Systems Of Equations Solving Two Variable Systems Of Equations By ... - Free Printable

To solve the given systems of linear equations, we will use algebraic methods such as substitution or elimination. Let's solve each system step by step.

---

Section A

#### 1)
\[
\begin{cases}
4x + y = 17 \\
2x + y = 9
\end{cases}
\]

Step 1: Subtract the second equation from the first to eliminate \( y \):
\[
(4x + y) - (2x + y) = 17 - 9
\]
\[
4x + y - 2x - y = 8
\]
\[
2x = 8
\]
\[
x = 4
\]

Step 2: Substitute \( x = 4 \) into the second equation:
\[
2(4) + y = 9
\]
\[
8 + y = 9
\]
\[
y = 1
\]

Solution:
\[
x = 4, \quad y = 1
\]

---

#### 2)
\[
\begin{cases}
2x + y = 7 \\
5x - y = 14
\end{cases}
\]

Step 1: Add the two equations to eliminate \( y \):
\[
(2x + y) + (5x - y) = 7 + 14
\]
\[
2x + y + 5x - y = 21
\]
\[
7x = 21
\]
\[
x = 3
\]

Step 2: Substitute \( x = 3 \) into the first equation:
\[
2(3) + y = 7
\]
\[
6 + y = 7
\]
\[
y = 1
\]

Solution:
\[
x = 3, \quad y = 1
\]

---

#### 3)
\[
\begin{cases}
3x + 2y = 19 \\
2x - 2y = 6
\end{cases}
\]

Step 1: Add the two equations to eliminate \( y \):
\[
(3x + 2y) + (2x - 2y) = 19 + 6
\]
\[
3x + 2y + 2x - 2y = 25
\]
\[
5x = 25
\]
\[
x = 5
\]

Step 2: Substitute \( x = 5 \) into the first equation:
\[
3(5) + 2y = 19
\]
\[
15 + 2y = 19
\]
\[
2y = 4
\]
\[
y = 2
\]

Solution:
\[
x = 5, \quad y = 2
\]

---

#### 4)
\[
\begin{cases}
3x - 4y = 17 \\
x - 4y = 3
\end{cases}
\]

Step 1: Subtract the second equation from the first to eliminate \( y \):
\[
(3x - 4y) - (x - 4y) = 17 - 3
\]
\[
3x - 4y - x + 4y = 14
\]
\[
2x = 14
\]
\[
x = 7
\]

Step 2: Substitute \( x = 7 \) into the second equation:
\[
7 - 4y = 3
\]
\[
-4y = 3 - 7
\]
\[
-4y = -4
\]
\[
y = 1
\]

Solution:
\[
x = 7, \quad y = 1
\]

---

#### 5)
\[
\begin{cases}
2x + 5y = 24 \\
2x + 3y = 16
\end{cases}
\]

Step 1: Subtract the second equation from the first to eliminate \( x \):
\[
(2x + 5y) - (2x + 3y) = 24 - 16
\]
\[
2x + 5y - 2x - 3y = 8
\]
\[
2y = 8
\]
\[
y = 4
\]

Step 2: Substitute \( y = 4 \) into the first equation:
\[
2x + 5(4) = 24
\]
\[
2x + 20 = 24
\]
\[
2x = 4
\]
\[
x = 2
\]

Solution:
\[
x = 2, \quad y = 4
\]

---

#### 6)
\[
\begin{cases}
4x - 3y = 7 \\
x + 3y = 13
\end{cases}
\]

Step 1: Add the two equations to eliminate \( y \):
\[
(4x - 3y) + (x + 3y) = 7 + 13
\]
\[
4x - 3y + x + 3y = 20
\]
\[
5x = 20
\]
\[
x = 4
\]

Step 2: Substitute \( x = 4 \) into the second equation:
\[
4 + 3y = 13
\]
\[
3y = 13 - 4
\]
\[
3y = 9
\]
\[
y = 3
\]

Solution:
\[
x = 4, \quad y = 3
\]

---

Section B

#### 1)
\[
\begin{cases}
2x + y = 4 \\
5x + 4y = 7
\end{cases}
\]

Step 1: Solve the first equation for \( y \):
\[
y = 4 - 2x
\]

Step 2: Substitute \( y = 4 - 2x \) into the second equation:
\[
5x + 4(4 - 2x) = 7
\]
\[
5x + 16 - 8x = 7
\]
\[
-3x + 16 = 7
\]
\[
-3x = 7 - 16
\]
\[
-3x = -9
\]
\[
x = 3
\]

Step 3: Substitute \( x = 3 \) back into \( y = 4 - 2x \):
\[
y = 4 - 2(3)
\]
\[
y = 4 - 6
\]
\[
y = -2
\]

Solution:
\[
x = 3, \quad y = -2
\]

