SOLUTION: SOLVE BY SUBSTITUTION METHOD :) x+4=-1/2+4x 2x+3y=1/2x+2y+1 solve by ADDITION METHOD :) 3Y-1 divided by 2= 3x+8 divided by 5 x+y divided by 2= 6+x-y divided by 2

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: SOLVE BY SUBSTITUTION METHOD :) x+4=-1/2+4x 2x+3y=1/2x+2y+1 solve by ADDITION METHOD :) 3Y-1 divided by 2= 3x+8 divided by 5 x+y divided by 2= 6+x-y divided by 2      Log On


   

Question 431949: SOLVE BY SUBSTITUTION METHOD :)
x+4=-1/2+4x
2x+3y=1/2x+2y+1
solve by ADDITION METHOD :)
3Y-1 divided by 2= 3x+8 divided by 5
x+y divided by 2= 6+x-y divided by 2
Answer by anshu_a(16) About Me  (Show Source):
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SOLVE BY SUBSTITUTION METHOD :)
x+4=-1/2+4x
4+1/2 = 4x-x
9/2 = 3x or x= 3/2
SUBTITUTE VALUE OF x
2x+3y=1/2x+2y+1
2*3/2 + 3y = 1/(2*3\2)+2y+1
3+3y = 1/3+2y+1
3y-2y = 1/3+1-3
y = 1/3-2 = -5/3
solve by ADDITION METHOD :)
3Y-1 divided by 2= 3x+8 divided by 5
x+y divided by 2= 6+x-y divided by 2
(3y-1)/2 = (3x+8)/5
or 5(3y-1) = 2(3x+8)
15y-5 = 6x+16
or 6x-15y = -21------------(1)
(x+y)/2 = (6+x-y)/2 or (x+y) = (6+x-y)
2y = 6 or y = 3
put value of y in (1) and get x