 4#

Now, let's substitute this value of #x# into the equation for #y#:

#=> y = 16 - 3 (4)#

#=> y = 16 - 12#

#=> y = 4#

Therefore, the solutions to the system of equations is #x = 4# and #y = 4#.