Solve the following system of equations 4x - 2y - z = -5, x - 3y + 2z = 3, 3x + y- 2z = -5.

Solution:

We will use the substitution method to find the values of x, y and z.

Let the equations 4x - 2y - z = -5 --- be (1)

⇒ x - 3y + 2z = 3 --- be (2)

⇒ 3x + y- 2z = -5 --- be (3)

By solving equation [1] for the variable z, we get

⇒ z = 4x - 2y + 5 --- be eq (4)

Substitute the value of z = 4x - 2y + 5 in equation [2]

⇒ x - 3y + 2 × (4x - 2y + 5 ) = 3

 ⇒ x - y = -1 --- be eq (5)

Substitute the value of z = 4x - 2y + 5 in equation [3]

⇒ 3x + y - 2 × (4x - 2y + 5) = -5

⇒ x + y = 1 --- be eq (6)

By adding equation [5] and [6] for the value of x, we get

⇒ x = 0

Substitute the value of x = 0 in equation [5]

⇒ x - y = -1

⇒ 0 - y = -1

⇒ y = 1

Substitute the values of x and y in equation [1], we get

⇒ 4(0) - 2(1) - z = -5

⇒ 0 - 2 - z = -5

⇒ -z = -5 + 2

⇒ -z = -3

⇒ z = 3

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Solve the following system of equations 4x - 2y - z = -5, x - 3y + 2z = 3, 3x + y- 2z = -5.

Summary:

By solving the system of linear equations 4x - 2y - z = -5, x - 3y + 2z = 3, 3x + y- 2z = -5.; we get (x, y, z) = (0, 1, 3).