Question

find a pair of numbers whose sum is 21 and 1 of them exceeds twice the other by 3

Answer 1

Hi ,

Let one number = x

Second number = 2x + 3

According to the problem given ,

Sum of these numbers = 21

x + 2x + 3 = 21

3x + 3 = 21

3x = 21 - 3

3x = 18

x = 18/3

x = 6

Therefore ,

Required pair ( x , 2x + 3 )

( 6, 2×6 + 3 ) ,

( 6 , 15 )


I hope this helps you.
:)

Answer 2

Heya,

Here's ur answer :-

Let one number be x and the other is y.

x + y = 21-------(¡)

x = 2y + 3

=> x - 2y = 3------(¡¡)

Subtracting (¡¡) from (¡),

x+y = 21
x-2y = 3
_______
3y = 18
=> y = 6

x + y = 21
=>x + 6 = 21
=> x = 15

So, the two numbers are 6 and 15

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Regards
Supersonu
Brainly Star