Graphing Linear Equations worksheet with tables and coordinate grids for plotting lines.

To solve the problem, we need to complete the tables for each linear equation by finding the corresponding \( y \)-values for given \( x \)-values and then plot the points on the graph. Let's go through each equation step by step.

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1) \( 2x - y = 7 \)

#### Step 1: Solve for \( y \)
Rearrange the equation to solve for \( y \):
\[
2x - y = 7 \implies y = 2x - 7
\]

#### Step 2: Complete the table
Use the equation \( y = 2x - 7 \) to find \( y \)-values for the given \( x \)-values:

| \( x \) | \( y = 2x - 7 \) |
|---------|------------------|
| 0 | \( 2(0) - 7 = -7 \) |
| 1 | \( 2(1) - 7 = -5 \) |
| 2 | \( 2(2) - 7 = -3 \) |
| 3 | \( 2(3) - 7 = -1 \) |
| 4 | \( 2(4) - 7 = 1 \) |

So, the completed table is:

| \( x \) | \( y \) |
|---------|---------|
| 0 | -7 |
| 1 | -5 |
| 2 | -3 |
| 3 | -1 |
| 4 | 1 |

#### Step 3: Plot the points
Plot the points \((0, -7)\), \((1, -5)\), \((2, -3)\), \((3, -1)\), and \((4, 1)\) on the graph and draw a straight line through them.

---

2) \( 2x - 8 = 4y \)

#### Step 1: Solve for \( y \)
Rearrange the equation to solve for \( y \):
\[
2x - 8 = 4y \implies y = \frac{2x - 8}{4} \implies y = \frac{1}{2}x - 2
\]

#### Step 2: Complete the table
Use the equation \( y = \frac{1}{2}x - 2 \) to find \( y \)-values for the given \( x \)-values:

| \( x \) | \( y = \frac{1}{2}x - 2 \) |
|---------|---------------------------|
| -8 | \( \frac{1}{2}(-8) - 2 = -4 - 2 = -6 \) |
| -6 | \( \frac{1}{2}(-6) - 2 = -3 - 2 = -5 \) |
| -4 | \( \frac{1}{2}(-4) - 2 = -2 - 2 = -4 \) |
| -2 | \( \frac{1}{2}(-2) - 2 = -1 - 2 = -3 \) |
| 0 | \( \frac{1}{2}(0) - 2 = 0 - 2 = -2 \) |

So, the completed table is:

| \( x \) | \( y \) |
|---------|---------|
| -8 | -6 |
| -6 | -5 |
| -4 | -4 |
| -2 | -3 |
| 0 | -2 |

#### Step 3: Plot the points
Plot the points \((-8, -6)\), \((-6, -5)\), \((-4, -4)\), \((-2, -3)\), and \((0, -2)\) on the graph and draw a straight line through them.

---

3) \( 3y = 12 - 2x \)

#### Step 1: Solve for \( y \)
Rearrange the equation to solve for \( y \):
\[
3y = 12 - 2x \implies y = \frac{12 - 2x}{3} \implies y = 4 - \frac{2}{3}x
\]

#### Step 2: Complete the table
Use the equation \( y = 4 - \frac{2}{3}x \) to find \( y \)-values for the given \( x \)-values:

| \( x \) | \( y = 4 - \frac{2}{3}x \) |
|---------|---------------------------|
| -12 | \( 4 - \frac{2}{3}(-12) = 4 + 8 = 12 \) |
| -6 | \( 4 - \frac{2}{3}(-6) = 4 + 4 = 8 \) |
| 0 | \( 4 - \frac{2}{3}(0) = 4 \) |
| 6 | \( 4 - \frac{2}{3}(6) = 4 - 4 = 0 \) |
| 12 | \( 4 - \frac{2}{3}(12) = 4 - 8 = -4 \) |

So, the completed table is:

| \( x \) | \( y \) |
|---------|---------|
| -12 | 12 |
| -6 | 8 |
| 0 | 4 |
| 6 | 0 |
| 12 | -4 |

#### Step 3: Plot the points
Plot the points \((-12, 12)\), \((-6, 8)\), \((0, 4)\), \((6, 0)\), and \((12, -4)\) on the graph and draw a straight line through them.

---

4) \( 15 = 10x + 5y \)

#### Step 1: Solve for \( y \)
Rearrange the equation to solve for \( y \):
\[
15 = 10x + 5y \implies 5y = 15 - 10x \implies y = 3 - 2x
\]

#### Step 2: Complete the table
Use the equation \( y = 3 - 2x \) to find \( y \)-values for the given \( x \)-values:

| \( x \) | \( y = 3 - 2x \) |
|---------|------------------|
| -2 | \( 3 - 2(-2) = 3 + 4 = 7 \) |
| -1 | \( 3 - 2(-1) = 3 + 2 = 5 \) |
| 0 | \( 3 - 2(0) = 3 \) |
| 2 | \( 3 - 2(2) = 3 - 4 = -1 \) |
| 3 | \( 3 - 2(3) = 3 - 6 = -3 \) |

So, the completed table is:

| \( x \) | \( y \) |
|---------|---------|
| -2 | 7 |
| -1 | 5 |
| 0 | 3 |
| 2 | -1 |
| 3 | -3 |

#### Step 3: Plot the points
Plot the points \((-2, 7)\), \((-1, 5)\), \((0, 3)\), \((2, -1)\), and \((3, -3)\) on the graph and draw a straight line through them.

