A table, an equation and a graph can describe exactly the same relationship while looking like three different topics. A learner who can substitute into a formula may still struggle to explain a gradient. A learner who can plot points may not recognise that the intersection of two lines is a solution satisfying two conditions at once. The missing connection is often meaning, not more plotting.
This Secondary 2 Mathematics Learning Guide develops coordinates, linear relationships, gradients, intercepts and graphical solutions as a connected system. You will move from words to tables, from tables to equations, from equations to graphs, and back to the original situation. The aim is to keep the relationship unchanged while choosing the representation that makes the next question easier.
Secondary Mathematics Hub: S1–S4 Capability Map · Secondary 2 Learning Guide, Batch 1, Guide 3. Companion guides cover algebraic control, proportion and rate, and geometry and similarity.
Course boundary. In the MOE G2 and G3 Mathematics syllabuses, basic linear graphs appear at different stages; Secondary Two includes simultaneous linear equations. This guide therefore combines foundation repair with current-level application. It does not imply that every idea here is first taught in Secondary 2. Follow your school’s current subject-level sequence. All models and questions below are original teaching examples.
Navigate: Coordinates and points · Tables and equations · Gradient and intercepts · Drawing and reading · Simultaneous equations · Graphs in context · Practice and answers · Teaching and transfer.
1. A coordinate pair is an ordered statement
The point (3, 5) says x = 3 and y = 5 at the same time. It is not an unordered collection of two numbers. On a standard Cartesian plane, locate the horizontal coordinate first, then the vertical coordinate. The point (5, 3) is generally somewhere else because exchanging the coordinates exchanges their roles.
In a context graph, the labels make this even more important. A point (3, 15) might mean three hours and fifteen kilometres, or three tickets and fifteen dollars. Without the axis labels, the numbers are not a complete description. Reading a graph starts with identifying what each coordinate measures.
Negative coordinates describe direction from the origin
On the usual orientation, (−2, 4) lies two units left and four units above the origin. The point (−2, −4) lies two units left and four units below. The signs are not optional decoration. They distinguish positions and therefore distinguish which relationships the points satisfy.
A useful practice task is to plot (2, 3), (−2, 3), (−2, −3) and (2, −3), then explain what changes each time. Keep the magnitudes fixed so that the sign change is the only new demand. Once the student can explain the locations, introduce different scales and contextual labels.
Being on a graph means satisfying its condition
A point belongs to the graph y = 2x + 3 when its coordinates make the equation true. The point (2, 7) belongs because 7 = 2(2) + 3. The point (2, 8) does not, even if an imprecise drawing makes it look close. Algebra can test membership exactly when exact coordinates and an exact equation are supplied.
The graph is therefore more than a picture of a formula. It is the set of coordinate pairs satisfying a relationship, subject to any domain restrictions. This interpretation becomes especially useful when two graphs meet: the shared point satisfies both relationships, not merely one.
For the standard coordinate and line vocabulary, see OpenStax: Linear Graphs Key Terms. The examples that follow build the connection from coordinates to meaning rather than assuming that drawing a line is the entire learning objective.
2. Words, tables, equations and graphs are four views
Suppose a fictional service charges 3 dollars initially and then 2 dollars for each unit of use. In words, there is a fixed starting amount and a constant additional charge. A table lists selected input-output pairs. The equation y = 2x + 3 compresses the entire stated rule. A graph makes the starting value and steady change visible.
None of these representations is the real service itself. Each is a model of the stated pricing rule. The model may have limits: perhaps x must be a whole number, or the rule applies only up to a stated capacity. The graph must respect those conditions rather than extend automatically across every possible real x.
| Input x | Calculation 2x + 3 | Output y | Point |
|---|---|---|---|
| 0 | 2(0) + 3 | 3 | (0, 3) |
| 1 | 2(1) + 3 | 5 | (1, 5) |
| 2 | 2(2) + 3 | 7 | (2, 7) |
| 3 | 2(3) + 3 | 9 | (3, 9) |
When x increases by 1, y increases by 2. When x increases by 3, y increases by 6. The change in output is a constant multiple of the change in input. The starting output is 3 rather than 0. Those two observations determine the structure of this line.
Do not mistake a table for a complete rule
A finite list of points can be consistent with more than one mathematical model. The pairs (0, 3) and (1, 5) lie on y = 2x + 3, but they also lie on some nonlinear graphs. To identify one straight line from two distinct points, you must know that the intended relationship is linear. The assumption is part of the reasoning.