---

#### 2)
\[
\begin{cases}
14x + 2y = 8 \\
x + y = 1
\end{cases}
\]

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

Step 2: Substitute \( y = 1 - x \) into the first equation:
\[
14x + 2(1 - x) = 8
\]
\[
14x + 2 - 2x = 8
\]
\[
12x + 2 = 8
\]
\[
12x = 6
\]
\[
x = \frac{1}{2}
\]

Step 3: Substitute \( x = \frac{1}{2} \) back into \( y = 1 - x \):
\[
y = 1 - \frac{1}{2}
\]
\[
y = \frac{1}{2}
\]

Solution:
\[
x = \frac{1}{2}, \quad y = \frac{1}{2}
\]

---

#### 3)
\[
\begin{cases}
7x + 8y = 3 \\
3x + 4y = 7
\end{cases}
\]

Step 1: Multiply the second equation by 2 to align coefficients of \( y \):
\[
2(3x + 4y) = 2(7)
\]
\[
6x + 8y = 14
\]

Step 2: Subtract the modified second equation from the first:
\[
(7x + 8y) - (6x + 8y) = 3 - 14
\]
\[
7x + 8y - 6x - 8y = -11
\]
\[
x = -11
\]

Step 3: Substitute \( x = -11 \) into the second original equation:
\[
3(-11) + 4y = 7
\]
\[
-33 + 4y = 7
\]
\[
4y = 7 + 33
\]
\[
4y = 40
\]
\[
y = 10
\]

Solution:
\[
x = -11, \quad y = 10
\]

---

#### 4)
\[
\begin{cases}
3x + 4y = 29 \\
4x - 2y = 2
\end{cases}
\]

Step 1: Multiply the second equation by 2 to align coefficients of \( y \):
\[
2(4x - 2y) = 2(2)
\]
\[
8x - 4y = 4
\]

Step 2: Add the modified second equation to the first:
\[
(3x + 4y) + (8x - 4y) = 29 + 4
\]
\[
3x + 4y + 8x - 4y = 33
\]
\[
11x = 33
\]
\[
x = 3
\]

Step 3: Substitute \( x = 3 \) into the first original equation:
\[
3(3) + 4y = 29
\]
\[
9 + 4y = 29
\]
\[
4y = 20
\]
\[
y = 5
\]

Solution:
\[
x = 3, \quad y = 5
\]

---

#### 5)
\[
\begin{cases}
6x - 5y = -1 \\
3x - y = -2
\end{cases}
\]

Step 1: Multiply the second equation by 5 to align coefficients of \( y \):
\[
5(3x - y) = 5(-2)
\]
\[
15x - 5y = -10
\]

Step 2: Subtract the modified second equation from the first:
\[
(6x - 5y) - (15x - 5y) = -1 - (-10)
\]
\[
6x - 5y - 15x + 5y = -1 + 10
\]
\[
-9x = 9
\]
\[
x = -1
\]

Step 3: Substitute \( x = -1 \) into the second original equation:
\[
3(-1) - y = -2
\]
\[
-3 - y = -2
\]
\[
-y = -2 + 3
\]
\[
-y = 1
\]
\[
y = -1
\]

Solution:
\[
x = -1, \quad y = -1
\]

---

#### 6)
\[
\begin{cases}
3x - 2y = 6 \\
5x + 6y = 38
\end{cases}
\]

Step 1: Multiply the first equation by 3 to align coefficients of \( y \):
\[
3(3x - 2y) = 3(6)
\]
\[
9x - 6y = 18
\]

Step 2: Add the modified first equation to the second:
\[
(9x - 6y) + (5x + 6y) = 18 + 38
\]
\[
9x - 6y + 5x + 6y = 56
\]
\[
14x = 56
\]
\[
x = 4
\]

Step 3: Substitute \( x = 4 \) into the first original equation:
\[
3(4) - 2y = 6
\]
\[
12 - 2y = 6
\]
\[
-2y = 6 - 12
\]
\[
-2y = -6
\]
\[
y = 3
\]

Solution:
\[
x = 4, \quad y = 3
\]

---

Final Answers:

\[
\boxed{
\begin{array}{ll}
\text{Section A:} & \text{Section B:} \\
1) \, x = 4, \, y = 1 & 1) \, x = 3, \, y = -2 \\
2) \, x = 3, \, y = 1 & 2) \, x = \frac{1}{2}, \, y = \frac{1}{2} \\
3) \, x = 5, \, y = 2 & 3) \, x = -11, \, y = 10 \\
4) \, x = 7, \, y = 1 & 4) \, x = 3, \, y = 5 \\
5) \, x = 2, \, y = 4 & 5) \, x = -1, \, y = -1 \\
6) \, x = 4, \, y = 3 & 6) \, x = 4, \, y = 3 \\
\end{array}
}
\]