Math worksheet for solving systems of linear equations using the elimination method, suitable for level 2 algebra students.

Let's solve each system of linear equations using the elimination method. The goal is to eliminate one variable by adding or subtracting the equations (after multiplying if necessary) to find the values of $x$ and $y$.

---

Problem 1:

$$
\begin{align*}
(1)\quad -7x - 2y &= -36 \\
(2)\quad -42x + 4y &= -152
\end{align*}
$$

Step 1: Make the coefficients of one variable the same. Let’s eliminate $y$.
Multiply equation (1) by 2:
$$
2(-7x - 2y) = 2(-36) \Rightarrow -14x - 4y = -72
$$

Now add this to equation (2):
$$
\begin{align*}
-14x - 4y &= -72 \\
-42x + 4y &= -152 \\
\hline
(-14x - 42x) + (-4y + 4y) &= -72 - 152 \\
-56x &= -224
\end{align*}
$$

Solve for $x$:
$$
x = \frac{-224}{-56} = 4
$$

Substitute $x = 4$ into equation (1):
$$
-7(4) - 2y = -36 \Rightarrow -28 - 2y = -36 \\
-2y = -8 \Rightarrow y = 4
$$

✔ Solution: $(x, y) = (4, 4)$

---

Problem 2:

$$
\begin{align*}
(1)\quad -x - 6y &= 49 \\
(2)\quad -4x - 2y &= 42
\end{align*}
$$

We can eliminate one variable. Let's eliminate $x$. Multiply equation (1) by $-4$:
$$
-4(-x - 6y) = -4(49) \Rightarrow 4x + 24y = -196
$$

Now add to equation (2):
$$
\begin{align*}
4x + 24y &= -196 \\
-4x - 2y &= 42 \\
\hline
(4x - 4x) + (24y - 2y) &= -196 + 42 \\
22y &= -154 \Rightarrow y = -7
\end{align*}
$$

Substitute $y = -7$ into equation (1):
$$
-x - 6(-7) = 49 \Rightarrow -x + 42 = 49 \Rightarrow -x = 7 \Rightarrow x = -7
$$

✔ Solution: $(x, y) = (-7, -7)$

---

Problem 3:

$$
\begin{align*}
(1)\quad -4x + 2y &= -2 \\
(2)\quad 20x + 6y &= -6
\end{align*}
$$

Eliminate $y$. Multiply equation (1) by 3:
$$
3(-4x + 2y) = 3(-2) \Rightarrow -12x + 6y = -6
$$

Now subtract from equation (2):
$$
\begin{align*}
20x + 6y &= -6 \\
-12x + 6y &= -6 \\
\hline
(20x + 12x) + (6y - 6y) &= -6 + 6 \\
32x &= 0 \Rightarrow x = 0
\end{align*}
$$

Substitute $x = 0$ into equation (1):
$$
-4(0) + 2y = -2 \Rightarrow 2y = -2 \Rightarrow y = -1
$$

✔ Solution: $(x, y) = (0, -1)$

---

Problem 4:

$$
\begin{align*}
(1)\quad -6x - y &= 23 \\
(2)\quad 12x - 6y &= -102
\end{align*}
$$

Eliminate $x$. Multiply equation (1) by 2:
$$
2(-6x - y) = 2(23) \Rightarrow -12x - 2y = 46
$$

Add to equation (2):
$$
\begin{align*}
-12x - 2y &= 46 \\
12x - 6y &= -102 \\
\hline
(-12x + 12x) + (-2y - 6y) &= 46 - 102 \\
-8y &= -56 \Rightarrow y = 7
\end{align*}
$$

Substitute $y = 7$ into equation (1):
$$
-6x - 7 = 23 \Rightarrow -6x = 30 \Rightarrow x = -5
$$

✔ Solution: $(x, y) = (-5, 7)$

---

Problem 5:

$$
\begin{align*}
(1)\quad -2x + 3y &= -21 \\
(2)\quad -6x + 7y &= -49
\end{align*}
$$

Eliminate $x$. Multiply equation (1) by 3:
$$
3(-2x + 3y) = 3(-21) \Rightarrow -6x + 9y = -63
$$

Now subtract equation (2):
$$
\begin{align*}
-6x + 9y &= -63 \\
-6x + 7y &= -49 \\
\hline
(-6x + 6x) + (9y - 7y) &= -63 + 49 \\
2y &= -14 \Rightarrow y = -7
\end{align*}
$$

Substitute $y = -7$ into equation (1):
$$
-2x + 3(-7) = -21 \Rightarrow -2x - 21 = -21 \Rightarrow -2x = 0 \Rightarrow x = 0
$$

✔ Solution: $(x, y) = (0, -7)$

---

Problem 6:

$$
\begin{align*}
(1)\quad 3x + 4y &= 8 \\
(2)\quad -4x - 12y &= -44
\end{align*}
$$

Eliminate $y$. Multiply equation (1) by 3:
$$
3(3x + 4y) = 3(8) \Rightarrow 9x + 12y = 24
$$

Now add to equation (2):
$$
\begin{align*}
9x + 12y &= 24 \\
-4x - 12y &= -44 \\
\hline
(9x - 4x) + (12y - 12y) &= 24 - 44 \\
5x &= -20 \Rightarrow x = -4
\end{align*}
$$

Substitute $x = -4$ into equation (1):
$$
3(-4) + 4y = 8 \Rightarrow -12 + 4y = 8 \Rightarrow 4y = 20 \Rightarrow y = 5
$$

✔ Solution: $(x, y) = (-4, 5)$

---

Problem 7:

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

Eliminate $y$. Subtract equation (2) from equation (1):
$$
(5x - y) - (x - y) = -1 - 3 \\
5x - y - x + y = -4 \Rightarrow 4x = -4 \Rightarrow x = -1
$$

Substitute $x = -1$ into equation (2):
$$
-1 - y = 3 \Rightarrow -y = 4 \Rightarrow y = -4
$$

✔ Solution: $(x, y) = (-1, -4)$

---

Problem 8:

$$
\begin{align*}
(1)\quad -x - 3y &= -1 \\
(2)\quad -4x + 6y &= -22
\end{align*}
$$

Eliminate $x$. Multiply equation (1) by $-4$:
$$
-4(-x - 3y) = -4(-1) \Rightarrow 4x + 12y = 4
$$

Now add to equation (2):
$$
\begin{align*}
4x + 12y &= 4 \\
-4x + 6y &= -22 \\
\hline
(4x - 4x) + (12y + 6y) &= 4 - 22 \\
18y &= -18 \Rightarrow y = -1
\end{align*}
$$

Substitute $y = -1$ into equation (1):
$$
-x - 3(-1) = -1 \Rightarrow -x + 3 = -1 \Rightarrow -x = -4 \Rightarrow x = 4
$$

✔ Solution: $(x, y) = (4, -1)$

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✔ Final Answers:

| Problem | Solution $(x, y)$ |
|--------|-------------------|
| 1 | $(4, 4)$ |
| 2 | $(-7, -7)$ |
| 3 | $(0, -1)$ |
| 4 | $(-5, 7)$ |
| 5 | $(0, -7)$ |
| 6 | $(-4, 5)$ |
| 7 | $(-1, -4)$ |
| 8 | $(4, -1)$ |

Let me know if you'd like these checked step-by-step or explained further!