To solve the given system of linear equations, we can use the method of elimination to find the values of [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex].
The given system of equations is:
[tex]\[
\begin{cases}
x - 5y + 4z = 27 \\
4x - 3y - z = 23 \\
3x + 3y - 6z = -9
\end{cases}
\][/tex]
Step 1: Eliminate [tex]\(x\)[/tex] from equations 2 and 3
We start by eliminating [tex]\(x\)[/tex] from equations (2) and (3) using equation (1).
Step 1a: Multiply equation (1) to match the coefficients of [tex]\(x\)[/tex]:
Equation (1) can be multiplied by 4 to help eliminate [tex]\(x\)[/tex] from equation (2):
[tex]\[ 4(x - 5y + 4z) = 4(27) \][/tex]
[tex]\[ 4x - 20y + 16z = 108 \][/tex] (Equation 4)
Subtract equation (4) from equation (2):
[tex]\[ (4x - 3y - z) - (4x - 20y + 16z) = 23 - 108 \][/tex]
[tex]\[ 4x - 3y - z - 4x + 20y - 16z = -85 \][/tex]
[tex]\[ 17y - 17z = -85 \][/tex]
[tex]\[ y - z = -5 \][/tex] (Equation 5)
Step 1b: Multiply equation (1) to match the coefficients of [tex]\(x\)[/tex]:
Equation (1) can be multiplied by 3 to help eliminate [tex]\(x\)[/tex] from equation (3):
[tex]\[ 3(x - 5y + 4z) = 3(27) \][/tex]
[tex]\[ 3x - 15y + 12z = 81 \][/tex] (Equation 6)
Subtract equation (6) from equation (3):
[tex]\[ (3x + 3y - 6z) - (3x - 15y + 12z) = -9 - 81 \][/tex]
[tex]\[ 3x + 3y - 6z - 3x + 15y - 12z = -90 \][/tex]
[tex]\[ 18y - 18z = -90 \][/tex]
[tex]\[ y - z = -5 \][/tex] (Equation 7)
Step 2: Solve for [tex]\(y\)[/tex] and [tex]\(z\)[/tex]
The resultant equation from Step 1 is [tex]\(y - z = -5\)[/tex]. Both are equivalent, stating:
[tex]\[ y - z = -5 \][/tex]
[tex]\[ y = z - 5 \][/tex] (Equation 8)
Step 3: Substitute [tex]\(y = z - 5\)[/tex] back into one of the original equations
Substitute [tex]\(y = z - 5\)[/tex] into Equation (1):
[tex]\[ x - 5(z - 5) + 4z = 27 \][/tex]
[tex]\[ x - 5z + 25 + 4z = 27 \][/tex]
[tex]\[ x - z + 25 = 27 \][/tex]
[tex]\[ x - z = 2 \][/tex]
[tex]\[ x = z + 2 \][/tex] (Equation 9)
Step 4: Substitute [tex]\(x\)[/tex] and [tex]\(y\)[/tex] in another original equation to solve for [tex]\(z\)[/tex]
Substitute [tex]\(x = z + 2\)[/tex] and [tex]\(y = z - 5\)[/tex] into Equation (2):
[tex]\[ 4(z + 2) - 3(z - 5) - z = 23 \][/tex]
[tex]\[ 4z + 8 - 3z + 15 - z = 23 \][/tex]
[tex]\[ 4z - 3z - z + 8 + 15 = 23 \][/tex]
[tex]\[ 0 = 0 \][/tex]
Since this equation is always true, we have infinitely many solutions.
The general solution for the system in terms of [tex]\(z\)[/tex] is:
[tex]\[
\begin{cases}
x = z + 2 \\
y = z - 5 \\
z = z
\end{cases}
\][/tex]
Therefore, the solution set for the system of equations [tex]\( x - 5y + 4z = 27\)[/tex], [tex]\(4x - 3y - z = 23\)[/tex], and [tex]\(3x + 3y - 6z = -9 \)[/tex] is:
[tex]\[
(x, y, z) = (z + 2, z - 5, z)
\][/tex]
where [tex]\(z\)[/tex] can be any real number.
