This worksheet helps students practice solving systems of linear equations algebraically through two sections of simultaneous equation problems.

Let's solve each system of linear equations algebraically, using either the substitution method or the elimination method. I'll go step-by-step for each problem.

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✔ Section A

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#### 1)
$$
\begin{align*}
(1)\quad 4x + y &= 17 \\
(2)\quad 2x + y &= 9
\end{align*}
$$

Step 1: Subtract equation (2) from equation (1):

$$
(4x + y) - (2x + y) = 17 - 9 \\
4x + y - 2x - y = 8 \\
2x = 8 \Rightarrow x = 4
$$

Step 2: Plug $ x = 4 $ into equation (2):

$$
2(4) + y = 9 \Rightarrow 8 + y = 9 \Rightarrow y = 1
$$

✔ Answer: $ x = 4, y = 1 $

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#### 2)
$$
\begin{align*}
(1)\quad 2x + y &= 7 \\
(2)\quad 5x - y &= 14
\end{align*}
$$

Step 1: Add equations (1) and (2):

$$
(2x + y) + (5x - y) = 7 + 14 \\
7x = 21 \Rightarrow x = 3
$$

Step 2: Plug $ x = 3 $ into (1):

$$
2(3) + y = 7 \Rightarrow 6 + y = 7 \Rightarrow y = 1
$$

✔ Answer: $ x = 3, y = 1 $

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#### 3)
$$
\begin{align*}
(1)\quad 3x + 2y &= 19 \\
(2)\quad 2x - 2y &= 6
\end{align*}
$$

Step 1: Add equations (1) and (2):

$$
(3x + 2y) + (2x - 2y) = 19 + 6 \\
5x = 25 \Rightarrow x = 5
$$

Step 2: Plug $ x = 5 $ into (2):

$$
2(5) - 2y = 6 \Rightarrow 10 - 2y = 6 \Rightarrow -2y = -4 \Rightarrow y = 2
$$

✔ Answer: $ x = 5, y = 2 $

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#### 4)
$$
\begin{align*}
(1)\quad 3x - 4y &= 17 \\
(2)\quad x - 4y &= 3
\end{align*}
$$

Step 1: Subtract (2) from (1):

$$
(3x - 4y) - (x - 4y) = 17 - 3 \\
3x - 4y - x + 4y = 14 \\
2x = 14 \Rightarrow x = 7
$$

Step 2: Plug $ x = 7 $ into (2):

$$
7 - 4y = 3 \Rightarrow -4y = -4 \Rightarrow y = 1
$$

✔ Answer: $ x = 7, y = 1 $

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#### 5)
$$
\begin{align*}
(1)\quad 2x + 5y &= 24 \\
(2)\quad 2x + 3y &= 16
\end{align*}
$$

Step 1: Subtract (2) from (1):

$$
(2x + 5y) - (2x + 3y) = 24 - 16 \\
2y = 8 \Rightarrow y = 4
$$

Step 2: Plug $ y = 4 $ into (2):

$$
2x + 3(4) = 16 \Rightarrow 2x + 12 = 16 \Rightarrow 2x = 4 \Rightarrow x = 2
$$

✔ Answer: $ x = 2, y = 4 $

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#### 6)
$$
\begin{align*}
(1)\quad 4x - 3y &= 7 \\
(2)\quad x + 3y &= 13
\end{align*}
$$

Step 1: Add equations (1) and (2):

$$
(4x - 3y) + (x + 3y) = 7 + 13 \\
5x = 20 \Rightarrow x = 4
$$

Step 2: Plug $ x = 4 $ into (2):

$$
4 + 3y = 13 \Rightarrow 3y = 9 \Rightarrow y = 3
$$

✔ Answer: $ x = 4, y = 3 $

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✔ Section B

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#### 1)
$$
\begin{align*}
(1)\quad 2x + y &= 4 \\
(2)\quad 5x + 4y &= 7
\end{align*}
$$

Step 1: Solve (1) for $ y $:

$$
y = 4 - 2x
$$

Step 2: Substitute into (2):

$$
5x + 4(4 - 2x) = 7 \\
5x + 16 - 8x = 7 \\
-3x + 16 = 7 \Rightarrow -3x = -9 \Rightarrow x = 3
$$

