Question

41x+19y-161=0 and 19x+41y-139=0. solve the linear equation​

Answer 1

Answer:

• x = 3
• y = 2

Explanation:

Given equation,

⇒ 41x + 19y = 161 . . . . . . (i)

⇒ 19x + 41y = 139 . . . . . .(ii)

Subtracting equation (i) by equation (ii) we get,

41x + 19y = 161

19x + 41y = 139

(-) (-) (-)

⇒ 22x - 22y = 22

⇒ 22(x - y) = 22

⇒ x - y = 22/22

⇒ x - y = 1

⇒ x = 1 + y . . . . . . . (iii)

Substituting the value of ‘x’ in equation (i) :-

⇒ 41x + 19y = 161

⇒ 41(1 + y) + 19y = 161

⇒ 41 + 41y + 19y = 161

⇒ 41 + 60y = 161

⇒ 60y = 161 - 41

⇒ 60y = 120

⇒ y = 120/60

⇒ y = 2

∴ y = 2

Again, substitute the value of ‘y’ in eq.(iii) :-

⇒ x = 1 + y

⇒ x = 1 + 2

⇒ x = 3

∴ x = 3