To solve the system of linear equations using the substitution method, let's follow these steps:

System of Equations:

1. $y = 4x - 3$
2. $y = 3x - 1$

Step 1: Set the equations equal to each other

Since both equations are already solved for $y$, we can set them equal to each other: $4x - 3 = 3x - 1$

Step 2: Solve for $x$

Now, solve for $x$: $4x - 3 = 3x - 1$ Subtract $3x$ from both sides: $x - 3 = -1$ Add 3 to both sides: $x = 2$

Step 3: Substitute $x = 2$ back into one of the original equations

Substitute $x = 2$ into the first equation $y = 4x - 3$: $y = 4(2) - 3$ $y = 8 - 3$ $y = 5$

So the solution to the system is $x = 2$ and $y = 5$.

Step 4: Check the solution

Substitute $x = 2$ and $y = 5$ into the second equation $y = 3x - 1$: $5 = 3(2) - 1$ $5 = 6 - 1$ $5 = 5$

Since both equations are satisfied, the solution set $(2, 5)$ is correct.

Solution Set:

$x = 2$, $y = 5$

Would you like more details or have any further questions?

Here are 5 related questions:

1. What is the graphical interpretation of this solution?
2. Can you solve the system using the elimination method?
3. What happens if we change one of the equations slightly? How would the solution change?
4. How can we verify solutions for larger systems of equations?
5. What if the two lines were parallel? How would we handle that case?

Tip: In systems of equations, using substitution works best when one equation is already solved for one variable, like in this case!