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Let's solve the given system of equations step by step.

The system of equations is:

\[
\begin{aligned}
1) \ 2x + 3y &= 7 \\
2) \ -2x + 4y &= 14
\end{aligned}
\]

### Step 1: Add the equations
To eliminate \(x\), add both equations:

\[
(2x + 3y) + (-2x + 4y) = 7 + 14
\]

This simplifies to:

\[
0x + 7y = 21 \quad \Rightarrow \quad 7y = 21
\]

### Step 2: Solve for \(y\)
Divide both sides by 7:

\[
y = \frac{21}{7} = 3
\]

### Step 3: Substitute \(y = 3\) into one of the original equations
Now substitute \(y = 3\) into the first equation \(2x + 3y = 7\):

\[
2x + 3(3) = 7 \quad \Rightarrow \quad 2x + 9 = 7
\]

### Step 4: Solve for \(x\)
Subtract 9 from both sides:

\[
2x = 7 - 9 \quad \Rightarrow \quad 2x = -2
\]

Divide by 2:

\[
x = \frac{-2}{2} = -1
\]

### Solution
The solution to the system is:

\[
x = -1, \quad y = 3
\]

### Verification
Let's substitute \(x = -1\) and \(y = 3\) into the second equation to check:

\[
-2(-1) + 4(3) = 2 + 12 = 14
\]

This is true, so the solution is correct.

---

Would you like further details on this solution? Here are some related questions for practice:

1. What is the elimination method and when is it useful?
2. How would the solution differ if both equations were dependent?
3. What if one of the equations was non-linear—how would you approach it?
4. Could substitution work here, and how would you apply it?
5. How do you check if a system of equations has no solution or infinite solutions?

**Tip:** Always verify your solutions by substituting them back into the original equations to avoid mistakes.

Solve each system of equations without graphing and show your reasoning. The system is: 2x + 3y = 7 and -2x + 4y = 14.

Learn how to solve the system of equations 2x + 3y = 7 and -2x + 4y = 14 using the elimination method. This problem demonstrates key algebraic concepts like adding equations to eliminate variables, substitution, and solving linear equations. Suitable for students in Grades 8-10.

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