B - e o WA€= 2 -xaer-=£ 4 LySRRo

1.	B - e o WA€= 2 -xaer-=£ 4 LySRRo
Answer» By elimination method x/2 + 2y /3 = - 1 ………..(1) x – y/3 = 3 ………..(2) Multiplying equation (1) by 2, we obtain x + 4y/3 = - 2 ………(3) x – y/3 = 3 ………..(2) Subtracting equation (2) from equation (3), we obtain 5y /3 = -5 Divide by 5 and multiply by 3 we get Y = -15/5 Y = - 3 Substituting in equation (2), we obtain x – y/3 = 3 ………..(2) x – (-3)/3 = 3 x + 1 = 3 x = 2 Hence our answer is x = 2 and y = −3 By substitution method x – y/3 = 3 ………..(2) Add y/3 both side we get x= 3 + y/3 ……(4) Plug this value in equation (1) we get x/2 + 2y /3 = - 1 ………..(1) (3+ y/3)/2 + 2y /3 = -1 3/2 + y /6 + 2y/3 = - 1 Multiply by 6 we get 9 + y + 4y = - 6 5y = -15 Y = - 3 Hence our answer is x = 2 and y = −3 Like my answer if you find it useful!

Answer»

By elimination method x/2 + 2y /3 = - 1 ………..(1)

x – y/3 = 3 ………..(2)

Multiplying equation (1) by 2, we obtain x + 4y/3 = - 2 ………(3)

x – y/3 = 3 ………..(2)

Subtracting equation (2) from equation (3),

we obtain 5y /3 = -5

Divide by 5 and multiply by 3 we get Y = -15/5

Y = - 3

Substituting in equation (2), we obtain x – y/3 = 3 ………..(2)

x – (-3)/3 = 3 x + 1 = 3 x = 2

Hence our answer is x = 2 and y = −3

By substitution method x – y/3 = 3 ………..(2)

Add y/3 both side we get x= 3 + y/3 ……(4)

Plug this value in equation (1) we get x/2 + 2y /3 = - 1 ………..(1)

(3+ y/3)/2 + 2y /3 = -1 3/2 + y /6 + 2y/3 = - 1

Multiply by 6 we get 9 + y + 4y = - 6

5y = -15

Y = - 3

Hence our answer is x = 2 and y = −3

Like my answer if you find it useful!

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