In a school exercise, that assumption may be stated explicitly or established by the task asking for a line. With measured data, an apparently straight pattern is evidence to examine, not a guarantee that the same rule holds outside the observed range. Distinguish an exact mathematical relation from an approximate model fitted to observations.
Worked example 1: unequal input steps need a rate comparison
A stated linear relationship has x-values 0, 1, 3, 6 and corresponding y-values 4, 6, 10, 16. Find its equation. The output differences are 2, 4 and 6, but the input differences are 1, 2 and 3. The ratios of changes are 2/1, 4/2 and 6/3, all equal to 2.
At x = 0, the output is 4. Therefore the equation is y = 2x + 4. Looking only for equal output differences would be misleading because the input steps are not equal. The constant feature is output change per unit input change, not the raw difference between every pair of listed outputs.
3. Gradient measures change per unit change
For a non-vertical straight line, gradient m = change in y divided by change in x. Given two distinct points with different x-coordinates, m = (y₂ − y₁)/(x₂ − x₁). The subtraction order must match in numerator and denominator. A gradient is a ratio of changes, not usually the ratio of the coordinates of one point.
For y = 2x + 3, using the point (2, 7) to calculate 7/2 would give 3.5, which is not the gradient. The line does not pass through the origin. Its gradient compares two points on the line, such as (0, 3) and (2, 7): (7 − 3)/(2 − 0) = 2.
Worked example 2: calculate gradient and recover the equation
A line passes through (2, 7) and (6, 15). Find its equation. The gradient is (15 − 7)/(6 − 2) = 8/4 = 2. Write y = 2x + c. Substitute either point, for example 7 = 2(2) + c, to obtain c = 3. Hence y = 2x + 3.
Check the second point separately: 2(6) + 3 = 15. This matters because a sign or copying error in the first calculation can otherwise survive into the final equation. One point determines c once the gradient is known; the other point supplies an independent check on the resulting line.
Worked example 3: a negative gradient describes falling output
A line passes through (−2, 7) and (4, −5). Find its gradient. Using the same point order in both differences gives (−5 − 7)/(4 − (−2)) = −12/6 = −2. As x increases by 1, y decreases by 2.
The equation is y = −2x + 3, since 7 = −2(−2) + 3. A common error is to write the numerator in one order and the denominator in the other, producing a positive answer. Reversing both differences is harmless because both signs change; reversing only one changes the result incorrectly.
Zero and undefined gradients are not the same
The horizontal line y = 4 has gradient zero because its vertical change is zero while its horizontal change can be non-zero. The vertical line x = −2 has undefined gradient because its horizontal change is zero. Dividing by zero does not produce a numerical gradient.
A vertical line is still a straight line and still represents a valid relation. It simply cannot be written as y = mx + c with a finite value of m. It also does not assign one unique y-value to each x in the usual function interpretation: at x = −2, many y-values occur. This is a limitation of a representation, not a failure of the line to exist.
The intercept is a point, and its coordinate has meaning
For y = mx + c, setting x = 0 gives y = c. The y-intercept is the point (0, c); the number c is its vertical coordinate. If the graph models a starting quantity, c may represent that initial amount. However, an intercept outside the model’s valid domain may have no practical interpretation in the situation.
The x-intercept is found by setting y = 0. For y = 2x + 3, this gives x = −3/2, so the point is (−1.5, 0). Do not swap the two intercepts or assume that the constant term in any equation is automatically an x-intercept. Setting the appropriate coordinate to zero is the dependable method.
Worked example 4: the equation may hide its gradient
Find the gradient and y-intercept of 3y = 6x − 9. Divide every term by 3 to obtain y = 2x − 3. The gradient is 2 and the y-intercept is (0, −3). Reading 6 directly as the gradient would ignore the coefficient of y.
For 2x + 3y = 12, rearrange to y = −(2/3)x + 4. The gradient is −2/3, the y-intercept is (0, 4), and the x-intercept is (6, 0). The intercept method can be convenient for plotting this line because both intercepts are simple integer points.
The connection among gradient, intercept and equation is developed in OpenStax: Find the Equation of a Line. Notice that some references use b where this guide uses c; the letter changes, but the role of the constant term does not.
