Algebra practice problems with handwritten solutions on a printed worksheet.

To solve the given system of equations, we will follow a step-by-step approach for each problem. Let's go through them one by one.

---

Problem 1:

Solve the system:
\[
\begin{cases}
x + y = 5 \\
2x - y = 4
\end{cases}
\]

#### Step 1: Add the two equations to eliminate \( y \).
\[
(x + y) + (2x - y) = 5 + 4
\]
\[
x + y + 2x - y = 9
\]
\[
3x = 9
\]

#### Step 2: Solve for \( x \).
\[
x = \frac{9}{3} = 3
\]

#### Step 3: Substitute \( x = 3 \) into the first equation to solve for \( y \).
\[
x + y = 5
\]
\[
3 + y = 5
\]
\[
y = 5 - 3
\]
\[
y = 2
\]

#### Final Answer:
\[
\boxed{(3, 2)}
\]

---

Problem 2:

Solve the system:
\[
\begin{cases}
3x + 2y = 12 \\
x - y = 1
\end{cases}
\]

#### Step 1: Solve the second equation for \( x \).
\[
x - y = 1
\]
\[
x = y + 1
\]

#### Step 2: Substitute \( x = y + 1 \) into the first equation.
\[
3x + 2y = 12
\]
\[
3(y + 1) + 2y = 12
\]
\[
3y + 3 + 2y = 12
\]
\[
5y + 3 = 12
\]

#### Step 3: Solve for \( y \).
\[
5y = 12 - 3
\]
\[
5y = 9
\]
\[
y = \frac{9}{5}
\]

#### Step 4: Substitute \( y = \frac{9}{5} \) back into \( x = y + 1 \).
\[
x = \frac{9}{5} + 1
\]
\[
x = \frac{9}{5} + \frac{5}{5}
\]
\[
x = \frac{14}{5}
\]

#### Final Answer:
\[
\boxed{\left( \frac{14}{5}, \frac{9}{5} \right)}
\]

---

Problem 3:

Solve the system:
\[
\begin{cases}
4x - 3y = 10 \\
2x + y = 5
\end{cases}
\]

#### Step 1: Multiply the second equation by 3 to align the coefficients of \( y \).
\[
2x + y = 5
\]
Multiply by 3:
\[
6x + 3y = 15
\]

#### Step 2: Add the modified second equation to the first equation to eliminate \( y \).
\[
(4x - 3y) + (6x + 3y) = 10 + 15
\]
\[
4x - 3y + 6x + 3y = 25
\]
\[
10x = 25
\]

#### Step 3: Solve for \( x \).
\[
x = \frac{25}{10} = \frac{5}{2}
\]

#### Step 4: Substitute \( x = \frac{5}{2} \) into the second equation to solve for \( y \).
\[
2x + y = 5
\]
\[
2\left(\frac{5}{2}\right) + y = 5
\]
\[
5 + y = 5
\]
\[
y = 5 - 5
\]
\[
y = 0
\]

#### Final Answer:
\[
\boxed{\left( \frac{5}{2}, 0 \right)}
\]

---

Problem 4:

Solve the system:
\[
\begin{cases}
5x + 2y = 17 \\
3x - 4y = -1
\end{cases}
\]

#### Step 1: Multiply the first equation by 2 to align the coefficients of \( y \).
\[
5x + 2y = 17
\]
Multiply by 2:
\[
10x + 4y = 34
\]

#### Step 2: Add the modified first equation to the second equation to eliminate \( y \).
\[
(10x + 4y) + (3x - 4y) = 34 + (-1)
\]
\[
10x + 4y + 3x - 4y = 33
\]
\[
13x = 33
\]

#### Step 3: Solve for \( x \).
\[
x = \frac{33}{13}
\]

#### Step 4: Substitute \( x = \frac{33}{13} \) into the first equation to solve for \( y \).
\[
5x + 2y = 17
\]
\[
5\left(\frac{33}{13}\right) + 2y = 17
\]
\[
\frac{165}{13} + 2y = 17
\]
Convert 17 to a fraction with denominator 13:
\[
\frac{165}{13} + 2y = \frac{221}{13}
\]
Subtract \(\frac{165}{13}\) from both sides:
\[
2y = \frac{221}{13} - \frac{165}{13}
\]
\[
2y = \frac{56}{13}
\]
Solve for \( y \):
\[
y = \frac{56}{13} \cdot \frac{1}{2}
\]
\[
y = \frac{28}{13}
\]

#### Final Answer:
\[
\boxed{\left( \frac{33}{13}, \frac{28}{13} \right)}
\]

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Final Answers:

1. \(\boxed{(3, 2)}\)
2. \(\boxed{\left( \frac{14}{5}, \frac{9}{5} \right)}\)
3. \(\boxed{\left( \frac{5}{2}, 0 \right)}\)
4. \(\boxed{\left( \frac{33}{13}, \frac{28}{13} \right)}\)