To solve the system of equations using elimination, we start with:

1. \( 2x = y + 3 \)
2. \( 5 - 3x = -y \)

Let's rewrite each equation in standard form, with \( x \) and \( y \) terms on one side and constants on the other.

### Step 1: Rewrite equations in standard form
From the first equation:
\[
2x - y = 3
\]

From the second equation:
\[
-3x + y = -5
\]

Now, our system is:
1. \( 2x - y = 3 \)
2. \( -3x + y = -5 \)

### Step 2: Add the two equations to eliminate \( y \)
Adding the equations:
\[
(2x - y) + (-3x + y) = 3 + (-5)
\]

This simplifies to:
\[
- x = -2
\]

Solving for \( x \):
\[
x = 2
\]

### Step 3: Substitute \( x = 2 \) into one of the original equations
Substitute \( x = 2 \) into the first equation:
\[
2(2) = y + 3
\]
\[
4 = y + 3
\]
\[
y = 1
\]

### Solution
The solution to the system is:
\[
x = 2, \quad y = 1
\]

### Verification
Substitute \( x = 2 \) and \( y = 1 \) back into the original equations to verify:

1. \( 2(2) = 1 + 3 \Rightarrow 4 = 4 \) (True)
2. \( 5 - 3(2) = -1 \Rightarrow -1 = -1 \) (True)

Both equations are satisfied, so the solution is correct:
\[
(x, y) = (2, 1)
\]

Would you like additional details on any step or have any questions?

---

1. What is elimination in solving systems of equations?
2. How does elimination differ from substitution?
3. Can we use elimination when neither variable is easily eliminated?
4. What methods can verify solutions to systems of equations?
5. How does elimination handle more complex systems?

**Tip:** For elimination, choose the variable that cancels out easily with minimal manipulation.

solve using elimination 2x = y + 3 and 5 - 3x = -y

Learn how to solve a system of linear equations, 2x = y + 3 and 5 - 3x = -y, using the elimination method. This step-by-step solution covers transforming equations to standard form and eliminating variables, ideal for students in Grades 8-10.

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