4. Draw a graph that preserves the numerical relationship
Start by labelling the axes and choosing scales that fit the required range. Mark the origin where relevant and show any break or truncated range clearly. Use equal numerical increments for equal physical intervals along each individual axis. The horizontal and vertical scales need not be the same, but each must be read correctly.
Calculate a small table of values, plot the points carefully and use a ruler for a straight-line relationship. Two distinct points determine a mathematical line, but a third calculated point is useful for detecting a plotting or substitution error. If the points do not align when the relationship is known to be linear, investigate before forcing a line through them.
Worked example 5: construct y = −x + 4
Choose x = 0, 2 and 4. The corresponding y-values are 4, 2 and 0, producing (0, 4), (2, 2) and (4, 0). On labelled axes with a suitable range, plot these points and draw the straight line. The intercepts supply two easy locations; the middle point checks the arithmetic and placement.
Before reading answers from the drawing, state its domain. For the unrestricted algebraic line, values beyond the plotted points are part of the relation. If the question restricts x to 0 ≤ x ≤ 4, only the corresponding line segment is relevant. A graph should show the conditions in the question rather than silently extending them.
Count coordinate units, not just squares
Suppose one horizontal square represents 2 units and one vertical square represents 5 units. A line rises two vertical squares over two horizontal squares. Its numerical gradient is 10/4 = 2.5, not 1. The physical rise and run each cover two squares, but the coordinate changes are different.
This is why the apparent steepness of two separate graphs cannot be compared reliably without their scales. The same equation can look steeper or flatter when the display proportions change. Gradient is defined by the coordinates; the drawing is a representation of that numerical relationship.
Read from the relevant axis to the line and back
To estimate y for a given x, begin at the x-value, move vertically to the graph, then horizontally to the y-axis. To estimate x for a given y, reverse the roles. A common reading error is to begin from the correct number on the wrong axis. Say what the requested quantity represents before tracing the route.
Where an exact equation is available, substitution gives an exact result subject to the exactness of the inputs. Reading from a hand-drawn graph usually gives an approximation. Do not report more precision than the scale and plotting justify. If the question specifies graphical reading, show or describe that method rather than presenting an unexplained exact algebraic value as though it came from the drawing.
Continuous quantities and whole-number counts need different treatment
A graph of cost against the number of identical notebooks may have meaningful points only at whole-number inputs. A point corresponding to 2.5 notebooks is not automatically a possible transaction. A graph of distance against elapsed time can usually use intermediate times within the stated interval. The underlying variable determines whether every point on a drawn line has a practical meaning.
A straight line may still be useful as a visual guide for a discrete model, but the permitted solutions must be restricted to the meaningful inputs. If an algebraic intersection gives 4.5 people, it is not a valid count merely because it lies on both extended lines. Return to the domain before accepting the answer.
5. Two lines create a simultaneous condition
A single linear equation in two variables usually allows many pairs. The equation x + y = 9 is satisfied by (0, 9), (4, 5) and many others. A second equation adds another condition. A simultaneous solution must satisfy both at the same time, so it belongs to both graphs.
This is the meaning of an intersection. It is not just where two pen strokes happen to cross. The coordinates at that location meet both mathematical rules. Graphical solving makes the shared condition visible; substitution and elimination can determine it exactly when the algebra is manageable.
Worked example 6: solve graphically and check algebraically
Solve x + y = 9 and 2x + y = 13. Rearrange to y = 9 − x and y = 13 − 2x. Plot both on the same axes. Their intersection is at (4, 5). Thus x = 4 and y = 5.
For an algebraic check, subtract the first original equation from the second: x = 4. Substitute into x + y = 9 to obtain y = 5. Finally test both original conditions: 4 + 5 = 9 and 2(4) + 5 = 13. Verifying only one equation would not establish that the pair is a simultaneous solution.
Worked example 7: two hypothetical charges become equal
Plan A costs 6 + 2t dollars and Plan B costs 2 + 3t dollars for t hours of use within their stated valid range. When do the costs match? Equate the expressions: 6 + 2t = 2 + 3t. This gives t = 4, and both costs are 14 dollars.
On a graph of cost against time, the point (4, 14) lies on both lines. For t below 4, Plan B is cheaper in this model; for t above 4, Plan A is cheaper. Test t = 2: A costs 10 and B costs 8. Test t = 6: A costs 18 and B costs 20. The intersection separates the comparison ranges.
