To solve the system of linear equations:

\[
\begin{aligned}
3y & = -x + 11 \quad \text{(Equation 1)} \\
x + 4y & = 14 \quad \text{(Equation 2)}
\end{aligned}
\]

we can use either substitution or elimination. Let's use substitution since Equation 1 is already solved for \( x \) in terms of \( y \).

### Step 1: Solve Equation 1 for \( x \)
First, rewrite Equation 1 to express \( x \) in terms of \( y \):
\[
3y = -x + 11
\]
\[
x = 11 - 3y
\]

### Step 2: Substitute into Equation 2
Now substitute \( x = 11 - 3y \) into Equation 2:
\[
(11 - 3y) + 4y = 14
\]
Simplify and solve for \( y \):
\[
11 - 3y + 4y = 14
\]
\[
11 + y = 14
\]
\[
y = 14 - 11
\]
\[
y = 3
\]

### Step 3: Solve for \( x \)
Now that we have \( y = 3 \), substitute it back into the expression for \( x \):
\[
x = 11 - 3(3)
\]
\[
x = 11 - 9
\]
\[
x = 2
\]

### Solution:
The solution to the system is \( x = 2 \) and \( y = 3 \). Therefore, the solution is the point \( (2, 3) \).

Would you like any further explanations or related questions to practice?

Here are some related questions to explore further:

1. How can you verify that the solution is correct?
2. What is the graphical interpretation of the solution?
3. How would you solve the system using the elimination method?
4. What would happen if both equations were multiplied by a constant?
5. Can you write the solution as a parameterization?

**Tip:** Always substitute the solution back into the original equations to verify its correctness.

Learn how to solve the system of linear equations 3y = -x + 11 and x + 4y = 14 step-by-step. This problem involves the substitution method and covers key algebraic concepts, including the solution of linear equations. Suitable for students in Grades 7-9.

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