1 and 20.
the 2 numbers are 51 and 19
72 plus 19 = 91
Name the two numbers x & y , you can form these equations:x + y = 21x - y = 19Solve the system by substitution:x - y = 19 --> x = 19 + y19 + y + y = 2119 + 2y = 212y = 2y = 1But, X + y = 21X + 1 = 21x = 20
17 and 19
1 and 20.
the 2 numbers are 51 and 19
16 and 35
Find 3 consecutive numbers where the product of the smaller two numbers is 19 less than the square of the largest number.
19 and -7
20 and 1. 20 + 1 = 21 20 - 1 = 19
72 plus 19 = 91
Surely the question is wasn't meant to say what it literally said, but a solution to its literal statementis still possible.The two consecutive numbers are: 18 and 19.The smaller two numbers are: 18 and 19.Their product is 342 .The largest number is 19.Its square is 361.342 is 19 less than 361.
The sum of two numbers witha difference of two equals forty can be set up and solved this way: x + (x - 2) = 40 2x - 2 = 40 2x = 42 x = 21 x - 2 = 19 The two numbers are 21 and 19.
How about: 14 and 11
4x4=16, and 5x5=25, so 4 and 5 are the whole numbers closest to the square root of 19. [And incidentally of all the square roots of all the whole numbers between 16 and 25.]
20 and 1 Seems to be a common problem. Add the sum to the difference and divide by 2 gives one number. Then start over and subtract the smaller of the two givens from the larger and then divide by 2 to get the other number. So here you have (21+19)/2 and (21-19)/2