Slope, linear equations and systems
Graphing proportional relationships, finding slope and the equation of a line, solving linear equations, and solving systems of two equations
About five lessons of 45 to 60 minutes
Two phone plans. Which one is actually cheaper?
Plan A costs $20 a month plus 5 cents a text. Plan B costs $10 a month plus 7 cents a text. For a light texter, B looks cheaper. For a heavy texter, A wins. Somewhere in between, the two plans cost exactly the same, a single number of texts where the two lines describing each plan cross.
Slope is the idea underneath all of this: how steeply a quantity changes. Every straight-line relationship, a phone bill, a savings account, a car's fuel level, can be described by its slope and starting value, solved for an unknown with a linear equation, or compared to another line by solving a system.
- Comparing two phone plansa system of two linear equations, solved by finding where they cost the same
- A taxi's fare: a flat fee plus a rate per kilometreslope is the per-kilometre rate, the y-intercept is the flat fee
- A pool filling at a constant ratea proportional relationship graphing as a line straight through the origin
- "I'm thinking of a number" style puzzlessolved with the same equation-solving moves used throughout this unit
What students will be able to do
Students will graph a proportional relationship and interpret its slope as the unit rate, find the slope and equation of a line from two points, solve one-variable linear equations including those with variables on both sides, and solve a system of two linear equations algebraically and by reasoning about where two lines meet.
- I can graph a proportional relationship and identify its slope as the unit rate.
- I can find the slope of a line from two points, and write its equation in the form y = mx + b.
- I can solve a linear equation with variables on both sides, including cases with no solution or infinitely many solutions.
- I can solve a system of two linear equations using substitution.
- I can explain why the solution of a system is the point where the two lines cross.
Standards this unit teaches
- 8.EE.B.5Common Core (US)Graph proportional relationships
Graph proportional relationships and interpret the unit rate as the slope of the line.
- 8.EE.B.6Common Core (US)Slope and the equation of a line
Use similar triangles to explain why slope is constant and derive the equations y equals mx and y equals mx plus b.
- 8.EE.C.7Common Core (US)Solve linear equations
Solve linear equations in one variable, including those with variables on both sides.
- 8.EE.C.8Common Core (US)Solve systems of equations
Analyse and solve pairs of simultaneous linear equations algebraically and by graphing.
Printable anchor chart
Print a wall poster to go with this unit, a code-drawn diagram students can point to during and after the lesson.
Prior knowledge
This unit builds on skills students should already have met. Revisit any that are shaky first.
- Grade 8 linear equations & expressions worksheetsolving one- and two-step equations, the direct foundation for section 3
- Ratios, rates & scale worksheetunit rates, which are exactly what slope measures in a proportional relationship
- Equation in the glossarya refresher on what it means for a value to make an equation true
Words to teach and display
- Proportional relationship
- a relationship of the form y = kx, whose graph is a straight line through the origin
- Slope
- how steeply a line rises or falls; the change in y divided by the change in x between any two points on it
- Unit rate
- the amount of one quantity per single unit of another; in a proportional relationship, this is the slope
- Linear equation
- an equation whose variable appears only to the first power, solved by undoing operations in reverse
- System of equations
- two or more equations considered together, whose solution is the point (or points) that make every equation true at once
- No solution / infinitely many solutions
- special outcomes of solving a linear equation: no value works (a false statement remains), or every value works (a true statement remains)
Teach it: concrete, pictorial, abstract
The lesson moves from things students can hold, to pictures and diagrams, to the written maths. The diagrams below are drawn from data, so they are accurate and print cleanly. Teach straight from them.
1. Graphing proportional relationships and slope as unit rate
ConcreteA proportional relationship has the form y = kx: double the input and the output doubles too, and its graph is always a straight line through the origin (0,0). The steepness of that line, its slope, is exactly the unit rate, k.
A car travels at a constant rate: 300 miles on 10 gallons of gas is a unit rate of 30 miles per gallon, so the relationship is y = 30x, where x is gallons and y is miles. Every point on its graph, (1,30), (2,60), (3,90), reflects that same steady rate.
A pool fills at a constant rate: after 4 minutes it holds 60 litres. Write the proportional relationship and find how much it holds after 7 minutes.
- Find the unit rate: 60 litres in 4 minutes is 60/4 = 15 litres per minute.
- Write the relationship: y = 15x, where x is minutes and y is litres.
- Substitute x = 7: y = 15(7) = 105.
Answer: y = 15x. After 7 minutes, the pool holds 105 litres.
- Why does a proportional relationship's graph always pass through the origin?
- If two proportional relationships have different unit rates, how does that show up on their graphs?
