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To solve the given system of equations:

\[
\begin{aligned}
5x - 2y &= 20 \\
7x + 4y &= 11
\end{aligned}
\]

### Step 1: Eliminate one variable
We will use the elimination method. To eliminate \(y\), we make the coefficients of \(y\) in both equations equal. Multiply the first equation by 2 and the second equation by 1 (to match the \(y\)-coefficients):

\[
\begin{aligned}
2(5x - 2y) &= 2(20) \implies 10x - 4y = 40, \\
1(7x + 4y) &= 1(11) \implies 7x + 4y = 11.
\end{aligned}
\]

### Step 2: Add the two equations
Adding the equations eliminates \(y\):

\[
(10x - 4y) + (7x + 4y) = 40 + 11 \implies 17x = 51.
\]

Solve for \(x\):

\[
x = \frac{51}{17} = 3.
\]

### Step 3: Substitute \(x = 3\) into one of the original equations
Use the first equation \(5x - 2y = 20\):

\[
5(3) - 2y = 20 \implies 15 - 2y = 20 \implies -2y = 5 \implies y = -\frac{5}{2}.
\]

### Final Answer:
\[
x = 3, \quad y = -\frac{5}{2}.
\]

Would you like me to explain any step in further detail or solve it using a different method?

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Here are 5 related questions to deepen your understanding:
1. How can the substitution method be used to solve this system of equations?
2. What happens if you eliminate \(x\) instead of \(y\)?
3. Can a system of equations have no solution or infinitely many solutions? How do we identify such cases?
4. How would graphing these two equations look, and where would the solution lie?
5. What are other examples of real-life problems modeled by linear systems?

**Tip:** Always double-check your solution by substituting both \(x\) and \(y\) back into the original equations.

Solve the system of equations: 5x - 2y = 20 and 7x + 4y = 11.

Learn how to solve the system of equations 5x - 2y = 20 and 7x + 4y = 11 using the elimination method. This solution shows step-by-step instructions and provides insights into algebraic concepts like variable elimination and solving linear systems. Suitable for students in Grades 7-9.

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