We will solve the given system of linear equations using the elimination method. Here are the equations:
[tex]\[
\left\{\begin{array}{l}
3x + y = 6 \\
4x - 3y = -5
\end{array}\right.
\][/tex]
### Step 1: Eliminate one variable
To eliminate one of the variables, we can manipulate the equations such that the coefficients of one of the variables become opposites. In this case, we can eliminate [tex]\(y\)[/tex] by making the coefficients of [tex]\(y\)[/tex] in the two equations opposites.
To do this, we will multiply the first equation by 3:
[tex]\[
3(3x + y) = 3 \cdot 6 \\
9x + 3y = 18
\][/tex]
Now we have:
[tex]\[
\left\{\begin{array}{l}
9x + 3y = 18 \\
4x - 3y = -5
\end{array}\right.
\][/tex]
### Step 2: Add the equations
Next, we add the two equations together to eliminate [tex]\(y\)[/tex]:
[tex]\[
(9x + 3y) + (4x - 3y) = 18 + (-5) \\
9x + 3y + 4x - 3y = 18 - 5 \\
13x = 13
\][/tex]
### Step 3: Solve for [tex]\(x\)[/tex]
Now solve for [tex]\(x\)[/tex]:
[tex]\[
13x = 13 \\
x = \frac{13}{13} \\
x = 1
\][/tex]
### Step 4: Substitute [tex]\(x\)[/tex] back into one of the equations
We now substitute [tex]\(x = 1\)[/tex] back into one of the original equations to solve for [tex]\(y\)[/tex]. We'll use the first equation:
[tex]\[
3x + y = 6 \\
3(1) + y = 6 \\
3 + y = 6 \\
y = 6 - 3 \\
y = 3
\][/tex]
### Conclusion
The solution to the system of equations is
[tex]\[
(x, y) = (1, 3)
\][/tex]
This means [tex]\(x = 1\)[/tex] and [tex]\(y = 3\)[/tex]. The elimination method works by removing one variable so that we can solve for the other variable directly, then substituting back to find the remaining variable. In this case, our solution is:
[tex]\[
\boxed{(1, 3)}
\][/tex]