---

5) \( 4y = -x + 12 \)

#### Step 1: Solve for \( y \)
Rearrange the equation to solve for \( y \):
\[
4y = -x + 12 \implies y = \frac{-x + 12}{4} \implies y = -\frac{1}{4}x + 3
\]

#### Step 2: Complete the table
Use the equation \( y = -\frac{1}{4}x + 3 \) to find \( y \)-values for the given \( x \)-values:

| \( x \) | \( y = -\frac{1}{4}x + 3 \) |
|---------|---------------------------|
| -8 | \( -\frac{1}{4}(-8) + 3 = 2 + 3 = 5 \) |
| -4 | \( -\frac{1}{4}(-4) + 3 = 1 + 3 = 4 \) |
| 0 | \( -\frac{1}{4}(0) + 3 = 3 \) |
| 4 | \( -\frac{1}{4}(4) + 3 = -1 + 3 = 2 \) |
| 8 | \( -\frac{1}{4}(8) + 3 = -2 + 3 = 1 \) |

So, the completed table is:

| \( x \) | \( y \) |
|---------|---------|
| -8 | 5 |
| -4 | 4 |
| 0 | 3 |
| 4 | 2 |
| 8 | 1 |

#### Step 3: Plot the points
Plot the points \((-8, 5)\), \((-4, 4)\), \((0, 3)\), \((4, 2)\), and \((8, 1)\) on the graph and draw a straight line through them.

---

6) \( x + 12 = 6y \)

#### Step 1: Solve for \( y \)
Rearrange the equation to solve for \( y \):
\[
x + 12 = 6y \implies y = \frac{x + 12}{6} \implies y = \frac{1}{6}x + 2
\]

#### Step 2: Complete the table
Use the equation \( y = \frac{1}{6}x + 2 \) to find \( y \)-values for the given \( x \)-values:

| \( x \) | \( y = \frac{1}{6}x + 2 \) |
|---------|---------------------------|
| -12 | \( \frac{1}{6}(-12) + 2 = -2 + 2 = 0 \) |
| -6 | \( \frac{1}{6}(-6) + 2 = -1 + 2 = 1 \) |
| 0 | \( \frac{1}{6}(0) + 2 = 2 \) |
| 6 | \( \frac{1}{6}(6) + 2 = 1 + 2 = 3 \) |
| 12 | \( \frac{1}{6}(12) + 2 = 2 + 2 = 4 \) |

So, the completed table is:

| \( x \) | \( y \) |
|---------|---------|
| -12 | 0 |
| -6 | 1 |
| 0 | 2 |
| 6 | 3 |
| 12 | 4 |

#### Step 3: Plot the points
Plot the points \((-12, 0)\), \((-6, 1)\), \((0, 2)\), \((6, 3)\), and \((12, 4)\) on the graph and draw a straight line through them.

---

7) \( -x + 9y = 9 \)

#### Step 1: Solve for \( y \)
Rearrange the equation to solve for \( y \):
\[
-x + 9y = 9 \implies 9y = x + 9 \implies y = \frac{x + 9}{9} \implies y = \frac{1}{9}x + 1
\]

#### Step 2: Complete the table
Use the equation \( y = \frac{1}{9}x + 1 \) to find \( y \)-values for the given \( x \)-values:

| \( x \) | \( y = \frac{1}{9}x + 1 \) |
|---------|---------------------------|
| 9 | \( \frac{1}{9}(9) + 1 = 1 + 1 = 2 \) |
| 18 | \( \frac{1}{9}(18) + 1 = 2 + 1 = 3 \) |
| 27 | \( \frac{1}{9}(27) + 1 = 3 + 1 = 4 \) |
| 36 | \( \frac{1}{9}(36) + 1 = 4 + 1 = 5 \) |
| 45 | \( \frac{1}{9}(45) + 1 = 5 + 1 = 6 \) |

So, the completed table is:

| \( x \) | \( y \) |
|---------|---------|
| 9 | 2 |
| 18 | 3 |
| 27 | 4 |
| 36 | 5 |
| 45 | 6 |

#### Step 3: Plot the points
Plot the points \((9, 2)\), \((18, 3)\), \((27, 4)\), \((36, 5)\), and \((45, 6)\) on the graph and draw a straight line through them.

