Question

please answer thsee linear equations​

Answer 1

Solutions:

(i) (0, 2)

(x,y) = (0,2)

Here, x=0 and y=2

Substituting the values of x and y in the equation x–2y = 4, we get,

x–2y = 4

⟹ 0 – (2×2) = 4

But, -4 4

(0, 2) is not a solution of the equation x–2y = 4

(ii) (2, 0)

(x,y) = (2, 0)

Here, x = 2 and y = 0

Substituting the values of x and y in the equation x -2y = 4, we get,

x -2y = 4

⟹ 2-(2×0) = 4

⟹ 2 -0 = 4

But, 2 ≠ 4

(2, 0) is not a solution of the equation x-2y = 4

(iii) (4, 0)

Solution:

(x,y) = (4, 0)

Here, x= 4 and y=0

Substituting the values of x and y in the equation x -2y = 4, we get,

x–2y = 4

⟹ 4 – 2×0 = 4

⟹ 4-0 = 4

⟹ 4 = 4

(4, 0) is a solution of the equation x–2y = 4

(iv) (√2,4√2)

Solution:

(x,y) = (√2,4√2)

Here, x = √2 and y = 4√2

Substituting the values of x and y in the equation x–2y = 4, we get,

x –2y = 4

⟹ √2-(2×4√2) = 4

√2-8√2 = 4

But, -7√2 ≠ 4

(√2,4√2) is not a solution of the equation x–2y = 4

(v) (1, 1)

Solution:

(x,y) = (1, 1)

Here, x= 1 and y= 1

Substituting the values of x and y in the equation x–2y = 4, we get,

x –2y = 4

⟹ 1 -(2×1) = 4

⟹ 1-2 = 4

But, -1 ≠ 4

(1, 1) is not a solution of the equation x–2y = 4

hope it helps you

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Answer 2

Step-by-step explanation:

community to property of addition subtraction multiplication division in rational numbers

show it by using egs in your copy

no net