https://file.aiquickdraw.com/mathbot/file/17261108755310lg0yiae.jpg

The image shows a system of linear equations:

\[
\begin{align*}
1. & \quad 2x + 3y = 2 \\
2. & \quad x - y = 1
\end{align*}
\]

Let's solve this system using the substitution or elimination method. I'll use substitution first:

From the second equation:

\[
x - y = 1 \implies x = y + 1
\]

Now, substitute \(x = y + 1\) into the first equation:

\[
2(y + 1) + 3y = 2
\]

Simplify and solve for \(y\):

\[
2y + 2 + 3y = 2 \implies 5y + 2 = 2 \implies 5y = 0 \implies y = 0
\]

Now, substitute \(y = 0\) back into \(x = y + 1\):

\[
x = 0 + 1 = 1
\]

So, the solution to the system is:

\[
x = 1, \quad y = 0
\]

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

Here are some related questions for further exploration:
1. How do you solve a system of equations using the elimination method?
2. What happens when a system of linear equations has no solution?
3. How can you verify the solution to a system of linear equations?
4. Can systems of linear equations have infinitely many solutions? If so, how?
5. How do graphing methods compare with algebraic methods for solving systems of equations?

**Tip:** Always double-check your solution by substituting it back into the original equations to ensure both equations are satisfied.

Solve the system of linear equations: 2x + 3y = 2 and x - y = 1.

Learn how to solve the system of linear equations 2x + 3y = 2 and x - y = 1 using the substitution method. Step-by-step breakdown included, suitable for students in Grades 7-9.

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