---

8) \( -15 = 5y - x \)

#### Step 1: Solve for \( y \)
Rearrange the equation to solve for \( y \):
\[
-15 = 5y - x \implies 5y = x - 15 \implies y = \frac{x - 15}{5} \implies y = \frac{1}{5}x - 3
\]

#### Step 2: Complete the table
Use the equation \( y = \frac{1}{5}x - 3 \) to find \( y \)-values for the given \( x \)-values:

| \( x \) | \( y = \frac{1}{5}x - 3 \) |
|---------|---------------------------|
| 5 | \( \frac{1}{5}(5) - 3 = 1 - 3 = -2 \) |
| 10 | \( \frac{1}{5}(10) - 3 = 2 - 3 = -1 \) |
| 15 | \( \frac{1}{5}(15) - 3 = 3 - 3 = 0 \) |
| 20 | \( \frac{1}{5}(20) - 3 = 4 - 3 = 1 \) |
| 25 | \( \frac{1}{5}(25) - 3 = 5 - 3 = 2 \) |

So, the completed table is:

| \( x \) | \( y \) |
|---------|---------|
| 5 | -2 |
| 10 | -1 |
| 15 | 0 |
| 20 | 1 |
| 25 | 2 |

#### Step 3: Plot the points
Plot the points \((5, -2)\), \((10, -1)\), \((15, 0)\), \((20, 1)\), and \((25, 2)\) on the graph and draw a straight line through them.

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9) \( 2y = -x + 8 \)

#### Step 1: Solve for \( y \)
Rearrange the equation to solve for \( y \):
\[
2y = -x + 8 \implies y = \frac{-x + 8}{2} \implies y = -\frac{1}{2}x + 4
\]

#### Step 2: Complete the table
Use the equation \( y = -\frac{1}{2}x + 4 \) to find \( y \)-values for the given \( x \)-values:

| \( x \) | \( y = -\frac{1}{2}x + 4 \) |
|---------|---------------------------|
| -2 | \( -\frac{1}{2}(-2) + 4 = 1 + 4 = 5 \) |
| 0 | \( -\frac{1}{2}(0) + 4 = 4 \) |
| 2 | \( -\frac{1}{2}(2) + 4 = -1 + 4 = 3 \) |
| 4 | \( -\frac{1}{2}(4) + 4 = -2 + 4 = 2 \) |
| 6 | \( -\frac{1}{2}(6) + 4 = -3 + 4 = 1 \) |

So, the completed table is:

| \( x \) | \( y \) |
|---------|---------|
| -2 | 5 |
| 0 | 4 |
| 2 | 3 |
| 4 | 2 |
| 6 | 1 |

#### Step 3: Plot the points
Plot the points \((-2, 5)\), \((0, 4)\), \((2, 3)\), \((4, 2)\), and \((6, 1)\) on the graph and draw a straight line through them.

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Final Answer

The completed tables and graphs are as follows:

1. \( 2x - y = 7 \)
- Table: \(\begin{array}{c|c} x & y \\ \hline 0 & -7 \\ 1 & -5 \\ 2 & -3 \\ 3 & -1 \\ 4 & 1 \end{array}\)

2. \( 2x - 8 = 4y \)
- Table: \(\begin{array}{c|c} x & y \\ \hline -8 & -6 \\ -6 & -5 \\ -4 & -4 \\ -2 & -3 \\ 0 & -2 \end{array}\)

3. \( 3y = 12 - 2x \)
- Table: \(\begin{array}{c|c} x & y \\ \hline -12 & 12 \\ -6 & 8 \\ 0 & 4 \\ 6 & 0 \\ 12 & -4 \end{array}\)

4. \( 15 = 10x + 5y \)
- Table: \(\begin{array}{c|c} x & y \\ \hline -2 & 7 \\ -1 & 5 \\ 0 & 3 \\ 2 & -1 \\ 3 & -3 \end{array}\)

5. \( 4y = -x + 12 \)
- Table: \(\begin{array}{c|c} x & y \\ \hline -8 & 5 \\ -4 & 4 \\ 0 & 3 \\ 4 & 2 \\ 8 & 1 \end{array}\)

6. \( x + 12 = 6y \)
- Table: \(\begin{array}{c|c} x & y \\ \hline -12 & 0 \\ -6 & 1 \\ 0 & 2 \\ 6 & 3 \\ 12 & 4 \end{array}\)

7. \( -x + 9y = 9 \)
- Table: \(\begin{array}{c|c} x & y \\ \hline 9 & 2 \\ 18 & 3 \\ 27 & 4 \\ 36 & 5 \\ 45 & 6 \end{array}\)

8. \( -15 = 5y - x \)
- Table: \(\begin{array}{c|c} x & y \\ \hline 5 & -2 \\ 10 & -1 \\ 15 & 0 \\ 20 & 1 \\ 25 & 2 \end{array}\)

9. \( 2y = -x + 8 \)
- Table: \(\begin{array}{c|c} x & y \\ \hline -2 & 5 \\ 0 & 4 \\ 2 & 3 \\ 4 & 2 \\ 6 & 1 \end{array}\)

For each equation, plot the points from the table on the graph and draw a straight line through them.

\boxed{\text{See the detailed steps above for each equation.}}