Grade 6 · MathematicsAbout 120 minutes
MathematicsGrade 6About 120 minutes

Connect changing quantities through an equation, table and graph

Translate the situation, justify operations or graph choices, then demonstrate independent solution and substitution checks.

Ohio learning standards this draft targets

math-2017:6.EE.6math-2017:6.EE.9

Read the standard text

math-2017:6.EE.6 Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set.

math-2017:6.EE.9 Use variables to represent two quantities in a real-world problem that change in relationship to one another; write an equation to express one quantity, thought of as the dependent variable, in terms of the other quantity, thought of as the independent variable. Analyze the relationship between the dependent and independent variables using graphs and tables, and relate these to the equation. For example, in a problem involving motion at constant speed, list and graph ordered pairs of distances and times, and write the equation d = 65t to represent the relationship between distance and time.

Provisional targets. Clause-by-clause educational review is still pending.

Study these connected lessons together: Solve addition and multiplication equations with nonnegative rational values, Represent strict inequalities and test their solutions.

Section 1 · LearnLearn the idea
For d = 1.5t, time in seconds is the horizontal input and distance in meters is the vertical output. Points (0,0), (2,3), (4,6), (6,9) follow the same constant-speed relationship, and the line through them covers the times in between.For d = 1.5t, time in seconds is the horizontal input and distance in meters is the vertical output. Points (0,0), (2,3), (4,6), (6,9) follow the same constant-speed relationship, and the line through them covers the times in between.Distance d = 1.5t01234560123456789Distance (meters)Time (seconds)(0,0), (2,3), (4,6), (6,9)
For d = 1.5t, time in seconds is the horizontal input and distance in meters is the vertical output. Points (0,0), (2,3), (4,6), (6,9) follow the same constant-speed relationship, and the line through them covers the times in between.

Learning goal: Identify independent and dependent quantities in a specified relationship, write the dependent quantity in terms of the independent one and connect an equation with a table and graph. Explain the assumptions and allowable values in the context.

A model vehicle moves at a constant speed of 1.5 meters per second, starting with zero distance traveled at time zero. Let t be elapsed seconds and d be distance traveled in meters. The relationship is d = 1.5t. Time is the independent variable selected as the input; distance is the dependent variable calculated from that time in this model.

The words independent and dependent describe roles in the chosen representation. They do not say that one quantity always causes the other. In this constant-speed situation, the model uses elapsed time to determine distance. The statement also assumes that the speed remains constant and measures distance from the chosen starting event.

Create a table: t = 0 gives d = 0; t = 2 gives d = 3; t = 4 gives d = 6; t = 6 gives d = 9. Each row satisfies the same equation. For example, at four seconds, 1.5 × 4 = 6 meters. The input and output units stay attached to the column headings.

Plot time on the horizontal axis and distance on the vertical axis. The ordered pair (4,6) means four seconds and six meters. The plotted pairs lie along a straight line through the origin because equal time increases produce equal distance increases at this constant speed. Here, the model permits intermediate times, so the connecting line represents meaningful predictions between the table's rows.

The graph's scales need not have matching numerical labels because seconds and meters are different units. They must each be consistent. In the diagram, time increases by two seconds between selected points and distance by three meters. A point's physical position is determined by both labeled scales, not by assuming that equal pixel distances always mean equal numerical values across different units.

Find an unlisted value using the equation: at five seconds, d = 1.5 × 5 = 7.5 meters. To find the time for twelve meters, solve 12 = 1.5t, giving t = 8 seconds. The graph and table can support the reasoning, but an equation gives exact values without relying on an approximate visual reading.

Changing the initial condition changes the equation. If a counter already reads four meters at time zero and then increases at 1.5 meters per second, its reading is r = 4 + 1.5t. At two seconds, r = 7. This graph would begin at (0,4), not the origin. The four-meter initial reading is not another speed; it is a starting amount added to the accumulated change.

Some contexts permit only separate whole-number inputs. If n counts complete packs, each containing six cards, total cards c = 6n has whole-number inputs n. A graph can show separate points for 0, 1, 2 and more packs. Connecting them as if 1.4 complete packs were an available count would need a different interpretation. Context determines whether intermediate values are meaningful.

Use a table to examine whether a proposed equation fits. A row (4,7) does not fit d = 1.5t, because four seconds predicts six meters. One mismatching row shows that the equation does not describe every supplied observation exactly. Possible causes include a changed speed, different starting condition or measurement error; the numbers alone do not identify which explanation is correct.

Keep the variables distinct. Writing t = 1.5d reverses the relationship and units. The correct reversed expression for this model is t = d/1.5. Saying the input and output names aloud before substitution helps avoid exchanging them.

Section 2 · ModelWork through the examples

Worked comparison: A second vehicle follows e = 2t. At six seconds, e = 12 meters while d = 9 meters for the first. Both use the same time input, so the comparison is meaningful. The second coefficient represents a greater constant speed in the same units.

Worked initial value: For r = 4 + 1.5t, substitute t = 0 to get r = 4. A graph beginning at zero would not satisfy the equation even if its later increments had the right rate. Check both the starting point and the change per input unit.

Section 3 · DoPractice, demonstrate, and explain

Guided connections: Build the constant-speed table, plot its ordered pairs and explain how each pair satisfies the equation. Identify independent and dependent variables with units. Predict one intermediate time and check an intentionally incorrect row.

Independent performance: Model one continuous relationship and one complete-object count relationship. For each, write an equation, create a table and draw a graph with labeled axes. Explain permissible inputs, the meaning of the coefficient and any starting amount. Use the model to find an unlisted output and recover an input from a given output. State the assumptions that make the prediction reasonable.

Section 4 · Check8-question lesson practice test

An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.

Problem 1

Identify independent and dependent variables in d = 1.5t.

Show worked solution

Time t is the input; distance d is the output determined by the constant-speed model.

Problem 2

Find d at t = 5.

Show worked solution

7.5 meters, from 1.5 × 5.

Problem 3

Find time for twelve meters at that speed.

Show worked solution

Eight seconds, from 12 ÷ 1.5.

Problem 4

Does (4,7) fit d = 1.5t?

Show worked solution

No. Four seconds predicts six meters.

Problem 5

What does four mean in r = 4 + 1.5t?

Show worked solution

The initial reading at time zero, not an additional speed.

Problem 6

Why might a pack-count graph use separate points?

Show worked solution

Complete pack counts allow whole-number inputs; intermediate fractional counts may not be feasible in that context.

Problem 7

A hypothetical pump fills an empty tank at a constant 2.5 liters per minute. Choose variables, name the independent and the dependent one, write an equation and list the table rows for 0, 2 and 4 minutes.

Show worked solution

With m for minutes (independent) and v for liters in the tank (dependent), v = 2.5m. Rows: (0, 0), (2, 5), (4, 10). Graphed with minutes on the horizontal axis, the points lie on a straight line through the origin.

Problem 8

Using that pump model, how long does the tank take to reach 20 liters, and why is it reasonable to connect the graph’s points here?

Show worked solution

8 minutes, from 20 ÷ 2.5. Time can take values between the table rows, so the points between them are meaningful as long as the rate stays constant.