The initial charge and hourly increase play different roles. Plan A begins higher but grows more slowly. Memorising that a higher line is more expensive at one location does not answer which plan is cheaper at every time. The whole relationship matters, including any limits or additional conditions not included in the simplified model.
No intersection and infinitely many intersections
The lines y = 2x + 3 and y = 2x − 1 have equal gradients and different intercepts. They are parallel and have no shared point, so the system has no solution. Algebra shows the same issue: equating them gives 3 = −1, an impossibility.
The equations y = 2x + 3 and 2y = 4x + 6 describe the same line. Every point on that line satisfies both equations, giving infinitely many real solutions unless extra restrictions are supplied. Two equations do not guarantee one unique answer; the two conditions may be independent, contradictory or equivalent.
Worked example 8: two equations describe ticket counts
A fictional event sells 36 tickets. Adult tickets cost 8 dollars and student tickets cost 5 dollars. The total is 228 dollars. Find the numbers sold. Let x be adult tickets and y be student tickets. Then x + y = 36 and 8x + 5y = 228.
Multiply the first equation by 5 to obtain 5x + 5y = 180. Subtracting gives 3x = 48, so x = 16 and y = 20. Both are non-negative integers. Check the counts and total: 16 + 20 = 36 and 8(16) + 5(20) = 228.
A graph can show the intersection, but the units differ from the previous cost-time example: both axes now represent ticket counts. The equation coefficients carry the dollars-per-ticket information. Labelling both axes cost would change the meaning of the variables and obscure the system you intended to solve.
6. Interpret the graph’s story without inventing one
Graph shape only becomes meaningful after the variables are known. A horizontal line on a distance-from-home graph means that distance is unchanged. It does not automatically prove that the object is stationary: it might move around a circle at a constant distance from home. A horizontal line on a cumulative-distance-travelled graph has a different implication under the usual idealised journey model.
Similarly, a descending line on a distance-from-station graph may mean returning towards the station. A descending line on a graph labelled total distance travelled would require investigation because cumulative travelled distance does not ordinarily decrease. Read the quantity before attaching a movement story to the shape.
Worked example 9: a piecewise distance-time graph
A model of total distance travelled uses straight segments through (0, 0), (2, 120), (5, 120) and (8, 300), where time is in minutes and distance in metres. During the first two minutes, speed is 120/2 = 60 m/min. From minute 2 to minute 5, total distance does not change, representing a three-minute stop in this idealised model. From minute 5 to minute 8, the speed is (300 − 120)/(8 − 5) = 60 m/min.
Average speed over the whole eight-minute journey is 300/8 = 37.5 m/min. Average speed while moving is 300/5 = 60 m/min. These answer different questions. The first includes the stop; the second excludes it. The graph makes the interval boundaries visible, but the denominator still has to match the requested average.
The height of the graph at minute 8 is 300 metres, not 300 metres per minute. Speed is represented by the gradient of this distance-time graph, not by its vertical coordinate. On a speed-time graph, the vertical coordinate would already be a speed. The same visual feature can carry a different meaning when the axis quantity changes.
Interpolation, extrapolation and model limits
Interpolation estimates within a known range. Extrapolation extends beyond it. If a stated constant-rate model is valid from t = 0 to t = 8, reading at t = 4 uses the model within its range. Predicting at t = 30 requires an additional assumption that the rule continues. A line on paper does not supply that assumption by itself.
For an exact unrestricted algebraic equation, evaluating another x is a mathematical operation. For a contextual model, the resulting number also needs a plausibility and domain check. A filling container cannot continue gaining water beyond its capacity under the same uncomplicated volume rule unless overflow is modelled. A number-of-customers variable cannot be negative merely because the extended line includes negative coordinates.
Worked example 10: a budget becomes a boundary
A hypothetical charge is C = 4 + 1.5n dollars for n items, with n a whole number from 0 to 20. The available amount is 25 dollars. What is the greatest permitted n? The condition is 4 + 1.5n ≤ 25, so 1.5n ≤ 21 and n ≤ 14. The greatest permitted count is 14 items.
Graphically, compare the charge line with the horizontal line C = 25 and read the crossing at n = 14. Then retain only whole-number inputs within the stated domain. The inequality describes the affordable region, not only the point at which the entire amount is used. A graph can reveal a boundary as well as a single equality solution.
What changes when m or c changes?
