Question

solve the system if possible


| x+y+z = 15
|
|
< 7x+y+z = 75
|
|
| 3x+y-z=29


a. x=10,y=2,z=3
b. x=–10,y=–2,z=–3
c. x=3,y=10,z=–3
d. x=12,y=9,z=–3
e. no solution; inconsistent system

Answer 1

try a.
10+2+3 = 15 yes
70+2+3 = 75 yes
30+2-3 = 29 yes
so a works, the end :)

Answer 2

x + y + z = 15

7 x + y + z = 75

3 x + y - z = 29


Multiply first equation by − 7

( x + y + z = 15 ) * ( - 7 )

-7 x - 7 y - 7 z = - 105

add this to the second equation

-7 x - 7 y - 7 z = - 105
+
7 x + y + z = 75
____________________

The result is:

- 6 y - 6 z = - 30

Divide both sides by 6

- y - z = - 5 Multiply both sides by - 1

y + z = 5

Replace y + z with 5 in first equation:

x + y + z = 15

x + 5 = 15 Subtract 5 to both sides

x + 5 - 5 = 15 - 5

x = 10

Replace x = 10 in second and third equation:


7 x + y + z = 75

7 * 10 + y + z = 75

70 + y + z = 75 Subtract 70 to both sides

70 + y + z - 70 = 75 - 70

y + z = 5


3 x + y - z = 29

3 * 10 + y - z = 29

30 + y - z = 29 Subtract 30 to both sides

30 + y - z - 30 = 29 - 30

y - z = - 1


Now you must solve system of 2 equations with 2 unknow:

y + z = 5

y - z = - 1


Add the first equation to the second equation

y + z = 5
+
y - z = - 1
_________

2 y = 4 Divide both sides by 2

y = 4 / 2

y = 2

Replace x = 10 and y = 2 into first equation:

x + y + z = 15

10 + 2 + z = 15

12 + z = 15 Subtract 12 to both sides

12 + z - 12 = 15 - 12

z = 3

The solutions are:

x = 10 y = 2 and z = 3