https://file.aiquickdraw.com/mathbot/file/17267471879742dz0zsjc.jpg

The system of equations given is:

\[
y = 6x
\]
\[
7x - 8y = -41
\]

We are to solve this system using substitution. Let's go through the steps:

1. From the first equation, we already have \( y = 6x \).
2. Now, substitute \( y = 6x \) into the second equation \( 7x - 8y = -41 \):

\[
7x - 8(6x) = -41
\]

Simplify:

\[
7x - 48x = -41
\]

\[
-41x = -41
\]

Now, solve for \( x \):

\[
x = \frac{-41}{-41} = 1
\]

3. Now that we know \( x = 1 \), substitute this value back into the first equation \( y = 6x \):

\[
y = 6(1) = 6
\]

Thus, the solution to the system is:

\[
x = 1, \quad y = 6
\]

Would you like more details or clarification on any step?

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Here are 5 related questions for further practice:
1. How would you solve the system \( y = 4x \) and \( 5x + 3y = 19 \) by substitution?
2. If you change \( y = 6x \) to \( y = 5x \), how would the solution change?
3. Can you solve a system where both equations are in terms of \( y \)?
4. What happens when substitution results in a contradiction (e.g., \( 0 = 5 \))?
5. Can you solve the system \( y = 2x + 1 \) and \( 3x + y = 7 \) using substitution?

**Tip:** In substitution, always aim to express one variable in terms of the other from the simpler equation for easier substitution!

Solve the system by substitution. y = 6x, 7x - 8y = -41

This page provides a step-by-step solution for solving the system of linear equations y = 6x and 7x - 8y = -41 using the substitution method. Ideal for students in Grades 8-10, this problem covers essential algebra concepts.

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