Compare y = 2x + 4 with y = 2x + 7. At every x, the second output is 3 greater. The gradient is unchanged, so the lines are parallel. Now compare y = 2x + 4 with y = 3x + 4. They share the point (0, 4), but their output difference is x rather than a fixed amount.
At x = 5, the second comparison differs by 5; at x = 10, it differs by 10. For negative x, the ordering reverses. This is why a larger gradient does not mean the line is above another line everywhere. Interpret the parameter and the input range together.
7. Practice: keep the same relationship across representations
Use graph paper for the drawing tasks. For the algebra tasks, show enough working to identify the gradient, intercept or simultaneous condition. When the question is contextual, state the relevant units and domain. Check answers with a different representation wherever possible: a plotted point can be tested by substitution, and an algebraic result can be located on a sketch.
Questions 1–5. 1. Describe the position of (−3, 2) relative to the origin. 2. Does (4, 11) lie on y = 2x + 3? 3. For y = 3x − 2, find y when x = −1, 0 and 2. 4. Find the gradient through (1, 4) and (5, 12). 5. Find the gradient through (−1, 6) and (3, −2).
Questions 6–10. 6. State the gradient and y-intercept of y = −4x + 9. 7. Find both intercepts of 2x + y = 10. 8. Find the equation of a line with gradient 3 passing through (2, 11). 9. Find the equation through (0, 5) and (4, 13). 10. Explain why x = 7 does not have gradient zero.
Questions 11–15. 11. Draw y = −x + 6 for 0 ≤ x ≤ 6 and mark its intercepts. 12. A graph rises three vertical squares over two horizontal squares. Each vertical square represents 4 units and each horizontal square represents 3 units. Find the gradient. 13. Solve x + y = 10 and 2x + y = 14. 14. Compare y = 2x + 1 with y = 2x + 5: can they meet? 15. Do y = 3x − 2 and 2y = 6x − 4 give one unique simultaneous solution?
Questions 16–20. 16. Two hypothetical costs are A = 5 + 4t and B = 11 + 2t. Find when they are equal and the common cost. 17. A distance-time segment runs from (2, 50) to (7, 200), with minutes and metres as units. Find its speed. 18. A charge C = 3 + 2n must not exceed 18, where n is a non-negative integer. Find the greatest possible n. 19. Is y = 4x + 2 directly proportional to x? 20. A data set contains only (0, 1) and (1, 3). Without a linearity condition, does it uniquely establish the rule y = 2x + 1?
Explained answers: questions 1–5
1. Three units left and two units up. 2. Yes: 2(4) + 3 = 11. 3. The outputs are −5, −2 and 4, giving points (−1, −5), (0, −2) and (2, 4). Keep the negative input inside brackets during substitution.
4. (12 − 4)/(5 − 1) = 2. 5. (−2 − 6)/(3 − (−1)) = −8/4 = −2. In each case, the numerator and denominator use the same order of points. The signs describe the direction of output change as x increases.
Explained answers: questions 6–10
6. Gradient −4; y-intercept (0, 9). 7. The x-intercept is (5, 0), and the y-intercept is (0, 10). Set one coordinate to zero at a time. 8. y = 3x + 5, because 11 = 3(2) + c gives c = 5.
9. The gradient is (13 − 5)/4 = 2, and the given point at x = 0 supplies the intercept, so y = 2x + 5. 10. x = 7 is vertical. Its horizontal change is zero, so the gradient quotient is undefined. A horizontal line such as y = 7 has gradient zero.
Explained answers: questions 11–15
11. The required segment joins (0, 6) and (6, 0) and passes through (3, 3). The intercepts are its endpoints. 12. The coordinate rise is 3 × 4 = 12 and the run is 2 × 3 = 6, so the gradient is 2, not 3/2.
13. Subtract the first equation from the second to obtain x = 4, then y = 6. 14. No. Equal gradients and different intercepts give distinct parallel lines. 15. No unique solution: the second equation is twice the first, so they describe the same line and have infinitely many real simultaneous solutions.
Explained answers: questions 16–20
16. 5 + 4t = 11 + 2t gives t = 3. The common cost is 17 dollars. 17. (200 − 50)/(7 − 2) = 30 m/min. 18. 3 + 2n ≤ 18 gives n ≤ 7.5. Since n is a whole number, the greatest permitted value is 7.
19. No. The constant 2 prevents the quotient y/x from remaining constant, and the line does not pass through the origin. 20. No. The proposed line fits the points, but other nonlinear relationships can also pass through them. The intended model needs an additional condition or justification.
