Question

3.
Check which of the following are solutions of the equation x - 2y = 4 and which are
not:
(1) (0,2)
(ii) (2,0)
(iii) (4,0)
(iv) (V2,412)
(0) (1,1)​

Answer 1

hi mate,

solution:

Method To check whether given pair of values is a solution of given equation or not:

 

Sometimes a pair of values is given and we have to check whether this pair is a solution of given linear  equation in two variables or not

 For this we put the given values in given linear equation. If we get LHS = RHS then this pair of values is a solution of given linear equation, otherwise not.

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Solution:

 

i)

Given equation is

x-2y=4

On putting x=0 & y=2 in LHS

LHS= x-2y

0-2×2= -4

-4≠4

LHS ≠RHS

 

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

 

ii)

Given equation is

x-2y=4

On putting x=2 & y=0 in LHS

LHS= x-2y

2-2×0= 2

2 ≠4

LHS ≠RHS

 

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

 

iii)

Given equation is

x-2y=4

On putting x=4 & y=0 in LHS

LHS= x-2y

4-2×0= 4-0=4

4=4

LHS =RHS

Hence, (4,0) is a solution of x-2y=4

 

Iv)

Given equation is

x-2y=4

On putting x=√2 & y=4√2 in LHS

LHS= x-2y

√2- 2×4√2= √2-8√2=7√2

7√2≠4

LHS ≠RHS

 

Hence, (√2,4√2) is not a solution of x-2y=4

 

V)

Given equation is

x-2y=4

On putting x=1 & y=1 in LHS

LHS= x-2y

1-2×1= 1-2

-1≠4

LHS ≠RHS

 

Hence, (1,1) is not a solution of x-2y=4

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i hope it helps you.

Answer 2

$\huge\underbrace\mathrm\pink{Solution}$

(i) (0,2) means x = 0 and y = 2

Puffing x = 0 and y = 2 in x – 2y = 4, we get

L.H.S. = 0 – 2(2) = -4.

But R.H.S. = 4

∴ L.H.S. ≠ R.H.S.

∴ x =0, y =2 is not a solution.