How do you solve #y = 3 - 1/2 x, 3x + 4y = 1#?

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Answer 1 · 2016-06-26T16:59:02

#x=-11, y = 8 1/2#

Substitution seems quite a good method to choose, because #y# is given as the subject in one of the equations.

#y = (3-1/2x)# can be substituted into #3x+4y =1#

ie. Use# (3-1/2x)# in the place of #y# in the second equation:

#3x+4(3-1/2x) =1" "# now there are only x-terms.

#3x +12 -2x = 1#

#x = -11#

Now that we have the value for #x#, use it to find the value of #y#.

#y = (3-1/2x) rArr = 3-1/2(-11)#

#y = 3+5 1/2 rArr y = 8 1/2#