2. Slope and the equation of a line
PictorialFor any two points on a straight line, the slope is (change in y) / (change in x). This ratio comes out the same no matter which two points you pick, because every little triangle formed by rise and run along the line is a scaled copy of every other, a similar triangle, so their side ratios match.
Points (1,3) and (4,9) give slope (9-3)/(4-1) = 6/3 = 2. To find the full equation y = mx + b, substitute one point into y = 2x + b and solve for b: 3 = 2(1) + b, so b = 1. The line is y = 2x + 1.
Find the equation of the line through (1,3) and (4,9).
- Find the slope: m = (9 - 3) / (4 - 1) = 6/3 = 2.
- Substitute the point (1,3) into y = 2x + b: 3 = 2(1) + b, so 3 = 2 + b.
- Solve for b: b = 1.
- Write the equation: y = 2x + 1.
Answer: y = 2x + 1.
- Why does the slope come out the same no matter which two points on the line you use?
- Once you know the slope, what is the one extra piece of information needed to write the full equation?
3. Solving linear equations with variables on both sides
AbstractWhen a variable appears on both sides of an equation, collect it onto one side first, by adding or subtracting a variable term from both sides, then solve as usual.
For 5x + 3 = 2x + 18: subtract 2x from both sides to get 3x + 3 = 18, then subtract 3 to get 3x = 15, then divide by 3 to get x = 5. Two special results can also happen: if every variable term cancels and a TRUE statement remains (like 6 = 6), every value of x works (infinitely many solutions); if a FALSE statement remains (like 5 = 2), no value of x works (no solution).
Solve 5x + 3 = 2x + 18, then check the answer.
- Subtract 2x from both sides: 5x - 2x + 3 = 2x - 2x + 18, so 3x + 3 = 18.
- Subtract 3 from both sides: 3x = 15.
- Divide both sides by 3: x = 5.
- Check: 5(5) + 3 = 25 + 3 = 28. 2(5) + 18 = 10 + 18 = 28. Both sides match.
Answer: x = 5.
- Why is it fine to subtract 2x from both sides, exactly like subtracting a number?
- If solving an equation ends with a false statement like 5 = 2, what does that tell you about the solution?
4. Solving systems of equations
AbstractTwo linear equations considered together form a system. Graphically, the solution is the point where the two lines CROSS, since that is the only (x,y) pair that makes both equations true at once. Algebraically, substitution finds that same point without graphing.
For y = x + 2 and y = 3x - 4: since both equal y, set them equal to each other: x + 2 = 3x - 4. Solve for x, then substitute back to find y. Back to the hook: Plan A costs 20 + 0.05m dollars and Plan B costs 10 + 0.07m dollars, where m is the number of texts; setting them equal finds the break-even number of texts where both plans cost the same.
Plan A costs $20 a month plus 5 cents a text: cost = 20 + 0.05m. Plan B costs $10 a month plus 7 cents a text: cost = 10 + 0.07m. Find the number of texts, m, where both plans cost the same, and what that cost is.
- Set the two expressions equal, since both represent the same cost: 20 + 0.05m = 10 + 0.07m.
- Subtract 0.05m from both sides: 20 = 10 + 0.02m.
- Subtract 10 from both sides: 10 = 0.02m.
- Divide both sides by 0.02: m = 500.
- Find the shared cost by substituting into either plan: 20 + 0.05(500) = 20 + 25 = 45. Check Plan B: 10 + 0.07(500) = 10 + 35 = 45. Both match.
Answer: The plans cost the same at 500 texts, both costing $45.
- Why must the solution to a system make BOTH original equations true, not just one?
- On a graph, what would it mean if two lines never crossed at all?
Common misconceptions and how to address them
MisconceptionWhen solving an equation with variables on both sides, you should move the variable to whichever side has more of it, but the operation direction does not matter.
Why it happens: Students focus on which side 'has more x's' rather than consistently subtracting the same variable term from both sides.
How to address it: Pick either variable term to eliminate and subtract it from BOTH sides, every time, the same balance rule used throughout equation solving. Show that starting from either side of 5x + 3 = 2x + 18 (subtracting 2x, or subtracting 5x instead) reaches the same solution.
MisconceptionA steeper line always represents a bigger starting value, not just a bigger rate.
Why it happens: Students conflate the visual impression of 'a bigger line' with either slope or intercept, without distinguishing the two properties.
How to address it: Steepness is ONLY about slope (the rate of change); where a line crosses the y-axis is a completely separate property (the initial value). Compare two lines that cross, one steeper but starting lower, to show these two properties vary independently.
MisconceptionReaching a result like '6 = 6' or '5 = 2' while solving an equation means you made a mistake.
Why it happens: Students expect every equation to end with 'x = a number' and assume any other outcome is an error to fix.
