two-variables linear equations (x=d) Math Worksheets 1Math ... - Free Printable

The image you've uploaded contains a worksheet titled "Two-Variables Linear Equations (x=d)", which includes 8 problems, each with two equations. The goal is to solve for both variables $ x $ and $ y $ in each system of equations.

Each problem has one equation that directly gives the value of $ x $ (e.g., $ 8x = 40 $), so we can solve for $ x $ first and then substitute it into the second equation to find $ y $. Let's go through each problem step-by-step to verify the solutions provided (in red) and explain how they were obtained.

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Problem 1:

$$
\begin{align*}
8x &= 40 \quad \text{(1)}\\
7x + 4y &= 59 \quad \text{(2)}
\end{align*}
$$

Step 1: Solve equation (1) for $ x $:
$$
x = \frac{40}{8} = 5
$$

Step 2: Substitute $ x = 5 $ into equation (2):
$$
7(5) + 4y = 59 \\
35 + 4y = 59 \\
4y = 24 \\
y = 6
$$

✔ Solution: $ x = 5, y = 6 $ — Correct

---

Problem 2:

$$
\begin{align*}
7x &= 21 \quad \text{(1)}\\
5x - 4y &= 11 \quad \text{(2)}
\end{align*}
$$

Step 1: Solve equation (1):
$$
x = \frac{21}{7} = 3
$$

Step 2: Plug $ x = 3 $ into equation (2):
$$
5(3) - 4y = 11 \\
15 - 4y = 11 \\
-4y = -4 \\
y = 1
$$

✔ Solution: $ x = 3, y = 1 $ — Correct

---

Problem 3:

$$
\begin{align*}
7x &= 7 \quad \text{(1)}\\
8x - 2y &= -8 \quad \text{(2)}
\end{align*}
$$

Step 1: Solve (1):
$$
x = \frac{7}{7} = 1
$$

Step 2: Plug $ x = 1 $ into (2):
$$
8(1) - 2y = -8 \\
8 - 2y = -8 \\
-2y = -16 \\
y = 8
$$

✔ Solution: $ x = 1, y = 8 $ — Correct

---

Problem 4:

$$
\begin{align*}
7x &= 35 \quad \text{(1)}\\
2x - 4y &= 6 \quad \text{(2)}
\end{align*}
$$

Step 1: Solve (1):
$$
x = \frac{35}{7} = 5
$$

Step 2: Plug $ x = 5 $ into (2):
$$
2(5) - 4y = 6 \\
10 - 4y = 6 \\
-4y = -4 \\
y = 1
$$

✔ Solution: $ x = 5, y = 1 $ — Correct

---

Problem 5:

$$
\begin{align*}
5x &= 5 \quad \text{(1)}\\
8x - 2y &= 0 \quad \text{(2)}
\end{align*}
$$

Step 1: Solve (1):
$$
x = \frac{5}{5} = 1
$$

Step 2: Plug $ x = 1 $ into (2):
$$
8(1) - 2y = 0 \\
8 - 2y = 0 \\
2y = 8 \\
y = 4
$$

✔ Solution: $ x = 1, y = 4 $ — Correct

---

Problem 6:

$$
\begin{align*}
2x &= 10 \quad \text{(1)}\\
6x + 5y &= 45 \quad \text{(2)}
\end{align*}
$$

Step 1: Solve (1):
$$
x = \frac{10}{2} = 5
$$

Step 2: Plug $ x = 5 $ into (2):
$$
6(5) + 5y = 45 \\
30 + 5y = 45 \\
5y = 15 \\
y = 3
$$

✔ Solution: $ x = 5, y = 3 $ — Correct

---

Problem 7:

$$
\begin{align*}
2x &= 10 \quad \text{(1)}\\
2x - 5y &= 5 \quad \text{(2)}
\end{align*}
$$

Step 1: Solve (1):
$$
x = \frac{10}{2} = 5
$$

Step 2: Plug $ x = 5 $ into (2):
$$
2(5) - 5y = 5 \\
10 - 5y = 5 \\
-5y = -5 \\
y = 1
$$

✔ Solution: $ x = 5, y = 1 $ — Correct

---

Problem 8:

$$
\begin{align*}
6x &= 18 \quad \text{(1)}\\
4x + 6y &= 66 \quad \text{(2)}
\end{align*}
$$

Step 1: Solve (1):
$$
x = \frac{18}{6} = 3
$$

Step 2: Plug $ x = 3 $ into (2):
$$
4(3) + 6y = 66 \\
12 + 6y = 66 \\
6y = 54 \\
y = 9
$$

✔ Solution: $ x = 3, y = 9 $ — Correct

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✔ Final Answer:

All 8 problems have been solved correctly using substitution method:

1. $ x = 5, y = 6 $
2. $ x = 3, y = 1 $
3. $ x = 1, y = 8 $
4. $ x = 5, y = 1 $
5. $ x = 1, y = 4 $
6. $ x = 5, y = 3 $
7. $ x = 5, y = 1 $
8. $ x = 3, y = 9 $

All answers match the ones shown in red on the worksheet.

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🔍 Summary of Method:

- Since each system has one equation where $ x $ is isolated (or easily solvable), we solve for $ x $ first.
- Then substitute $ x $ into the second equation to find $ y $.
- This is called the substitution method for solving systems of linear equations.

Let me know if you'd like a visual explanation or graphing version! 🎉