Question

Solve for x and y: 47x+31y=63 ; 31x+47y=15

Answer 1

when 47x+31y=63,
47x=(63-31y)

and when 31x+47y=15,
31x=(15-47y)

therefore,
47x+31x=(63-31y)+(15-47y)
78x=78-78y
78x+78y=78
(It is the same as - )
x+y=1
so the values are,
x=1-y
y=1-x

IM A GENIUS

Answer 2

Answer:

47x+31y=63(eqn 1)

31x+47y=15(eqn 2)

adding both eqn,we get

78x+78y=78

dividing this eqn by 78,we get

x+y=1 (be it eqn 3)

subtracting eqn 1 and eqn 2,we get

16x-16y=48

dividing this eqn by 16, we get

x-y=3(be it eqn 4)

using elimination method for eqn 3and eqn4

x+y=1

x-y=3

2x=4

x=2

putting value of x in x+y=1,by solving we get

y=-(negative)1

therefore

x=2

y=-1