Question

solve the follwing pair of equations-47x+31y=63 and 31x+47y=15.

Answer 1

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

Answer 2

Answer:

Solving of equations we get X=2;Y=-1

Step-by-step explanation:

Given:

$&47 \mathrm{x}+31 \mathrm{y}=63...................(\mathrm{i}) \\&31 \mathrm{x}+47 \mathrm{y}=15.......................(ii) \end{aligned}$

1. Multiplying (i) by 31 and multiplying (ii) by 47 we get

$1457 \mathrm{x}+961 \mathrm{y}=1953..................................--(iii)\\1457 \mathrm{x}+2209 \mathrm{y}=705 \text ..................................{--(iv) }$

     2. Subtracting (iv) from (iii) we get

$&-1248 y=1248 \\y=-1\\putting y=-1 in (i)47 \mathrm{x}+31 \times-1=63 \\ 47 \mathrm{x}=94 \\\mathrm{x}=2\\\\ Therefore \mathrm{x}=2 and \mathrm{y}=-1$.