How do you solve #x + 4y = 12# and #2x - 3y = 6# using substitution?

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Answer 1 · 2018-03-18T06:29:35

#x=60/11andy=18/11#

Here,#x+4y=12...to(1)and2x-3y=6...to(2)#

From #(1) color(red)(x=12-4y)...to(3).#

substituting value of x in #equ^n(2)#,we get

#2color(red)((12-4y))-3y=6=>24-8y-3y=6#

#=>24-6=8y+3y=>18=11y=>y=18/11#

From(3),we have

#x=12-4(18/11)=(132-72)/11=60/11#

Check the answer:

from (1) #x+4y=12#

#=>LHS=60/11+4(18/11)=60/11+72/11=132/11=12=RHS#

graph{(x+4y-12)(2x-3y-6)=0 [-10, 10, -5, 5]}