Step 3: Plug back:

$$
y = 4 - 2(3) = 4 - 6 = -2
$$

✔ Answer: $ x = 3, y = -2 $

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#### 2)
$$
\begin{align*}
(1)\quad 14x + 2y &= 8 \\
(2)\quad x + y &= 1
\end{align*}
$$

Step 1: Solve (2) for $ y $: $ y = 1 - x $

Step 2: Plug into (1):

$$
14x + 2(1 - x) = 8 \\
14x + 2 - 2x = 8 \\
12x + 2 = 8 \Rightarrow 12x = 6 \Rightarrow x = 0.5
$$

Step 3: $ y = 1 - 0.5 = 0.5 $

✔ Answer: $ x = 0.5, y = 0.5 $ or $ x = \frac{1}{2}, y = \frac{1}{2} $

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#### 3)
$$
\begin{align*}
(1)\quad 7x + 8y &= 3 \\
(2)\quad 3x + 4y &= 7
\end{align*}
$$

Step 1: Multiply (2) by 2 to match coefficients of $ y $:

$$
2 \times (3x + 4y) = 2 \times 7 \Rightarrow 6x + 8y = 14
$$

Now subtract from (1):

$$
(7x + 8y) - (6x + 8y) = 3 - 14 \\
x = -11
$$

Step 2: Plug $ x = -11 $ into (2):

$$
3(-11) + 4y = 7 \Rightarrow -33 + 4y = 7 \Rightarrow 4y = 40 \Rightarrow y = 10
$$

✔ Answer: $ x = -11, y = 10 $

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#### 4)
$$
\begin{align*}
(1)\quad 3x + 4y &= 29 \\
(2)\quad 4x - 2y &= 2
\end{align*}
$$

Step 1: Multiply (2) by 2 to eliminate $ y $:

$$
2 \times (4x - 2y) = 2 \times 2 \Rightarrow 8x - 4y = 4
$$

Now add to (1):

$$
(3x + 4y) + (8x - 4y) = 29 + 4 \\
11x = 33 \Rightarrow x = 3
$$

Step 2: Plug $ x = 3 $ into (2):

$$
4(3) - 2y = 2 \Rightarrow 12 - 2y = 2 \Rightarrow -2y = -10 \Rightarrow y = 5
$$

✔ Answer: $ x = 3, y = 5 $

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#### 5)
$$
\begin{align*}
(1)\quad 6x - 5y &= -1 \\
(2)\quad 3x - y &= -2
\end{align*}
$$

Step 1: Solve (2) for $ y $: $ y = 3x + 2 $

Step 2: Plug into (1):

$$
6x - 5(3x + 2) = -1 \\
6x - 15x - 10 = -1 \\
-9x - 10 = -1 \Rightarrow -9x = 9 \Rightarrow x = -1
$$

Step 3: $ y = 3(-1) + 2 = -3 + 2 = -1 $

✔ Answer: $ x = -1, y = -1 $

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#### 6)
$$
\begin{align*}
(1)\quad 3x - 2y &= 6 \\
(2)\quad 5x + 6y &= 38
\end{align*}
$$

Step 1: Multiply (1) by 3 to match $ y $-coefficients:

$$
3 \times (3x - 2y) = 3 \times 6 \Rightarrow 9x - 6y = 18
$$

Now add to (2):

$$
(9x - 6y) + (5x + 6y) = 18 + 38 \\
14x = 56 \Rightarrow x = 4
$$

Step 2: Plug $ x = 4 $ into (1):

$$
3(4) - 2y = 6 \Rightarrow 12 - 2y = 6 \Rightarrow -2y = -6 \Rightarrow y = 3
$$

✔ Answer: $ x = 4, y = 3 $

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✔ Final Answers Summary

#### Section A
1. $ x = 4, y = 1 $
2. $ x = 3, y = 1 $
3. $ x = 5, y = 2 $
4. $ x = 7, y = 1 $
5. $ x = 2, y = 4 $
6. $ x = 4, y = 3 $

#### Section B
1. $ x = 3, y = -2 $
2. $ x = \frac{1}{2}, y = \frac{1}{2} $
3. $ x = -11, y = 10 $
4. $ x = 3, y = 5 $
5. $ x = -1, y = -1 $
6. $ x = 4, y = 3 $

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Let me know if you'd like these checked with substitution or graphing!