Question

use back-substitution to solve the system of linear equations.\

$$\begin{cases}4x - 3y - 2z = 48\\\\6y - 5z = -21\\\\z = -3\\end{cases}$$

\\(x, y, z) = \left(\boxed{\quad}\
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Explanation:

Step1: Substitute \( z = -3 \) into the second equation

We have the second equation \( 6y - 5z = -21 \). Substitute \( z = -3 \) into it:
\( 6y - 5(-3) = -21 \)
Simplify the left - hand side: \( 6y+15=-21 \)

Step2: Solve for \( y \)

Subtract 15 from both sides of the equation \( 6y + 15=-21 \):
\( 6y=-21 - 15 \)
\( 6y=-36 \)
Divide both sides by 6: \( y=\frac{-36}{6}=-6 \)

Step3: Substitute \( y = - 6 \) and \( z=-3 \) into the first equation

The first equation is \( 4x-3y - 2z = 48 \). Substitute \( y=-6 \) and \( z = - 3 \):
\( 4x-3(-6)-2(-3)=48 \)
Simplify the left - hand side: \( 4x + 18+6=48 \)
\( 4x+24 = 48 \)

Step4: Solve for \( x \)

Subtract 24 from both sides of the equation \( 4x + 24=48 \):
\( 4x=48 - 24 \)
\( 4x=24 \)
Divide both sides by 4: \( x=\frac{24}{4}=6 \)

Answer:

\((6, - 6, - 3)\)