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You provided a system of equations:

\[
\begin{aligned}
2x + 3y &= 16 \\
6x - 5y &= 20
\end{aligned}
\]

### Solving the system:
We can solve this system using either substitution or elimination. Let's use the **elimination method** to eliminate one variable and solve for the other.

1. **Multiply the first equation by 3** to align the coefficients of \(x\) in both equations:
   \[
   6x + 9y = 48
   \]

   Now, the system is:
   \[
   \begin{aligned}
   6x + 9y &= 48 \\
   6x - 5y &= 20
   \end{aligned}
   \]

2. **Subtract the second equation from the first**:
   \[
   (6x + 9y) - (6x - 5y) = 48 - 20
   \]
   Simplifying:
   \[
   14y = 28
   \]
   Therefore:
   \[
   y = \frac{28}{14} = 2
   \]

3. **Substitute \(y = 2\) into the first equation** to solve for \(x\):
   \[
   2x + 3(2) = 16
   \]
   Simplifying:
   \[
   2x + 6 = 16
   \]
   \[
   2x = 10 \quad \Rightarrow \quad x = \frac{10}{2} = 5
   \]

### Final solution:
\[
x = 5, \quad y = 2
\]

Would you like any further details or explanation on the steps?

---

Here are 5 related questions to deepen your understanding:
1. How would the solution change if one of the equations was nonlinear?
2. What other methods can be used to solve a system of equations?
3. Can a system of linear equations have no solution? Why?
4. How would you solve a system with three variables and three equations?
5. What is the graphical interpretation of solving two linear equations?

**Tip:** When solving by elimination, ensure the coefficients of one variable are aligned before subtracting or adding equations.

Solve the system of equations: 2x + 3y = 16 and 6x - 5y = 20.

Learn how to solve a system of linear equations using the elimination method. This example includes the steps of aligning coefficients, subtracting equations, and solving for both variables. Perfect for students in Grades 8-10 learning algebra and linear equations.

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