How to address it: These are valid, meaningful outcomes: a true statement with no variable left (like 6 = 6) means every x works (infinitely many solutions); a false statement (like 5 = 2) means no x works (no solution). Both are correct answers, not mistakes.
MisconceptionSolving a system means solving each equation completely separately for x, with no connection between them.
Why it happens: Students have practised solving single equations so much that they do not see why the two equations in a system need to be combined.
How to address it: The whole point of substitution is that BOTH equations describe the same y (or the same relationship), so setting the two expressions for y equal to each other uses information from both equations at once, which is exactly why it finds the point where both are true.
MisconceptionForgetting to distribute across parentheses before combining like terms, such as treating 2(x + 3) as 2x + 3.
Why it happens: Students apply the multiplication to only the first term inside the parentheses and stop.
How to address it: Every term inside the parentheses must be multiplied by the number outside: 2(x + 3) = 2x + 6, not 2x + 3. Rewrite the distributed form as its own explicit step before combining anything else.
Guided practice (with answers)
1. A runner covers 5 metres every second at a constant pace. Write the proportional relationship and find the distance after 12 seconds.
Answer: y = 5x, and after 12 seconds: y = 5(12) = 60 metres.
2. Find the slope of the line through (2,4) and (5,13).
Answer: 3, because (13-4)/(5-2) = 9/3 = 3.
3. Solve 4x - 7 = x + 8.
Answer: x = 5, because subtracting x from both sides gives 3x - 7 = 8, then adding 7 gives 3x = 15, then dividing by 3 gives x = 5.
4. Solve 3(x + 2) = 3x + 6, and explain the result.
Answer: Every value of x works (infinitely many solutions), because distributing gives 3x + 6 = 3x + 6, which is true no matter what x is.
5. Solve the system y = 2x + 1 and y = 5x - 8.
Answer: x = 3, y = 7, because setting 2x + 1 = 5x - 8 gives 9 = 3x, so x = 3, and substituting back gives y = 2(3) + 1 = 7 (checked in the other equation: 5(3) - 8 = 7, which matches).
Independent practice worksheets
Practise slope, linear equations and systems with computed, never-wrong answer keys.
Differentiation
- Start slope calculations with points that already have a positive, whole-number slope before introducing negative or fractional slopes.
- Provide a 'balance scale' visual reminder that whatever is done to one side of an equation must be done to the other, before tackling variables on both sides.
- For systems, always write out both original equations side by side, and underline the shared variable (y) that lets substitution work.
- Use a two-column table (Plan A cost, Plan B cost) filled in row by row for small numbers of texts, so students can SEE the costs meet before solving algebraically.
- Explore what a system with no solution looks like graphically (parallel lines) and algebraically (a false statement when solved).
- Explore what a system with infinitely many solutions looks like (the same line written two different ways).
- Solve a system using elimination (adding or subtracting the two equations) instead of substitution, and compare the two methods.
- Investigate real break-even problems with three or more options (three phone plans, three delivery services) and determine which is cheapest across different ranges of usage.
Assessment: exit ticket
A three-question exit ticket sampling slope, solving equations, and systems.
1. Find the slope of the line through (0,2) and (3,11).
Answer: 3, because (11-2)/(3-0) = 9/3 = 3.
2. Solve 6x + 5 = 3x + 20.
Answer: x = 5, because subtracting 3x gives 3x + 5 = 20, then subtracting 5 gives 3x = 15, then dividing by 3 gives x = 5.
3. Solve the system y = x + 4 and y = 2x + 1.
Answer: x = 3, y = 7, because setting x + 4 = 2x + 1 gives 3 = x, so x = 3, and y = 3 + 4 = 7 (checked: 2(3) + 1 = 7, which matches).
Teacher notes and timings
- Rough timing across five lessons: Lesson 1 proportional relationships (section 1), Lesson 2 slope and the equation of a line (section 2), Lesson 3 solving equations with variables on both sides (section 3), Lesson 4 systems of equations (section 4), Lesson 5 mixed practice plus the exit ticket.
- This unit assumes comfort with one- and two-step equation solving (Grade 7-8 linear equations). Revisit that first if isolating a single variable term already feels shaky before adding the 'variables on both sides' complication.
- Language to repeat: slope is rise over run, the same for any two points on a line; whatever you do to one side of an equation, do to the other; a system's solution is the point that satisfies BOTH equations, exactly where two lines cross.
- Curriculum note: 8.EE.B.5 and 8.EE.B.6 (Common Core) cover proportional relationships and slope; 8.EE.C.7 covers solving linear equations; 8.EE.C.8 covers systems. All four are grouped here since they form one connected story, from a line's slope, to solving for an unknown point on it, to comparing two lines at once.
- Print and present both work: use the Print button for a clean handout, or work the phone-plan break-even problem live on the board, letting students predict which plan wins before and after the crossover point.