Question

Solve the following system of linear equations :
         41x+19y-161=0 and 19x+41y-139=0. ​

Answer 1

Answer:

x=3 , y=2

Step-by-step explanation:

add both equation,we get

60x+60y-300=0

x+y=5-(1)

Subtract both equation, we get

22x-22y-22=0

x-y=1-(2)

Then add equation 1 and 2, we get

2x=6

x=3

put value of x in equation (1),we get

3+y=5

y=2

Therefore, x=3 , y=2