To solve the system of equations:

\[
x - 7y = -37 \tag{1}
\]
\[
4x - 3y = 7 \tag{2}
\]

We can use either the substitution method or the elimination method. Here, we'll use the substitution method.

### Step 1: Solve Equation (1) for \( x \)
From equation (1):

\[
x = 7y - 37 \tag{3}
\]

### Step 2: Substitute Equation (3) into Equation (2)
Substitute \( x = 7y - 37 \) into equation (2):

\[
4(7y - 37) - 3y = 7
\]

Simplify:

\[
28y - 148 - 3y = 7
\]

\[
25y - 148 = 7
\]

### Step 3: Solve for \( y \)
Add 148 to both sides:

\[
25y = 155
\]

Now, divide by 25:

\[
y = \frac{155}{25} = 6.2
\]

### Step 4: Substitute \( y = 6.2 \) back into Equation (3)
Substitute \( y = 6.2 \) into equation (3) to find \( x \):

\[
x = 7(6.2) - 37 = 43.4 - 37 = 6.4
\]

### Solution
The solution to the system is:

\[
x = 6.4, \quad y = 6.2
\]

Would you like more details on this process or have any questions?

Here are 5 related questions to extend your understanding:
1. How can you check if a solution satisfies both equations?
2. What is the elimination method, and how does it differ from substitution?
3. How would you solve the same system graphically?
4. Can this system of equations have a different type of solution (e.g., no solution or infinitely many solutions)?
5. How would you solve the system if the coefficients of \( x \) and \( y \) were different?

**Tip:** Always check your solution by substituting the values of \( x \) and \( y \) back into the original equations to verify they hold true.

Learn how to solve the system of linear equations x - 7y = -37 and 4x - 3y = 7 using the substitution method. This detailed solution is suitable for students in Grades 8-10, covering core algebra concepts like solving for variables and checking solutions.

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