Worksheet for solving systems of linear equations using the substitution method.

Let's solve each system of linear equations using the substitution method, as instructed.

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Problem 1:

$$
\begin{align*}
(1)\quad & -2x + y = 5 \\
(2)\quad & 6x - 3y = -9
\end{align*}
$$

Step 1: Solve one equation for one variable.
From (1):
$$
-2x + y = 5 \Rightarrow y = 2x + 5
$$

Step 2: Substitute into the other equation.
Substitute $ y = 2x + 5 $ into (2):
$$
6x - 3(2x + 5) = -9 \\
6x - 6x - 15 = -9 \\
-15 = -9 \quad \text{✘ Contradiction!}
$$

Wait — that can't be right. Let's double-check:
$$
6x - 3(2x + 5) = -9 \\
6x - 6x - 15 = -9 \\
0x - 15 = -9 \\
-15 = -9 \quad \text{False}
$$

This means no solution — the lines are parallel.

But let’s check if we made a mistake in substitution.

Wait — actually, let's try solving it differently.

Alternatively, maybe simplify the second equation:

Equation (2): $ 6x - 3y = -9 $

Divide entire equation by 3:
$$
2x - y = -3
$$

Now we have:
$$
(1)\quad -2x + y = 5 \\
(2)\quad 2x - y = -3
$$

Add both equations:
$$
(-2x + y) + (2x - y) = 5 + (-3) \\
0 = 2 \quad \text{✘ Contradiction}
$$

So again, no solution → inconsistent system.

✔ Answer: No solution (parallel lines)

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Problem 2:

$$
\begin{align*}
(1)\quad & y = x + 4 \\
(2)\quad & 3x - 4y = -19
\end{align*}
$$

Step 1: Since $ y = x + 4 $, substitute into (2):

$$
3x - 4(x + 4) = -19 \\
3x - 4x - 16 = -19 \\
-x - 16 = -19 \\
-x = -3 \Rightarrow x = 3
$$

Now plug back into (1):
$ y = 3 + 4 = 7 $

✔ Answer: $ x = 3, y = 7 $

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Problem 3:

$$
\begin{align*}
(1)\quad & -4x + y = 6 \\
(2)\quad & -5x - y = 21
\end{align*}
$$

Solve (1) for $ y $:
$$
y = 4x + 6
$$

Substitute into (2):
$$
-5x - (4x + 6) = 21 \\
-5x - 4x - 6 = 21 \\
-9x - 6 = 21 \\
-9x = 27 \Rightarrow x = -3
$$

Now find $ y $:
$ y = 4(-3) + 6 = -12 + 6 = -6 $

✔ Answer: $ x = -3, y = -6 $

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Problem 4:

$$
\begin{align*}
(1)\quad & 6x + 4y = -8 \\
(2)\quad & -6x + y = 28
\end{align*}
$$

Solve (2) for $ y $:
$$
y = 6x + 28
$$

Substitute into (1):
$$
6x + 4(6x + 28) = -8 \\
6x + 24x + 112 = -8 \\
30x + 112 = -8 \\
30x = -120 \Rightarrow x = -4
$$

Now find $ y $:
$ y = 6(-4) + 28 = -24 + 28 = 4 $

✔ Answer: $ x = -4, y = 4 $

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Problem 5:

$$
\begin{align*}
(1)\quad & x = -2y \\
(2)\quad & x - 9 = 9
\end{align*}
$$

From (2):
$ x = 9 + 9 = 18 $

Now use (1):
$ 18 = -2y \Rightarrow y = -9 $

✔ Answer: $ x = 18, y = -9 $

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Problem 6:

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

Solve (1) for $ x $:
$ x = 3 - y $

Substitute into (2):
$$
(3 - y) - y = -3 \\
3 - 2y = -3 \\
-2y = -6 \Rightarrow y = 3
$$

Then $ x = 3 - 3 = 0 $

✔ Answer: $ x = 0, y = 3 $

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Problem 7:

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

Substitute (1) into (2):
$$
-2(y - 4) + 3y = 6 \\
-2y + 8 + 3y = 6 \\
y + 8 = 6 \Rightarrow y = -2
$$

Now $ x = -2 - 4 = -6 $

✔ Answer: $ x = -6, y = -2 $

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Problem 8:

$$
\begin{align*}
(1)\quad & -5x + 20 = 5y \\
(2)\quad & -2x + y = 7
\end{align*}
$$

First, simplify (1):
Divide both sides by 5:
$$
- x + 4 = y \Rightarrow y = -x + 4
$$

Now substitute into (2):
$$
-2x + (-x + 4) = 7 \\
-2x - x + 4 = 7 \\
-3x + 4 = 7 \\
-3x = 3 \Rightarrow x = -1
$$

Now find $ y $:
$ y = -(-1) + 4 = 1 + 4 = 5 $

✔ Answer: $ x = -1, y = 5 $

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✔ Final Answers:

| Problem | Solution |
|--------|----------|
| 1 | No solution (inconsistent) |
| 2 | $ x = 3, y = 7 $ |
| 3 | $ x = -3, y = -6 $ |
| 4 | $ x = -4, y = 4 $ |
| 5 | $ x = 18, y = -9 $ |
| 6 | $ x = 0, y = 3 $ |
| 7 | $ x = -6, y = -2 $ |
| 8 | $ x = -1, y = 5 $ |

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