Certainly! Let's solve the given system of linear equations:
[tex]\[
\left\{
\begin{array}{l}
x + y + z = 3 \quad \quad (1)\\
2x - y + 3z = 5 \quad \quad (2)
\end{array}
\right.
\][/tex]
We'll use the substitution or elimination method to find the values of the variables [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex].
### Step 1: Solve for one variable
First, let's solve equation (1) for [tex]\( x \)[/tex]:
[tex]\[
x = 3 - y - z \quad \quad (3)
\][/tex]
### Step 2: Substitute the expression into the second equation
Next, we substitute the expression for [tex]\( x \)[/tex] from equation (3) into equation (2):
[tex]\[
2(3 - y - z) - y + 3z = 5
\][/tex]
### Step 3: Distribute and simplify
Now, distribute and simplify the equation:
[tex]\[
6 - 2y - 2z - y + 3z = 5
\][/tex]
Combine like terms:
[tex]\[
6 - 3y + z = 5
\][/tex]
Isolate the term with [tex]\(y\)[/tex] on one side:
[tex]\[
6 - 5 = 3y - z
\][/tex]
Thus,
[tex]\[
1 = 3y - z \quad \quad (4)
\][/tex]
### Step 4: Solve for [tex]\( y \)[/tex] in terms of [tex]\( z \)[/tex]
From equation (4), solve for [tex]\( y \)[/tex]:
[tex]\[
3y = z + 1
\][/tex]
Divide by 3:
[tex]\[
y = \frac{z + 1}{3} \quad \quad (5)
\][/tex]
### Step 5: Substitute [tex]\( y \)[/tex] back into equation (3)
Now, substitute the expression for [tex]\( y \)[/tex] from equation (5) back into equation (3):
[tex]\[
x = 3 - \frac{z + 1}{3} - z
\][/tex]
### Step 6: Simplify this expression
Simplify the equation:
[tex]\[
x = 3 - \frac{z + 1}{3} - z
\][/tex]
Find a common denominator for the fraction:
[tex]\[
x = 3 - \frac{z + 1}{3} - z = 3 - \frac{z + 1}{3} - z = 3 - \frac{z + 1 + 3z}{3} = 3 - \frac{4z + 1}{3}
\][/tex]
Simplify further:
[tex]\[
x = 3 - \left( \frac{4z + 1}{3} \right)
\][/tex]
[tex]\[
x = \frac{9}{3} - \frac{4z + 1}{3}
\][/tex]
Combine the fractions:
[tex]\[
x = \frac{9 - 4z - 1}{3} = \frac{8 - 4z}{3}
\][/tex]
Thus,
[tex]\[
x = \frac{8}{3} - \frac{4z}{3}
\][/tex]
### Final expressions for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]
Thus, we have the solutions for [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in terms of [tex]\( z \)[/tex]:
[tex]\[
x = \frac{8}{3} - \frac{4z}{3}
\][/tex]
[tex]\[
y = \frac{z + 1}{3}
\][/tex]
So, the solution to the system of equations is:
[tex]\[
\boxed{x = \frac{8}{3} - \frac{4z}{3}, \; y = \frac{z + 1}{3}}
\][/tex]
Since the system has three variables but only two equations, [tex]\( z \)[/tex] is a free parameter and can take any real number value. Hence, [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are expressed in terms of [tex]\( z \)[/tex].
