Math, asked by bhargu, 1 year ago

solve the following pair of linear equations by the substitution method:
(i)3x-y=3
   9x-3y=9
(ii)0.2x+0.3y=1.3
   0.4+0.5y=2.3
(iii)√2x+√3y=0
     √3x-√8y=0
(iv)3x/2-5y/2=-2
    x/3+y/2=13/6
(v)2x+2y=12
   2x+3y=13
(vi)2(x+y)-3(x-y)=7
    3(x-2y)+5(2x+y)=1


Snaky: tenth completed...u??
Snaky: reply
bhargu: 10th
Snaky: k...
bhargu: u just do 5 and 6
Snaky: see pro 2...is there x beside 0.4 or not!!
bhargu: and for rest u just simply write the solutions
Snaky: k...
bhargu: pl. fast
bhargu: i have to check my answers it is becoming to late

Answers

Answered by Snaky
24
1)two eq.ns are same so x and y can be any value.
2)x=0.8 and y=3.8
3)x=0 and y=0
4)x=47/19 and y=51/19
5)2x+2y=12
(-)2x+3y=13
       -y=-1
       y=1
substitute y=1 in eq.1
2x+2(1)=12
2x=12-2=10
x=5
 x=5 and y=1
6)-x+5y=7
  13x+1y=1
multiply eq.2 with 5
-x+5y=7
65x-5y=5
64x=12 and x=3/16
substitute x=3/16 in eq.1 ,we get
y=23/16
x=3/16 and y=23/16

manitkapoor2: great well done!!
bhargu: thanks a lot
bhargu: can u tell me what is ur mother tongue
Snaky: thnx....
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