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
try a.
10+2+3 = 15 yes
70+2+3 = 75 yes
30+2-3 = 29 yes
so a works, the end :)
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
To solve this system of linear equations, we can use the method of elimination or substitution. Let's use the method of elimination:
Step 1: Start by eliminating the z variable. We can eliminate z by subtracting the second equation from the first equation:
(x + y + z) - (7x + y + z) = 15 - 75
Simplifying, we get:
-6x = -60
x = 10
Step 2: Substitute the value of x into two of the equations to solve for y and z. Let's substitute x = 10 into the first and third equations:
Equation 1: x + y + z = 15
10 + y + z = 15
y + z = 5 ---- (Equation 4)
Equation 3: 3x + y - z = 29
3(10) + y - z = 29
30 + y - z = 29
y - z = -1 ---- (Equation 5)
Step 3: Add equations 4 and 5 to eliminate y:
(y + z) + (y - z) = 5 + (-1)
Simplifying, we get:
2y = 4
y = 2
Step 4: Substitute the values of x and y into one of the equations to solve for z. Let's substitute x = 10 and y = 2 into equation 4:
y + z = 5
2 + z = 5
z = 3
Therefore, the solution to the system of equations is x = 10, y = 2, and z = 3.
Option (a) x = 10, y = 2, z = 3 is the correct solution.