Worksheet for practicing graphing linear equations by completing function tables and plotting points on coordinate grids.

Explanation:
We are given six linear equations. For each, we need to:
1. Fill in the missing y-values in the table by plugging the given x-values into the equation.
2. Then (though not required for the answer here), you would graph the points and draw the line.

Let’s compute each table carefully.

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(1) Equation: $ y = 4x $

Given x-values:
- $ x = -1\frac{1}{2} = -\frac{3}{2} $ → $ y = 4 \cdot (-\frac{3}{2}) = -6 $
- $ x = \frac{1}{2} $ → $ y = 4 \cdot \frac{1}{2} = 2 $
- $ x = 0 $ → $ y = 4 \cdot 0 = 0 $
- $ x = 1 $ → $ y = 4 \cdot 1 = 4 $

So y-values: –6, 2, 0, 4

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(2) Equation: $ y = x + 6 $

x-values: –4, –3, –2, 1
- $ x = -4 $ → $ y = -4 + 6 = 2 $
- $ x = -3 $ → $ y = -3 + 6 = 3 $
- $ x = -2 $ → $ y = -2 + 6 = 4 $
- $ x = 1 $ → $ y = 1 + 6 = 7 $

y-values: 2, 3, 4, 7

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(3) Equation: $ y = \frac{1}{3}x - 4 $

x-values: –7, –5, –1, 1
- $ x = -7 $: $ y = \frac{1}{3}(-7) - 4 = -\frac{7}{3} - 4 = -\frac{7}{3} - \frac{12}{3} = -\frac{19}{3} = -6\frac{1}{3} $
- $ x = -5 $: $ y = \frac{1}{3}(-5) - 4 = -\frac{5}{3} - 4 = -\frac{5}{3} - \frac{12}{3} = -\frac{17}{3} = -5\frac{2}{3} $
- $ x = -1 $: $ y = \frac{1}{3}(-1) - 4 = -\frac{1}{3} - 4 = -\frac{13}{3} = -4\frac{1}{3} $
- $ x = 1 $: $ y = \frac{1}{3}(1) - 4 = \frac{1}{3} - 4 = \frac{1}{3} - \frac{12}{3} = -\frac{11}{3} = -3\frac{2}{3} $

So y-values: –6⅓, –5⅔, –4⅓, –3⅔
(Or as improper fractions: –19/3, –17/3, –13/3, –11/3)

But since the worksheet likely expects decimals or fractions, and the inputs are integers, fractions are fine. Let’s keep as mixed numbers unless told otherwise.

Wait — double-check: maybe they expect decimal approximations? But the worksheet uses fractions like $ \frac{1}{2} $, so mixed numbers or improper fractions are acceptable. However, let’s verify if any simplification is possible — no, these are correct.

Hold on — maybe the problem expects exact values, and we can write them as fractions. But for consistency with typical middle-school worksheets, they often accept simplified fractions.

Let me recompute quickly:

- $ x = -7 $: $ \frac{-7}{3} - 4 = \frac{-7 - 12}{3} = \frac{-19}{3} $
- $ x = -5 $: $ \frac{-5 - 12}{3} = \frac{-17}{3} $
- $ x = -1 $: $ \frac{-1 - 12}{3} = \frac{-13}{3} $
- $ x = 1 $: $ \frac{1 - 12}{3} = \frac{-11}{3} $

Yes.

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(4) Equation: $ y = \frac{1}{4}x + 1 $

x-values: –6, –4, 0, 5
- $ x = -6 $: $ y = \frac{1}{4}(-6) + 1 = -\frac{6}{4} + 1 = -\frac{3}{2} + 1 = -\frac{1}{2} $
- $ x = -4 $: $ y = \frac{1}{4}(-4) + 1 = -1 + 1 = 0 $
- $ x = 0 $: $ y = 0 + 1 = 1 $
- $ x = 5 $: $ y = \frac{5}{4} + 1 = \frac{5}{4} + \frac{4}{4} = \frac{9}{4} = 2\frac{1}{4} $

So y-values: –½, 0, 1, 2¼

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(5) Equation: $ y = \frac{1}{2}x - 6 $

x-values: 1, 2, 3, 6
- $ x = 1 $: $ y = \frac{1}{2}(1) - 6 = 0.5 - 6 = -5.5 = -\frac{11}{2} $
- $ x = 2 $: $ y = 1 - 6 = -5 $
- $ x = 3 $: $ y = \frac{3}{2} - 6 = 1.5 - 6 = -4.5 = -\frac{9}{2} $
- $ x = 6 $: $ y = 3 - 6 = -3 $

So y-values: –5.5 (or –11/2), –5, –4.5 (or –9/2), –3

Since other problems used fractions, better to use fractions:
–11/2, –5, –9/2, –3

But –5 and –3 are integers; that’s fine.

