Question

1. solve the system of equations using substitution.

$x = 5$
$3x - 3y = 21$
$x + 2y + 3z = 7$
options:
(5, -2, 2)
(5, 13, -12)
(5, -12, 13)
(5, 2, -2)

Explanation:

Step1: Substitute \( x = 5 \) into the second equation

The second equation is \( 3x - 3y = 21 \). Substitute \( x = 5 \):
\( 3(5) - 3y = 21 \)
\( 15 - 3y = 21 \)

Step2: Solve for \( y \)

Subtract 15 from both sides:
\( -3y = 21 - 15 \)
\( -3y = 6 \)
Divide by -3:
\( y = \frac{6}{-3} = -2 \)

Step3: Substitute \( x = 5 \) and \( y = -2 \) into the third equation

The third equation is \( x + 2y + 3z = 7 \). Substitute \( x = 5 \), \( y = -2 \):
\( 5 + 2(-2) + 3z = 7 \)
\( 5 - 4 + 3z = 7 \)
\( 1 + 3z = 7 \)

Step4: Solve for \( z \)

Subtract 1 from both sides:
\( 3z = 7 - 1 \)
\( 3z = 6 \)
Divide by 3:
\( z = \frac{6}{3} = 2 \)

Answer:

\((5, -2, 2)\) (matching the first option in the choices: \((5, -2, 2)\))