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We are given the following simultaneous equations:

\[
3x + 4y = 7 \tag{1}
\]
\[
2x - 6y = -17 \tag{2}
\]

Let's solve these step by step using substitution or elimination. I'll use the **elimination method**.

### Step 1: Make the coefficients of \(x\) or \(y\) equal
We'll eliminate \(x\) by finding a common multiple of the coefficients of \(x\) in both equations. The least common multiple of 3 and 2 is 6, so we multiply each equation by the necessary factors to get 6 as the coefficient of \(x\).

Multiply equation (1) by 2:
\[
2(3x + 4y) = 2(7) \implies 6x + 8y = 14 \tag{3}
\]

Multiply equation (2) by 3:
\[
3(2x - 6y) = 3(-17) \implies 6x - 18y = -51 \tag{4}
\]

### Step 2: Subtract the equations to eliminate \(x\)
Now subtract equation (4) from equation (3):
\[
(6x + 8y) - (6x - 18y) = 14 - (-51)
\]
This simplifies to:
\[
6x + 8y - 6x + 18y = 14 + 51
\]
\[
26y = 65
\]

### Step 3: Solve for \(y\)
\[
y = \frac{65}{26} = 2.5
\]

### Step 4: Substitute \(y = 2.5\) into one of the original equations
Substitute \(y = 2.5\) into equation (1):
\[
3x + 4(2.5) = 7
\]
\[
3x + 10 = 7
\]
\[
3x = 7 - 10 = -3
\]
\[
x = \frac{-3}{3} = -1
\]

### Final solution:
The solution to the system of equations is:
\[
x = -1, \quad y = 2.5
\]

Let me know if you want more details or have any questions.

Here are five related questions to practice:
1. What are simultaneous equations, and how are they generally solved?
2. Can you solve the system \( 4x + 3y = 11 \) and \( 2x - y = 5 \)?
3. What is the substitution method in solving simultaneous equations?
4. How can simultaneous equations be solved graphically?
5. What happens if the two simultaneous equations represent parallel lines?

**Tip**: In elimination, ensure you carefully match coefficients to avoid errors in subtraction!

Use algebra to solve the simultaneous equations: 3x + 4y = 7 and 2x - 6y = -17

Learn how to solve the simultaneous equations 3x + 4y = 7 and 2x - 6y = -17 using the elimination method. This step-by-step guide covers the key algebraic concepts and formulas required for students in Grades 9-11.

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