7x-15y=2; x+2y=3

7x-15y=2   ...1x+2y=3   ...2From eqn1 , we get 7x=2+15yx=2+ 15 y7Substituting the value of x in eqn2 , we get2+ 15 y7+2y=32+15y+14y7=32+29y=2129 y =19y=1929.so, x=2+ 15 19297=58+285203x=343203=4929  so , value of x=4929 and y=1929

  • 3

You can do this by using elimination, substitution or cross multiplication method.

  • 3
  1. eqn says-7x-15y+2 nd 2nd eqn says :x+2y =3
    multpy 2nd one wid 7 .  nd den subtrct d eqn u got fm d 1st eqn . ----> 7x-15y-(7x+14y) =2-21 which is eql to -29y =-19 hnce y =19/29 put d value of y in any of d eqnz nd u ll get x.
  • 4

  by using elimination method-

  7x - 15y = 2  --------------- (1)

  x + 2y = 3  -----------------(2)

  7(x) + 7(2y)  = 3*7   (multiplying the both numbers by 7) ------ (3)

  subtracting the equation (3) from (1)  

  -29y = -19

  y = 19/29

  putting the value of y in equation (2) 

  x + 2 * 19/29 = 3

  x  =  29*3/19*2 = 87/38 ANS.

  • -2

17x-15y=2

x+2y=3

x=3-2y¨

{by sustitution method}

17{3-2y}-15y=2

51-34y-15y=2

51-2=15y+34y

49=49y

y=1

x=3-2y

x=3-2{1}

x=1

hence x,y=1

  • -2

s-t=3, s/3=t/2=6

  • 2
7x-15y=2..........(i)
x+2y=3.................(ii)
multiplying (ii) by 7 and using ELIMINATION METHOD IN i AND ii
 7x - 15y=  2
​ 7x +14y=+21
-     -         -     
-29y=-19
y=19/29
PUTTING IN ii
x=3-2(19)
          (29)
x=   3-38  
           29
x=87-38
     29
x=49
   29
 
  • -1
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