Equivalent Linear Representations: Extended Worked Examples

This record is for Grade 8 learners studying linear relations in the Canadian curriculum. It is designed to be used alongside the overview and lesson records for this topic rather than replacing them:

• See [rea.m08.algebra.patterns-and-relations.representation-changes.overview] for the unit purpose and scope.
• See [rea.m08.algebra.patterns-and-relations.representation-changes.lesson] for the main teaching sequence and core explanations.
• If you want more context-based equation setup, see [rea.m08.algebra.solving-linear-equations.linear-modeling.worked-examples].

Here the goal is narrower: take one linear relation and show it correctly in another form.

What must stay the same?

When two representations are equivalent, they describe the same relation. That means all of these must stay unchanged:

• the variable meanings n- the constant rate of change
• the starting value
• the ordered pairs that belong to the relation
• the situation, if there is a context

A quick checkpoint:

• If the table increases by 3 each time x increases by 1, the equation must have slope 3.
• If the graph crosses the y-axis at 5, the equation must have constant term +5.
• If a word rule says “starts with 12 and adds 4 each step,” then the table, graph, and equation must all match that.

A reliable procedure

When changing representations, use this order:

1. Identify the variables and what they mean.
2. Find the starting value.
3. Find the rate of change.
4. Write the relation in a form such as y = mx + b.
5. Check the new representation against at least two known points.

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Worked Example 1: Words to equation and table

Problem. A plant is 8 cm tall at the start. It grows 2 cm each week. Write an equation and make a table for weeks 0 to 4.

Step 1: Choose variables.

Let w be the number of weeks. Let h be the plant height in centimetres.

Step 2: Find the starting value.

“At the start” means when w = 0, the height is 8. So the starting value is 8.

Step 3: Find the rate of change.

The plant grows 2 cm each week, so the rate of change is 2.

Step 4: Write the equation.

h = 2w + 8

Step 5: Build the table.

w	0	1	2	3	4
h	8	10	12	14	16

Commentary.

The key choice was recognizing that “at the start” means the value when the input is zero. Many errors happen when students use 8 as the slope. Here 8 is not “how much it grows”; it is where the relation begins.

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Worked Example 2: Table to equation

Problem. Find the equation for this table.

x	0	1	2	3
y	5	9	13	17

Step 1: Find the rate of change.

As x increases by 1, y increases by 4. So the rate of change is 4.

Step 2: Find the starting value.

When x = 0, y = 5. So the starting value is 5.

Step 3: Write the equation.

y = 4x + 5

Step 4: Check.

If x = 2, then y = 4(2) + 5 = 13, which matches the table.

Commentary.

Because the table includes x = 0, the starting value is easy to read directly. In Grade 8, this is the fastest route whenever the zero-input row is given.

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Worked Example 3: Table to words and equation when x = 0 is missing

Problem. Write a word rule and an equation for this table.

x	2	3	4	5
y	11	14	17	20

Step 1: Find the rate of change.

Each time x goes up by 1, y goes up by 3. So the slope is 3.

Step 2: Use a known pair to find the starting value.

We know the equation has the form y = 3x + b. Use the point (2, 11):

11 = 3(2) + b

11 = 6 + b

b = 5

So the equation is:

y = 3x + 5

Step 3: Write the word rule.

“Multiply the input by 3, then add 5.”

Commentary.

Since the table does not show x = 0, the starting value is not visible right away. The useful choice is to find the rate first, then substitute a known point. This avoids guessing.

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Worked Example 4: Equation to table and words

Problem. The relation is y = -2x + 7. Make a table for x = 0, 1, 2, 3 and describe the relation in words.

Step 1: Substitute each x-value.

For x = 0: y = -2(0) + 7 = 7

For x = 1: y = -2(1) + 7 = 5

For x = 2: y = -2(2) + 7 = 3

For x = 3: y = -2(3) + 7 = 1

x	0	1	2	3
y	7	5	3	1

Step 2: Write the word description.

“Start at 7 and decrease by 2 for each increase of 1 in x.”

Commentary.

A negative slope means the outputs get smaller as the inputs get larger. The choice that matters here is noticing that “-2” means “decrease by 2,” not “add negative 2 and ignore the direction.”

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Worked Example 5: Graph description to equation

Problem. A line passes through (0, -4) and (3, 2). Write its equation.

Step 1: Find the rate of change.

From (0, -4) to (3, 2), the rise is 2 - (-4) = 6 and the run is 3 - 0 = 3.

So the slope is:

m = 6 / 3 = 2

Step 2: Find the starting value.

Because the line passes through (0, -4), the y-intercept is -4.

Step 3: Write the equation.

y = 2x - 4

Step 4: Check with the second point.

If x = 3, then y = 2(3) - 4 = 2, which matches.

Commentary.

This example shows why the point with x = 0 is powerful: it gives the starting value directly. Once the slope is known, the equation follows immediately.

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Worked Example 6: Context to graph-ready table

Problem. A taxi charges a $6 starting fee and then $3 per kilometre. Create an equation and a table that could be used to graph the relation for k = 0, 1, 2, 3, 4.

Step 1: Choose variables.

Let k be the number of kilometres. Let C be the total cost in dollars.

Step 2: Identify the starting fee and rate.

Starting fee: 6 Rate: 3 dollars per kilometre

Step 3: Write the equation.

C = 3k + 6

Step 4: Make the table.

k	0	1	2	3	4
C	6	9	12	15	18

Step 5: Describe the graph.

