 Hi, and welcome to the session. Let's discuss the following question. The question says, check which of the following are solutions of the equation x minus 2y equals to 4 and which are not. First order pair is 0, 2. Second is 2, 0. Third is 4, 0. Fourth is 2, 2, 4, 2, and fifth is 1, 1. In this question we have to check which of the ordered pair is the solution of the given linear equation x minus 2y equals to 4. We know that solution means pair of values 1 for x and 1 for y would satisfy the given equation. So in this question we will put the value of x and y in the given equation. If LHS is equal to RHS then this implies that the given ordered pair is the solution of the given equation. But if LHS is not equal to RHS then this implies that given ordered pair is not a solution of this equation. So let's now proceed on with our solution. First ordered pair is 0, 2. Equal to 0 and y is equal to 2. So let's now substitute the value of x and y in the given equation x minus 2y equals to 4. LHS is equal to x minus 2y. Substitute the value of x and y. x is equal to 0 and y is equal to 2. 0 minus 4 is equal to minus 4. But our RHS is equal to 4. So LHS is not equal to RHS and this implies 0, 2 is not a solution of the given linear equation. So now the given ordered pair is not a solution of the given linear equation. This completes the first part. Again ordered pair given to us is 2, 0. Now here x is equal to 2 and y is equal to 0. So now we will again substitute the value of x and y in x minus 2y equals to 4. LHS of this equation is x minus 2y. Now substitute the value of x and y. x is equal to 2 and y is equal to 0. So we get 2 minus 0 and 2 minus 0 is equal to 2. Right. But our RHS is equal to 4. Therefore LHS is not equal to RHS. Right. And this implies that 2, 0 is not a solution of the given equation x minus 2y equals to 4. So now the given ordered pair is not a solution of the given linear equation. So this completes the second part. Ordered pair is 4, 0. Now here x is equal to 4 and y is equal to 0. Right. So now substitute the value of x and y in x minus 2y equals to 4. LHS of this equation is x minus 2y. Right. x is equal to 4 and y is equal to 0. So we get 4 minus 0 and 4 minus 0 is equal to 4. Our RHS is also equal to 4. Therefore LHS is equal to RHS and this implies that 4, 0 is a solution of the given linear equation. So our answer is yes, since 4, 0 is a solution of the given linear equation. So this completes the third part. Fourth ordered pair is root 2, 4 root 2. Here x is equal to root 2 and y is equal to 4 root 2. So let's now substitute the value of x and y in x minus 2y equals to 4. Now LHS of this equation is x minus 2y. x is equal to root 2 and y is equal to 4 root 2. And this is equal to root 2 minus 8 root 2. Right. And this is equal to minus 7 root 2. LHS is equal to 4. Therefore LHS is not equal to RHS. Right. And this implies that root 2, 4 root 2 is not a solution of the given linear equation. So our answer is no. Since the given ordered pair is not a solution of the given linear equation. So this completes the fourth part. And the last ordered pair is 1, 1. Here x is equal to 1 and y is also equal to 1. Right. So now we will substitute the value of and y x minus 2y equals to 4. LHS of this equation is x minus 2y. So now substitute x equals to 1 and y equals to 1. And this is equal to 1 minus 2. So our LHS is equal to minus 1. But our RHS is equal to 4. Therefore LHS is not equal to RHS. And this implies that the given ordered pair 1, 1 is not a solution of the given equation. So no. The given ordered pair is not a solution of the given linear equation. So this completes the last part. Bye and take care. Hope you have enjoyed this session.