Certainly! Let's solve the given system of linear equations step-by-step:
The system of equations is:
[tex]\[
\left\{
\begin{array}{l}
3x + y = 6 \\
x - 3y = -1
\end{array}
\right.
\][/tex]
### Step 1: Solve one equation for one variable
Let's solve the second equation for [tex]\(x\)[/tex]:
[tex]\[
x - 3y = -1
\][/tex]
Add [tex]\(3y\)[/tex] to both sides:
[tex]\[
x = 3y - 1
\][/tex]
### Step 2: Substitute this expression into the other equation
Now substitute [tex]\(x = 3y - 1\)[/tex] into the first equation:
[tex]\[
3(3y - 1) + y = 6
\][/tex]
Distribute the 3:
[tex]\[
9y - 3 + y = 6
\][/tex]
Combine like terms:
[tex]\[
10y - 3 = 6
\][/tex]
Add 3 to both sides:
[tex]\[
10y = 9
\][/tex]
Divide by 10:
[tex]\[
y = \frac{9}{10}
\][/tex]
### Step 3: Substitute the value of [tex]\(y\)[/tex] back into the expression for [tex]\(x\)[/tex]
Now that we have [tex]\(y\)[/tex], substitute [tex]\(y = \frac{9}{10}\)[/tex] back into [tex]\(x = 3y - 1\)[/tex]:
[tex]\[
x = 3\left(\frac{9}{10}\right) - 1
\][/tex]
Multiply:
[tex]\[
x = \frac{27}{10} - 1
\][/tex]
Convert 1 to a fraction with the same denominator:
[tex]\[
x = \frac{27}{10} - \frac{10}{10}
\][/tex]
Subtract the fractions:
[tex]\[
x = \frac{17}{10}
\][/tex]
### Step 4: State the solution as an ordered pair
The solution to the system of equations is:
[tex]\[
\left( x, y \right) = \left( \frac{17}{10}, \frac{9}{10} \right)
\][/tex]
Therefore, the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations are [tex]\(x = \frac{17}{10}\)[/tex] and [tex]\(y = \frac{9}{10}\)[/tex].
