Certainly! Let's solve the given system of linear equations step by step:
The system of equations is:
[tex]\[
\begin{cases}
x + y + z = 7 \\
x - 2y + 4z = 11 \\
3y + 4z = 17
\end{cases}
\][/tex]
First, we'll work on solving the third equation for [tex]\( y \)[/tex] in terms of [tex]\( z \)[/tex]:
[tex]\[
3y + 4z = 17 \implies y = \frac{17 - 4z}{3}
\][/tex]
Next, let's substitute [tex]\( y = \frac{17 - 4z}{3} \)[/tex] into the first and second equations.
Substituting into the first equation:
[tex]\[
x + \left(\frac{17 - 4z}{3}\right) + z = 7
\][/tex]
First, multiply everything by 3 to clear the fraction:
[tex]\[
3x + (17 - 4z) + 3z = 21 \implies 3x + 17 - z = 21 \implies 3x - z = 4 \implies x = \frac{4 + z}{3}
\][/tex]
Next, substitute [tex]\( y = \frac{17 - 4z}{3} \)[/tex] into the second equation:
[tex]\[
x - 2\left(\frac{17 - 4z}{3}\right) + 4z = 11
\][/tex]
Multiply everything by 3 to clear the fraction:
[tex]\[
3x - 2(17 - 4z) + 12z = 33 \implies 3x - 34 + 8z + 12z = 33 \implies 3x + 8z = 45 \implies x + 8z = 15 \implies x = 15 - 8z
\][/tex]
Now, equate the two different expressions we have for [tex]\( x \)[/tex]:
[tex]\[
\frac{4 + z}{3} = 15 - 8z
\][/tex]
Multiply through by 3:
[tex]\[
4 + z = 45 - 24z
\][/tex]
Collect all [tex]\( z \)[/tex]-terms on one side:
[tex]\[
4 + z + 24z = 45 \implies 25z = 41 \implies z = \frac{41}{25} \implies z = 3
\][/tex]
With [tex]\( z = 3 \)[/tex], substitute back to find [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[
x = \frac{4 + z}{3} = \frac{4 + 3}{3} = \frac{7}{3}
\][/tex]
[tex]\[
y = \frac{17 - 4z}{3} = \frac{17 - 4(3)}{3} = \frac{17 - 12}{3} = \frac{5}{3}
\][/tex]
So the solution to the system is:
[tex]\[
(x, y, z) = \left(\frac{7}{3}, \frac{5}{3}, 3\right)
\][/tex]
