Question

solve with eliminated method
47x+31=63,31x+47y=15​

Answer 1

Step-by-step explanation:

Given -

• 47x + 31y = 63
• 31x + 47y = 15

To Find -

• Value of x and y by elimination method

Now,

→ [ 47x + 31y = 63 ] × 31

[ 31x + 47y = 15 ] × 47

→ 1457x + 961y = 1953

1457x + 2209y = 705

(-) (-) (-)

__________________

→ -1248y = 1248

→ y = -1

Now,

Substituting the value of y on 47x + 31y = 63, we get :-

→ 47x + 31(-1) = 63

→ 47x = 63 + 31

→ 47x = 94

→ x = 2

Hence,

The value of x is 2 and y is -1.

Answer 2

ANSWER:-

47x+31y=63-----(1)

31x+47y=15-------(2)

adding both eqn,we get

• 78x+78y=78

dividing this eqn by 78,we get

x+y=1 - - - - - - (3)

subtracting eqn 1 and eqn 2,we get

• 16x-16y=48

dividing this eqn by 16, we get

x-y=3------(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