solve the system 4x + y = 2 and x - y = 3

To solve the system of equations:

\[
\begin{aligned}
1. & \quad 4x + y = 2 \\
2. & \quad x - y = 3
\end{aligned}
\]

### Step 1: Solve the second equation for \( y \).
Given \( x - y = 3 \), solve for \( y \):

\[
y = x - 3
\]

### Step 2: Substitute this expression for \( y \) into the first equation.
Substitute \( y = x - 3 \) into the first equation \( 4x + y = 2 \):

\[
4x + (x - 3) = 2
\]

### Step 3: Simplify and solve for \( x \).
Combine like terms:

\[
5x - 3 = 2
\]

Add 3 to both sides:

\[
5x = 5
\]

Divide by 5:

\[
x = 1
\]

### Step 4: Substitute \( x = 1 \) back into the expression for \( y \).
Using the equation \( y = x - 3 \):

\[
y = 1 - 3 = -2
\]

### Solution:
The solution to the system is:

\[
x = 1, \quad y = -2
\]

So the solution is the point \( (1, -2) \).

Would you like more details or have any questions?

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Here are 5 related questions:
1. How can you verify the solution to a system of equations?
2. What is the graphical interpretation of the solution \( (1, -2) \)?
3. How do you solve a system of equations using the elimination method?
4. What does it mean if a system of equations has no solution?
5. How can systems of equations be used in real-world problems?

**Tip:** Always check your solution by substituting it back into the original equations to ensure it satisfies both.

Learn how to solve the system of equations 4x + y = 2 and x - y = 3 with a detailed step-by-step solution. This problem covers key algebraic concepts, including systems of linear equations and the substitution method. Suitable for students in Grades 7-9.

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