Simultaneous equations | Maths Teaching - Free Printable

To solve the given simultaneous equations, we will use methods such as substitution, elimination, or a combination of both. Let's solve each pair step by step.

---

1)

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

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

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

#### Solution:
\[
(x, y) = (4, 1)
\]

---

2)

\[
\begin{cases}
5x + 2y = 13 \\
x + 2y = 9
\end{cases}
\]

#### Step 1: Subtract the second equation from the first.
\[
(5x + 2y) - (x + 2y) = 13 - 9
\]
\[
5x + 2y - x - 2y = 4
\]
\[
4x = 4
\]
\[
x = 1
\]

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

#### Solution:
\[
(x, y) = (1, 4)
\]

---

3)

\[
\begin{cases}
3x + 2y = 11 \\
2x - 2y = 14
\end{cases}
\]

#### Step 1: Add the two equations.
\[
(3x + 2y) + (2x - 2y) = 11 + 14
\]
\[
3x + 2y + 2x - 2y = 25
\]
\[
5x = 25
\]
\[
x = 5
\]

#### Step 2: Substitute \( x = 5 \) into the first equation.
\[
3x + 2y = 11
\]
\[
3(5) + 2y = 11
\]
\[
15 + 2y = 11
\]
\[
2y = -4
\]
\[
y = -2
\]

#### Solution:
\[
(x, y) = (5, -2)
\]

---

4)

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

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

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

#### Solution:
\[
(x, y) = (7, 1)
\]

---

5)

\[
\begin{cases}
2x + 5y = 37 \\
y = 11 - 2x
\end{cases}
\]

#### Step 1: Substitute \( y = 11 - 2x \) into the first equation.
\[
2x + 5(11 - 2x) = 37
\]
\[
2x + 55 - 10x = 37
\]
\[
-8x + 55 = 37
\]
\[
-8x = 37 - 55
\]
\[
-8x = -18
\]
\[
x = \frac{18}{8} = \frac{9}{4}
\]

#### Step 2: Substitute \( x = \frac{9}{4} \) into \( y = 11 - 2x \).
\[
y = 11 - 2\left(\frac{9}{4}\right)
\]
\[
y = 11 - \frac{18}{4}
\]
\[
y = 11 - \frac{9}{2}
\]
\[
y = \frac{22}{2} - \frac{9}{2}
\]
\[
y = \frac{13}{2}
\]

#### Solution:
\[
(x, y) = \left(\frac{9}{4}, \frac{13}{2}\right)
\]

---

6)

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

#### Step 1: Substitute \( x = 13 - 3y \) into the first equation.
\[
4(13 - 3y) - 3y = 7
\]
\[
52 - 12y - 3y = 7
\]
\[
52 - 15y = 7
\]
\[
-15y = 7 - 52
\]
\[
-15y = -45
\]
\[
y = 3
\]

#### Step 2: Substitute \( y = 3 \) into \( x = 13 - 3y \).
\[
x = 13 - 3(3)
\]
\[
x = 13 - 9
\]
\[
x = 4
\]

#### Solution:
\[
(x, y) = (4, 3)
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
1) & \ (4, 1) \\
2) & \ (1, 4) \\
3) & \ (5, -2) \\
4) & \ (7, 1) \\
5) & \ \left(\frac{9}{4}, \frac{13}{2}\right) \\
6) & \ (4, 3)
\end{aligned}
}
\]