The graph is a straight line crossing the vertical axis at 6 and rising 3 for every 1 to the right.

Commentary.

This is a partial variation: it does not start at zero. A common mistake is C = 3k. That would ignore the starting fee and represent a different taxi rule.

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Worked Example 7: Detecting a non-equivalent representation

Problem. The word rule says: “Start at 4 and add 5 each step.” A student writes the equation y = 4x + 5. Is it equivalent? If not, correct it.

Step 1: Translate the words carefully.

“Start at 4” means the starting value is 4. “Add 5 each step” means the rate of change is 5.

So the correct equation should be:

y = 5x + 4

Step 2: Explain why the student equation is not equivalent.

The student wrote y = 4x + 5. In that equation:

• the rate of change is 4
• the starting value is 5

Those are reversed, so the student changed the relation.

Step 3: Check with a table.

Correct relation y = 5x + 4:

x	0	1	2
y	4	9	14

Student relation y = 4x + 5:

x	0	1	2
y	5	9	13

The tables are different, so the representations are not equivalent.

Commentary.

This is the most common translation error in this topic. The repair is simple: ask, “What happens at input 0?” That identifies the constant term.

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Worked Example 8: Fill in a missing representation set

Problem. Complete the missing table, equation, and word rule for a line with slope -3 and y-intercept 10.

Step 1: Write the equation from the graph information.

Slope -3 means m = -3. Y-intercept 10 means b = 10.

Equation:

y = -3x + 10

Step 2: Make a table.

For x = 0: y = 10

For x = 1: y = 7

For x = 2: y = 4

For x = 3: y = 1

x	0	1	2	3
y	10	7	4	1

Step 3: Write the word rule.

“Start at 10 and subtract 3 for each increase of 1 in x.”

Commentary.

This example runs in the opposite direction from earlier ones: instead of extracting slope and intercept from a table, we generate the table from slope and intercept. That is a good equivalence check because every form should tell the same story.

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Worked Example 9: Challenge with fractional rate

Problem. A candle is 12 cm tall and burns down by 0.5 cm each hour. Write an equation and find the height after 7 hours.

Step 1: Choose variables.

Let t be time in hours. Let h be height in centimetres.

Step 2: Identify the starting value and rate.

Starting height: 12 Rate of change: -0.5 cm per hour because the candle gets shorter.

Step 3: Write the equation.

h = -0.5t + 12

Step 4: Find the height after 7 hours.

h = -0.5(7) + 12

h = -3.5 + 12

h = 8.5

So after 7 hours, the candle is 8.5 cm tall.

Commentary.

The challenge is not the algebra; it is preserving meaning. Because the candle is shrinking, the slope must be negative. If a student writes +0.5, the equation describes growth instead of burning.

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Worked Example 10: Decide whether two representations are equivalent

Problem. Are these two representations equivalent?

• Equation: y = 2x + 1
• Table:

x	0	1	2	3
y	1	3	5	8

Step 1: Check the table's rate of change.

From 1 to 3 is +2. From 3 to 5 is +2. From 5 to 8 is +3.

The change is not constant all the way through.

Step 2: Compare with the equation.

The equation y = 2x + 1 should produce:

• x = 0 gives 1
• x = 1 gives 3
• x = 2 gives 5
• x = 3 gives 7

But the table gives 8 when x = 3.

So the representations are not equivalent.

Step 3: Repair the mismatch.

Either:

• correct the table entry from 8 to 7, or
• change the equation to match a different relation

Since only one table value is wrong, the simplest repair is to change 8 to 7.

Commentary.

Equivalent representations must agree on every point, not just the first few. A good habit is to test several values instead of stopping after one match.

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Common misconceptions and repairs

1. Mixing up slope and starting value

Error: From “start at 6 and add 4,” writing y = 6x + 4.

Repair: Ask, “What is the output when the input is 0?” That number is the constant term.

2. Forgetting the relation must stay the same

Error: Changing the table pattern while rewriting the equation.

Repair: Check at least two ordered pairs. If the points change, the representation is not equivalent.

3. Missing negative slope

Error: Writing y = 3x + 10 when the amount decreases by 3 each step.

Repair: Use language carefully: “decreases by 3” means slope -3.

4. Assuming every linear relation starts at 0

Error: Turning “$5 initial charge plus $2 each hour” into C = 2h.

Repair: Look for words like “initial,” “starting,” “base,” or “fixed fee.” These signal a nonzero intercept.

Short practice set

1. A gym charges a $12 signup fee and $4 per visit. Write an equation.
2. Find the equation for the table: (0, 9), (1, 7), (2, 5), (3, 3).
3. Does the table (0, 2), (1, 5), (2, 8) match the equation y = 3x + 2?
4. Write a word rule for y = 6x - 1.
5. A line has slope 1.5 and y-intercept -2. Write its equation.

Answers

1. C = 4v + 12 if v is the number of visits.
2. y = -2x + 9.
3. Yes. The equation gives 2, 5, 8 for x = 0, 1, 2.
4. “Multiply the input by 6, then subtract 1,” or “start at -1 and increase by 6 each step.”
5. y = 1.5x - 2.

Final takeaway

Equivalent linear representations are not just different-looking answers. They are different ways to express the same constant change, same starting value, and same set of points. In Grade 8, the most reliable strategy is to identify the slope and intercept first, then verify them across the new representation.