8. Repair the exact representation that failed
If the learner plots (2, 7) at (7, 2), practise coordinate order rather than more gradient calculations. If the points are plotted correctly but the line equation is wrong, inspect slope calculation and substitution. If the equation is correct but the story is misread, return to labels, units and domain. Different failures can produce an equally wrong final answer without sharing the same cause.
A useful diagnostic uses one relationship, y = 2x + 3, in four forms. Ask the student to explain it in words, complete a table, draw it and recover the equation from two points. Compare where the reasoning first becomes unstable. The task is deliberately connected so that the difficulty cannot be hidden by moving to an unrelated example each time.
A suggested lesson sequence for mixed readiness
Begin with coordinate meaning and point membership. Follow with a simple linear table whose x-values increase by equal steps. Then vary the x-step sizes so the learner has to calculate change per unit change. Introduce y = mx + c only after the gradient and starting value have been identified in examples. This order is a teaching proposal, not a claim that every student must follow the same lesson script.
Next reverse the route: provide an equation and ask for the table, then supply two points and ask for the equation. Finally introduce a second line and ask what their shared point means. A learner who needs support can use simple integer gradients; a learner ready for greater depth can work with negative coordinates, fractional gradients or a domain that excludes an otherwise valid algebraic intersection.
Make the student choose an efficient route
For 2x + 3y = 12, intercepts are convenient for drawing. For a line through two given points, calculating gradient and then substituting is direct. For two equations with matching y-coefficients, elimination may be efficient. For a question explicitly asking for a graphical estimate, the graph is the required representation even when algebra could also solve it.
The best explanation is not simply this is the method my teacher showed. It is this information makes this route suitable. Ask the learner what the chosen representation reveals and what it hides. A graph shows comparison ranges and shape clearly; an equation can deliver exact values; a table makes selected pairings easy to inspect.
An error record for graphs should include the axes
Copying only a corrected numerical gradient may not repair a scale-reading error. Record the axis scales, the two selected points, the coordinate differences and the final quotient. If a story was misread, record the exact vertical quantity: distance from home, total distance travelled or speed. Those labels determine which interpretation is valid.
On a later return, change the scale or context while preserving the same mathematical structure. A student who learned only the visual angle of the previous line may struggle; a student who uses coordinate differences can recover the gradient. This makes the return task a test of the relationship rather than recognition of an old picture.
Questions parents can use without drawing the graph for the learner
Ask what each axis measures, what one square represents and what a chosen point says in a sentence. Ask whether the answer should be a coordinate pair or one coordinate. For an intersection, ask which two conditions the answer satisfies. These questions make the student’s thinking visible without supplying the calculation.
Do not judge understanding by visual neatness alone. A beautiful line on unlabelled axes may communicate very little. An imperfect sketch with correct labels, a sound equation and a clear domain may reveal stronger reasoning. Accurate drawing still matters, but it must support rather than replace mathematical meaning.
9. A final transfer challenge
A fictional water container begins with 12 litres and gains water at a constant 3 litres per minute for the first eight minutes, without leakage or overflow. Write an equation, identify the gradient and intercept, find the amount after five minutes, and find when the amount first reaches 30 litres. State the valid time domain.
The equation is V = 12 + 3t for 0 ≤ t ≤ 8. The gradient is 3 L/min and the vertical intercept is (0, 12), representing the starting volume. At t = 5, V = 27 L. Setting V = 30 gives t = 6 minutes, which lies inside the stated domain.
Now change only the starting volume to 18 litres. The new graph is V = 18 + 3t, parallel to the first, and it reaches 30 litres after four minutes. Change only the inflow rate instead to 6 L/min while retaining the original 12 litres: V = 12 + 6t, reaching 30 litres after three minutes. Each condition changes a different part of the equation and a different feature of the graph.
This is the point of connected learning. The equation is not a code to memorise, the table is not busywork, and the graph is not decoration. Each representation allows a different question to be answered while preserving the same quantitative relationship.
Return to algebraic structural control when rearrangement or substitution needs repair. Continue to ratio, proportion, rate and percentage to distinguish constant rate from direct proportion. Use geometry and similarity when coordinates help describe spatial conditions.
Read the axes. Preserve the ordered pair. Calculate change in coordinate units. Interpret the answer inside the original conditions.
