Question

Solve the following system of equations.

4x – 2y – z = –5
x – 3y + 2z = 3
3x + y – 2z = –5


(0, 1, –3)
(0, 1, 3)
(0, –1, –3)
(0, –1, 3)


Solve the following system of equations.

4x – 2y – z = –5
x – 3y + 2z = 3
3x + y – 2z = –5


(0, 1, –3)
(0, 1, 3)
(0, –1, –3)
(0, –1, 3)

@Mathematics

Answer 1

Still 0,1,3

Answer 2

You see that you can trial and error here, right?

Answer 3

By substitution we get
-10y + 9z = 17
By the options it is clear that 9z is the largest number. As the result is a positive number, we can conclude that 9z should be positive and -10y should be negative. So, z should be positive and y should also be positive to make -10y positive. So the only option is 0,1,3, because it only has y and z positive.

Answer 4

Sorry, " So, z should be positive and y should also be positive to make -10y negative"