1. Solve.
2x+3y+4z=2
5x-2y+3z=0
x-5y-2z=-4
A.
(3, 4, 2)
B.
(-2 ,1, 2)
C.
(2, 2, -2)
D.
(1, 1, -2)
The school subject is math, not college.
You could either substitute the 4 sets of values into one equation or start solving:
1. 2x+3y+4z=2
2. 5x-2y+3z=0
3. x-5y-2z=-4
Equ 1 x 2
4x+6y+8z=4
Equ 2 x 3
15x-6y+9z=0
add 19x+17z=4
from by inspection z must be opposite sign to x and x equals z so B) and C) are possible answers.
You could either continue solving the equations or substitute the two sets in say equation 1.
To solve the given system of equations:
Step 1: Write the system of equations in matrix form. The matrix form of the given system is:
[ 2 3 4 | 2 ]
[ 5 -2 3 | 0 ]
[ 1 -5 -2 | -4 ]
Step 2: Use row reduction techniques to transform the augmented matrix into row echelon form or reduced row echelon form.
Performing row operations to simplify the matrix, we get:
[ 1 -5 -2 | -4 ]
[ 0 17 11 | 22 ]
[ 0 0 39 | 78 ]
Step 3: Solve the simplified matrix for the values of x, y, and z.
From the last row of the simplified matrix, we have 39z = 78, which implies z = 2.
Substituting z = 2 in the second row, we get 17y + 11(2) = 22. Simplifying, we have 17y + 22 = 22, and thus 17y = 0. This implies y = 0.
Again substituting z = 2 and y = 0, we can solve the first row equation 1x - 5(0) - 2(2) = -4. Simplifying, we get x - 4 = -4, and hence x = 0.
So, the solution to the given system of equations is (x, y, z) = (0, 0, 2).
Therefore, the correct answer is:
C. (0, 0, 2)