The sum of three numbers is 13. The sum of twice the first number, 3 times the second number, and 4 times the third number is 36. The difference between 6 times the first number and the second number is 40. Find the three numbers.

Solution

T h e s u m o f t h ree n u mb ers i s 13. T h e s u m o f tw i ce t h e f i rs t ​ n u mb er, 3 t im es t h e seco n d ​ n u mb er, an d 4 t im es t h e t hi r d n u mb er i s 36. T h e d i ff ere n ce b e tw ee n 6 t im es t h e f i rs t n u mb er an d t h e seco n d n u mb er i s 40. F in d t h e t h ree n u mb ers .

The numbers : 7, 2, 4

Solution steps

Represent the numbers with

x, y, z

number 1 : x

number 2 : y

number 3 : z

A.T.Q.:

x + y + z = 13

2 x + 3 y + 4 z = 36

6 x - y = 40

x + y + z = 13, 2 x + 3 y + 4 z = 36, 6 x - y = 40 : x = 7, y = 2, z = 4

The numbers : 7, 2, 4