Question

solve each system of 3 equations

1. ( x + y + z = 0 )

( 2x + y - 2z = -8 )
( -x + 0y + 4z = 10 )

1. ( x + 2y + z = 7 )

( x - 2y - 4z = 0 )
( 2x - y + 4z = -3 )

1. ( 2x + 2y + z = 10 )

( x - 2y + 3z = 13 )
( x - y + 3z = 12 )

Explanation:

Problem 6

Step1: Solve the third equation for \( x \)

The third equation is \( -x + 4z = 10 \). We can rewrite it as \( x = 4z - 10 \).

Step2: Substitute \( x = 4z - 10 \) into the first and second equations

• Substitute into the first equation \( x + y + z = 0 \):

\( (4z - 10) + y + z = 0 \)
Simplify: \( 5z + y - 10 = 0 \) ⇒ \( y = 10 - 5z \)

• Substitute into the second equation \( 2x + y - 2z = -8 \):

\( 2(4z - 10) + y - 2z = -8 \)
Simplify: \( 8z - 20 + y - 2z = -8 \) ⇒ \( 6z + y - 20 = -8 \) ⇒ \( 6z + y = 12 \)

Step3: Substitute \( y = 10 - 5z \) into \( 6z + y = 12 \)

\( 6z + (10 - 5z) = 12 \)
Simplify: \( z + 10 = 12 \) ⇒ \( z = 2 \)

Step4: Find \( x \) and \( y \) using \( z = 2 \)

• Find \( x \): \( x = 4(2) - 10 = 8 - 10 = -2 \)
• Find \( y \): \( y = 10 - 5(2) = 10 - 10 = 0 \)

Step1: Add the first and second equations to eliminate \( y \)

First equation: \( x + 2y + z = 7 \)
Second equation: \( x - 2y - 4z = 0 \)
Adding them: \( (x + x) + (2y - 2y) + (z - 4z) = 7 + 0 \)
Simplify: \( 2x - 3z = 7 \) (Equation 4)

Step2: Multiply the second equation by \( \frac{1}{2} \) and add to the third equation to eliminate \( y \)

Second equation multiplied by \( \frac{1}{2} \): \( \frac{1}{2}x - y - 2z = 0 \)
Third equation: \( 2x - y + 4z = -3 \)
Subtract the modified second equation from the third equation: \( (2x - \frac{1}{2}x) + (-y + y) + (4z + 2z) = -3 - 0 \)
Simplify: \( \frac{3}{2}x + 6z = -3 \)
Multiply through by 2: \( 3x + 12z = -6 \) ⇒ \( x + 4z = -2 \) (Equation 5)

Step3: Solve Equation 4 and Equation 5 simultaneously

Equation 4: \( 2x - 3z = 7 \)
Equation 5: \( x + 4z = -2 \) ⇒ \( x = -2 - 4z \)
Substitute \( x = -2 - 4z \) into Equation 4:
\( 2(-2 - 4z) - 3z = 7 \)
Simplify: \( -4 - 8z - 3z = 7 \) ⇒ \( -4 - 11z = 7 \) ⇒ \( -11z = 11 \) ⇒ \( z = -1 \)

Step4: Find \( x \) and \( y \) using \( z = -1 \)

• Find \( x \): \( x = -2 - 4(-1) = -2 + 4 = 2 \)
• Substitute \( x = 2 \) and \( z = -1 \) into the first equation \( x + 2y + z = 7 \):

\( 2 + 2y - 1 = 7 \) ⇒ \( 1 + 2y = 7 \) ⇒ \( 2y = 6 \) ⇒ \( y = 3 \)

Step1: Subtract the second equation from the third equation to eliminate \( x \) and \( z \)

Third equation: \( x - y + 3z = 12 \)
Second equation: \( x - 2y + 3z = 13 \)
Subtract: \( (x - x) + (-y + 2y) + (3z - 3z) = 12 - 13 \)
Simplify: \( y = -1 \)

Step2: Substitute \( y = -1 \) into the first and second equations

• First equation: \( 2x + 2(-1) + z = 10 \) ⇒ \( 2x + z - 2 = 10 \) ⇒ \( 2x + z = 12 \) (Equation 6)
• Second equation: \( x - 2(-1) + 3z = 13 \) ⇒ \( x + 2 + 3z = 13 \) ⇒ \( x + 3z = 11 \) (Equation 7)

Step3: Solve Equation 7 for \( x \) and substitute into Equation 6

From Equation 7: \( x = 11 - 3z \)
Substitute into Equation 6: \( 2(11 - 3z) + z = 12 \)
Simplify: \( 22 - 6z + z = 12 \) ⇒ \( 22 - 5z = 12 \) ⇒ \( -5z = -10 \) ⇒ \( z = 2 \)

Step4: Find \( x \) using \( z = 2 \)

\( x = 11 - 3(2) = 11 - 6 = 5 \)

Answer:

\( x = -2 \), \( y = 0 \), \( z = 2 \)

Problem 7