The sum of three numbers is 19. The sum of twice the first number, 3 times the second number, and 4 times the third number is 59. The:difterence between 6 times the first number and the second number is 31 . Find the three numbers.:first number:

Lösung

T h e s u m o f t h ree n u mb ers i s 19. 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 59. T h e : d i f t 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 31 . F in d t h e t h ree n u mb ers . : f i rs t n u mb er :

The numbers : \frac{51 + 2229}{12}, \frac{51 - 2229}{6}, \frac{2229 - 153}{12} + 19

Schritte zur Lösung

Represent the numbers with

x, y, z

number 1 : x

number 2 : y

number 3 : z

A.T.Q.:

x + y + z = 19

2 x + 3 y + 4 z = 59

6 x y = 31

Löse das Gleichungssystem:

x + y + z = 19, 2 x + 3 y + 4 z = 59, 6 x y = 31 : (x = \frac{51 + 2229}{12}, y = \frac{51 - 2229}{6}, z = \frac{2229 - 153}{12} + 19)

The numbers : \frac{51 + 2229}{12}, \frac{51 - 2229}{6}, \frac{2229 - 153}{12} + 19