You have a system of two linear equations:

\[
\begin{aligned}
1. & \quad 2x + 3y = 7 \\
2. & \quad 4x - y = 5
\end{aligned}
\]

Let's solve this system using the substitution or elimination method. I'll use the substitution method here:

### Step 1: Solve the second equation for \( y \)
Starting with the second equation:

\[
4x - y = 5
\]

Solve for \( y \):

\[
y = 4x - 5
\]

### Step 2: Substitute \( y = 4x - 5 \) into the first equation
Substitute this expression for \( y \) into the first equation:

\[
2x + 3(4x - 5) = 7
\]

### Step 3: Solve for \( x \)
Expand and simplify:

\[
2x + 12x - 15 = 7
\]

Combine like terms:

\[
14x - 15 = 7
\]

Add 15 to both sides:

\[
14x = 22
\]

Divide by 14:

\[
x = \frac{22}{14} = \frac{11}{7}
\]

### Step 4: Substitute \( x = \frac{11}{7} \) back into \( y = 4x - 5 \)
Now, substitute \( x = \frac{11}{7} \) back into the equation for \( y \):

\[
y = 4\left(\frac{11}{7}\right) - 5 = \frac{44}{7} - \frac{35}{7} = \frac{9}{7}
\]

### Final Solution
The solution to the system of equations is:

\[
x = \frac{11}{7}, \quad y = \frac{9}{7}
\]

Would you like a detailed explanation of any step, or do you have any questions?

Here are 5 related questions to expand on this topic:

1. How would the solution change if the second equation were \( 4x + y = 5 \) instead?
2. Can you solve the system using the elimination method and compare the steps?
3. What does the solution \( x = \frac{11}{7}, y = \frac{9}{7} \) represent graphically?
4. How would you solve the system if both equations were multiplied by different constants?
5. Can you solve a system of three equations with three variables using a similar method?

**Tip:** When solving systems of equations, it’s helpful to check your solution by substituting the values of \( x \) and \( y \) back into the original equations to ensure both are satisfied.

Learn how to solve a system of linear equations 2x + 3y = 7 and 4x - y = 5 using the substitution method. Detailed step-by-step solution provided. Suitable for high school students in Grades 9-12.

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