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To solve the system of equations by substitution:

1. **Equations**:  
   \[
   3x + y = 5 \tag{1}
   \]  
   \[
   5x - 4y = -3 \tag{2}
   \]

---

### Step 1: Solve for \( y \) in terms of \( x \) from Equation (1):
\[
3x + y = 5
\]
\[
y = 5 - 3x \tag{3}
\]

---

### Step 2: Substitute \( y \) into Equation (2):
Replace \( y \) with \( 5 - 3x \) in Equation (2):

\[
5x - 4y = -3
\]
\[
5x - 4(5 - 3x) = -3
\]

Simplify:
\[
5x - 20 + 12x = -3
\]
Combine like terms:
\[
17x - 20 = -3
\]
Add 20 to both sides:
\[
17x = 17
\]
Solve for \( x \):
\[
x = 1
\]

---

### Step 3: Solve for \( y \) using \( x = 1 \) in Equation (3):
\[
y = 5 - 3x
\]
\[
y = 5 - 3(1)
\]
\[
y = 5 - 3 = 2
\]

---

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

---

### Verification:
1. Substitute \( x = 1 \) and \( y = 2 \) into the original equations:
   - \( 3x + y = 3(1) + 2 = 5 \) ✅
   - \( 5x - 4y = 5(1) - 4(2) = 5 - 8 = -3 \) ✅

Thus, the solution is correct.

---

Let me know if you'd like further clarification!

### Related Questions:
1. How can the substitution method be used for larger systems of equations?
2. What are the differences between substitution and elimination methods?
3. How do you check if a solution satisfies both equations in a system?
4. How does substitution work when variables have fractions?
5. Can a system of equations have no solution or infinitely many solutions? How?

**Tip**: Always verify your solution by substituting it back into the original equations.

Solve the system of equations by substitution:
1. 3x + y = 5
2. 5x - 4y = -3

Learn how to solve the system of equations 3x + y = 5 and 5x - 4y = -3 using the substitution method. Follow a step-by-step approach to solve for x and y, verify the solution, and understand key algebraic concepts for Grades 8-10.

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