Find the solutions to x and y if (x+y=3) and (x-y=1)

You can solve the system of equations using the method of elimination or substitution. Here, I'll use the **elimination method**.

The system of equations is:
1. \( x + y = 3 \)
2. \( x - y = 1 \)

### Step 1: Add the two equations together
By adding the two equations, we eliminate \(y\):

\[
(x + y) + (x - y) = 3 + 1
\]

This simplifies to:

\[
2x = 4
\]

### Step 2: Solve for \(x\)
\[
x = \frac{4}{2} = 2
\]

### Step 3: Substitute \(x = 2\) into one of the original equations
Let's substitute \(x = 2\) into the first equation:

\[
2 + y = 3
\]

### Step 4: Solve for \(y\)
\[
y = 3 - 2 = 1
\]

### Solution:
The solutions are:
\[
x = 2, \quad y = 1
\]

Would you like further details or have any questions?

Here are five related questions for practice:
1. Solve the system: \(x + 2y = 5\) and \(x - 2y = 1\).
2. Find \(x\) and \(y\) for \(x + y = 7\) and \(x - y = 5\).
3. Use substitution to solve \(3x + y = 9\) and \(x - y = 2\).
4. Solve the system: \(2x + 3y = 12\) and \(4x - 3y = 6\).
5. Solve \(x + y = 4\) and \(2x - y = 3\).

**Tip:** When solving systems of equations, always check your solutions by substituting them back into the original equations.

Find the solutions to x and y if (x + y = 3) and (x - y = 1).

Learn how to solve the system of linear equations x + y = 3 and x - y = 1 using the elimination method. This problem involves key algebraic techniques such as eliminating a variable and substitution. Suitable for students in Grades 6-8.

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