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(6) Equation: $ y = \frac{1}{3}x - 6 $

x-values: –3, 1, 2, 7
- $ x = -3 $: $ y = \frac{1}{3}(-3) - 6 = -1 - 6 = -7 $
- $ x = 1 $: $ y = \frac{1}{3} - 6 = \frac{1 - 18}{3} = -\frac{17}{3} = -5\frac{2}{3} $
- $ x = 2 $: $ y = \frac{2}{3} - 6 = \frac{2 - 18}{3} = -\frac{16}{3} = -5\frac{1}{3} $
- $ x = 7 $: $ y = \frac{7}{3} - 6 = \frac{7 - 18}{3} = -\frac{11}{3} = -3\frac{2}{3} $

So y-values: –7, –17/3, –16/3, –11/3

Now, the question says: “Complete the function table and graph the line…” but since the user only asks to solve the problem (and options say ExplainAnswer: false), and we’re to give only the final answer — but what exactly is being asked?

Looking back: The task is to fill in the tables. Since there are 6 problems, and each has 4 blanks, the final answer should be the completed y-values for each table.

But the instruction says: “ONLY THEN provide the final answer”, and “Final Answer: [Provide only the final correct answer]”.

Given that this is a worksheet with 6 tables, the expected output is likely the filled-in y-values in order.

Let me list them clearly per problem:

(1) y = 4x → y = –6, 2, 0, 4
(2) y = x + 6 → y = 2, 3, 4, 7
(3) y = (1/3)x – 4 → y = –19/3, –17/3, –13/3, –11/3
(4) y = (1/4)x + 1 → y = –1/2, 0, 1, 9/4
(5) y = (1/2)x – 6 → y = –11/2, –5, –9/2, –3
(6) y = (1/3)x – 6 → y = –7, –17/3, –16/3, –11/3

To match typical worksheet expectations, they may prefer mixed numbers for improper fractions >1 in magnitude, but many worksheets accept improper fractions.

However, let’s check if any of these can be simplified or if I made arithmetic errors.

Recheck (3):
x = –7: (–7)/3 – 4 = –7/3 – 12/3 = –19/3 ✔
x = –5: –5/3 – 12/3 = –17/3 ✔
x = –1: –1/3 – 12/3 = –13/3 ✔
x = 1: 1/3 – 12/3 = –11/3 ✔

(6):
x = –3: –1 – 6 = –7 ✔
x = 1: 1/3 – 6 = –17/3 ✔
x = 2: 2/3 – 6 = –16/3 ✔
x = 7: 7/3 – 18/3 = –11/3 ✔

All correct.

Since the user didn’t specify format, and to keep it clean, I’ll present the y-values as simplified fractions (improper where numerator > denominator, but that’s standard in algebra).

But note: In many middle-school worksheets, they write mixed numbers. Let’s convert improper fractions with absolute value >1 to mixed numbers:

(3):
–19/3 = –6 1/3
–17/3 = –5 2/3
–13/3 = –4 1/3
–11/3 = –3 2/3

(4):
–1/2 stays, 9/4 = 2 1/4

(5):
–11/2 = –5 1/2
–9/2 = –4 1/2

(6):
–17/3 = –5 2/3
–16/3 = –5 1/3
–11/3 = –3 2/3

That’s probably what the worksheet expects.

So final filled tables:

(1) –6, 2, 0, 4
(2) 2, 3, 4, 7
(3) –6⅓, –5⅔, –4⅓, –3⅔
(4) –½, 0, 1, 2¼
(5) –5½, –5, –4½, –3
(6) –7, –5⅔, –5⅓, –3⅔

Now, since the instruction says “Provide only the final correct answer”, and there are multiple parts, I will list them clearly in order.

Final Answer:
(1) y = –6, 2, 0, 4
(2) y = 2, 3, 4, 7
(3) y = –6⅓, –5⅔, –4⅓, –3⅔
(4) y = –½, 0, 1, 2¼
(5) y = –5½, –5, –4½, –3
(6) y = –7, –5⅔, –5⅓, –3⅔