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Let's solve the system of equations using the substitution method:

Given the system:
\[
x = 4y - 3
\]
\[
x = 2y - 1
\]

### Step 1: Set the equations equal to each other
Since both expressions are equal to \( x \), we can set them equal to each other:
\[
4y - 3 = 2y - 1
\]

### Step 2: Solve for \( y \)
Move the terms involving \( y \) to one side:
\[
4y - 2y = -1 + 3
\]
\[
2y = 2
\]
\[
y = 1
\]

### Step 3: Substitute \( y = 1 \) into one of the original equations to find \( x \)
Using \( x = 4y - 3 \):
\[
x = 4(1) - 3 = 4 - 3 = 1
\]

### Solution:
The solution is \( (1, 1) \).

Would you like more details or have any questions?

Here are five related questions to deepen understanding:
1. What happens if both equations have no solution?
2. Can the substitution method always be applied to any system of equations?
3. How does the solution change if the system represents parallel lines?
4. What is the graphical interpretation of the solution to this system?
5. How does the substitution method differ from the elimination method?

**Tip:** Always simplify the equations as much as possible before starting the substitution process to avoid unnecessary complications.

Solve the system of equations by the substitution method. x = 4y - 3 and x = 2y - 1

Learn how to solve the system of equations x = 4y - 3 and x = 2y - 1 using the substitution method. This algebraic method helps find the intersection of two linear equations. Suitable for students in Grades 8-10.

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