
Write down a pair of integers whose
a) Sum is -3.
b) Difference is -5.
c) Difference is 2.
d) Sum is 0.
Answer
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Hint: We will fix one integer in every of the given options and then assume another integer to be any one of the variables like a or b. Then, we will put in the values as per given conditions and find the number which the variable takes and thus we have a pair of integers.
Complete step-by-step answer:
Let us solve the first option first of all.
Option A: sum is -3
Let us fix one integer to be 1. Now, let another integer be a.
Now, we have the sum of these both integers to be equal to -3.
$ \Rightarrow 1 + a = - 3$
Now, taking 1 from addition in LHS to subtraction in RHS, we will get:-
$ \Rightarrow a = - 3 - 1 = - 4$
Thus, we have our two integers 1 and -4.
$\therefore $ The required pair of integers whose sum will be -3 is 1 and -4.
Option B: difference is -5
Let us fix one integer to be 1. Now, let another integer be b.
Now, we have the difference of these both integers to be equal to -5.
$ \Rightarrow 1 - b = - 5$
Now, taking 1 from addition in LHS to subtraction in RHS, we will get:-
$ \Rightarrow - b = - 5 - 1 = - 6$
$ \Rightarrow b = 6$
Thus, we have our two integers 1 and 6.
$\therefore $ The required pair of integers whose difference will be -5 is 1 and 6 respectively.
Option C: difference is 2
Let us fix one integer to be 1. Now, let another integer be c.
Now, we have the difference of these both integers to be equal to 2.
$ \Rightarrow 1 - c = 2$
Now, taking 1 from addition in LHS to subtraction in RHS, we will get:-
$ \Rightarrow - c = 2 - 1 = 1$
$ \Rightarrow c = - 1$
Thus, we have our two integers 1 and -1.
$\therefore $ The required pair of integers whose difference will be 2 is 1 and -1 respectively.
Option D: sum is 0
Let us fix one integer to be 1. Now, let another integer be d.
Now, we have the sum of these both integers to be equal to 0.
$ \Rightarrow 1 + d = 0$
Now, taking 1 from addition in LHS to subtraction in RHS, we will get:-
$ \Rightarrow d = - 1$
Thus, we have our two integers 1 and -1.
$\therefore $ The required pair of integers whose sum will be 0 is 1 and -1.
Note: The students must note that they can fix any number (not necessarily equal to 1) for the calculations. But they must be aware of 0 for the fourth part because a sum of two numbers among which one is 0 is only possible if the other number is 0 as well. So, then both the numbers will be 0 and it would not be distinct digits (if required distinct digits).
We can here do this using fixing numbers because Integers are closed under addition and subtraction.
Complete step-by-step answer:
Let us solve the first option first of all.
Option A: sum is -3
Let us fix one integer to be 1. Now, let another integer be a.
Now, we have the sum of these both integers to be equal to -3.
$ \Rightarrow 1 + a = - 3$
Now, taking 1 from addition in LHS to subtraction in RHS, we will get:-
$ \Rightarrow a = - 3 - 1 = - 4$
Thus, we have our two integers 1 and -4.
$\therefore $ The required pair of integers whose sum will be -3 is 1 and -4.
Option B: difference is -5
Let us fix one integer to be 1. Now, let another integer be b.
Now, we have the difference of these both integers to be equal to -5.
$ \Rightarrow 1 - b = - 5$
Now, taking 1 from addition in LHS to subtraction in RHS, we will get:-
$ \Rightarrow - b = - 5 - 1 = - 6$
$ \Rightarrow b = 6$
Thus, we have our two integers 1 and 6.
$\therefore $ The required pair of integers whose difference will be -5 is 1 and 6 respectively.
Option C: difference is 2
Let us fix one integer to be 1. Now, let another integer be c.
Now, we have the difference of these both integers to be equal to 2.
$ \Rightarrow 1 - c = 2$
Now, taking 1 from addition in LHS to subtraction in RHS, we will get:-
$ \Rightarrow - c = 2 - 1 = 1$
$ \Rightarrow c = - 1$
Thus, we have our two integers 1 and -1.
$\therefore $ The required pair of integers whose difference will be 2 is 1 and -1 respectively.
Option D: sum is 0
Let us fix one integer to be 1. Now, let another integer be d.
Now, we have the sum of these both integers to be equal to 0.
$ \Rightarrow 1 + d = 0$
Now, taking 1 from addition in LHS to subtraction in RHS, we will get:-
$ \Rightarrow d = - 1$
Thus, we have our two integers 1 and -1.
$\therefore $ The required pair of integers whose sum will be 0 is 1 and -1.
Note: The students must note that they can fix any number (not necessarily equal to 1) for the calculations. But they must be aware of 0 for the fourth part because a sum of two numbers among which one is 0 is only possible if the other number is 0 as well. So, then both the numbers will be 0 and it would not be distinct digits (if required distinct digits).
We can here do this using fixing numbers because Integers are closed under addition and subtraction.
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