The sum of two numbers is less than 19. The second number is 7 more than the first. What is the possible?

Question

value for the first of the two numbers?

Answer 1

let one number x then other be y
x+y=19
x=19-y............................................(1)
y-x=7................................................(2)
y-19+y=7
2y=26
y=13
x=6

Answer 2

Two numbers are x and (x + 7)
Now their sum is less than 19
=> x + x + 7 < 19
=> 2x < 12
=> x < 6 so
x = 5, 4, 3, 2, 1 and second one is
x + 7 = 5+7, 4+7, 3+7, 2+7, 1+7
Numbers are
5, 12 or 4, 11 or 3, 10 or 2, 9 or 1, 8.

Answer 3

set up the equations based on the information x + y = -25 2x-7=y then you definately replace interior the 2nd equation for the y interior the 1st x + 2x -7 = -25 simplify 3x - 7 = -25 remedy 3x - 7 = -25 +7 +7 upload 7 3x = -18 ---- ----- divide with the aid of 3 3 3 x= -6 Now plug decrease back in for y x +y = 25 -6 +y = -25 +6 +6 upload 6 y= -19 x= -6 y= -19

Answer 4

well the first number could be 1, 2, 3, 4 or 5 and the second number could be either 8, 9, 10, 11 or 12

Answer 5

Assuming the largest value of the sum at most 19, by using algebra, the equation is as follows:

x + (x + 7) < 19
2x + 7 < 19
2x < 19 - 7
2x = 12
x < 6
Hence: The first number is, x < 6

Answer 6

Let the first number = x
the second number = x + 7

x + (x + 7) < 19
x + x + 7 < 19
2x + 7 < 19
2x < 12
x < 6

The largest integer x can be is 5 ;
the second number = 5 + 7 = 12

Answer 7

Let the numbers be x and y
x + y < 19
x + (x + 7) < 19
2x + 7 < 19
2x < 12
x < 6
Assume Natural numbers and thus:-
x could be 1 , 2 , 3 , 4 , 5

Answer 8

6 and 13

Answer 9

assume the two numbers are x and y
x + y < 19
y = 7 + x

substituting...
x + (7+x) <19
2x +7 <19
2x < 12
x < 6

the possibilities of the first number is any number less then 6

Answer 10

x+y =< 19 y=x +7
.
x+x+7 =19
2x =26
x=13 and ,,,,,,y=13-7=6,,
.

Answer 11

The 1st number: n
The 2nd number n+7

n+n+7 < 19

2n + 7 < 19 // - 7

2n < 12

n<6

{(1,8),(2,9),(3,10),(4,11),(5,12)}