Get Ready: Module 2 — Exploring Constant Change
Everyone starts somewhere. Before we explore lines and rates of change, let's build the floor before the building — one small skill at a time. Go at your own pace; there's no rush and nothing to prove here.
Skills to Build First
Five small building blocks. Master these and Module 2 will feel like the next natural step, not a wall.
Plotting Points on the Coordinate Plane
What it is & why you need it: A point is written \((x,\,y)\). The first number tells you how far to go left/right; the second tells you how far up/down. In Module 2, every line is just a trail of points — if you can plot points, you can graph a line.
Worked example: plot \((3,\,2)\)
- Start at the origin \((0,\,0)\) — the center where the two axes cross.
- The first number is \(3\). Move 3 units right along the x-axis (right because it's positive).
- The second number is \(2\). From there, move 2 units up (up because it's positive).
- Make a dot. That dot is the point \((3,\,2)\). Done!
Remember the order: \((x,\,y) = (\text{across},\ \text{up})\) — "you crawl across the floor before you climb up the stairs."
1. Plot the point \((-2,\,4)\). Which direction do you move first, and how far?
2. A point sits 5 to the left and 3 down from the origin. Write its coordinates.
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2. Left means negative \(x\): \(-5\). Down means negative \(y\): \(-3\). So the point is \((-5,\,-3)\).
Slope: Steepness as Rise over Run
What it is & why you need it: Slope measures how steep a line is — how much it climbs for each step you take to the right. We measure it as \(\text{slope} = \dfrac{\text{rise}}{\text{run}}\) (up-or-down over left-or-right). Module 2 is literally called "Exploring Constant Change," and slope is that constant change.
Worked example: find the slope between \((1,\,2)\) and \((4,\,8)\)
- Find the rise (vertical change): subtract the \(y\)-values. \(8 - 2 = 6\). The line goes up 6.
- Find the run (horizontal change): subtract the \(x\)-values in the same order. \(4 - 1 = 3\). The line goes right 3.
- Make the ratio: \(\text{slope} = \dfrac{\text{rise}}{\text{run}} = \dfrac{6}{3}\).
- Simplify: \(\dfrac{6}{3} = 2\). The line rises 2 units for every 1 unit right.
Quick read: a positive slope climbs uphill (left to right), a negative slope goes downhill, and a bigger number means a steeper climb.
1. A line goes through \((0,\,1)\) and \((2,\,7)\). What is its slope?
2. A ramp rises 4 feet over a horizontal run of 8 feet. What is its slope, simplified?
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2. Slope \(= \dfrac{\text{rise}}{\text{run}} = \dfrac{4}{8} = \dfrac{1}{2}\). The ramp rises 1 foot for every 2 feet across — a gentle slope.
Solving One- and Two-Step Equations
What it is & why you need it: Solving an equation means finding the value of the unknown (usually \(x\)) that makes it true. The golden rule: whatever you do to one side, do to the other — like keeping a balance scale even. In Module 2 you'll rearrange line equations and find intercepts, which is exactly this skill.
Worked example: solve \(2x + 3 = 11\)
- Goal: get \(x\) alone. First undo the \(+3\) by subtracting 3 from both sides: \(2x + 3 - 3 = 11 - 3\).
- Simplify: \(2x = 8\).
- Now undo the \(\times 2\) by dividing both sides by 2: \(\dfrac{2x}{2} = \dfrac{8}{2}\).
- Simplify: \(x = 4\). Check: \(2(4) + 3 = 8 + 3 = 11\). True!
Undo operations in reverse order: peel off addition/subtraction first, then multiplication/division — the opposite of the order of operations.
1. Solve the one-step equation \(x - 5 = 9\).
2. Solve the two-step equation \(3x - 4 = 14\).
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2. Add 4 to both sides: \(3x = 18\). Divide both sides by 3: \(x = 6\). Check: \(3(6) - 4 = 18 - 4 = 14\). True.
Evaluating Expressions
What it is & why you need it: Evaluating means plugging a number in for the variable and simplifying to a single value. In Module 2 you'll plug \(x\)-values into \(y = mx + b\) to build a table of points — that's evaluating, over and over.
Worked example: evaluate \(3x + 5\) when \(x = 4\)
- Rewrite the expression and substitute the value: replace \(x\) with \(4\), using parentheses: \(3(4) + 5\).
- Do multiplication first (order of operations): \(3 \times 4 = 12\), giving \(12 + 5\).
- Then add: \(12 + 5 = 17\).
- So when \(x = 4\), the expression equals \(17\).
Tip: always wrap the substituted number in parentheses — it keeps multiplication and negative signs from tripping you up.
1. Evaluate \(2x - 1\) when \(x = 6\).
2. Evaluate \(-2x + 7\) when \(x = -3\).
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2. Substitute: \(-2(-3) + 7\). A negative times a negative is positive: \(-2 \times -3 = 6\), giving \(6 + 7 = 13\). So the value is \(13\).
Proportional Relationships \(y = kx\)
What it is & why you need it: Two quantities are proportional when one is always a constant number times the other: \(y = kx\), where \(k\) is the constant of proportionality (the unit rate). This is the simplest line of all — it passes through the origin — and it's the doorway to the full \(y = mx + b\) lines in Module 2. Here \(k\) plays the same role the slope \(m\) will.
Worked example: apples cost \(\$2\) each — write and use \(y = kx\)
- Find the constant of proportionality \(k\): the cost per apple is \(\$2\), so \(k = 2\).
- Write the relationship with \(y\) = total cost and \(x\) = number of apples: \(y = 2x\).
- Use it — how much for 5 apples? Substitute \(x = 5\): \(y = 2(5)\).
- Simplify: \(y = 10\). Five apples cost \(\$10\).
To find \(k\) from a table or point, divide: \(k = \dfrac{y}{x}\). For example, if \(8\) apples cost \(\$16\), then \(k = \dfrac{16}{8} = 2\).
1. A car travels at a constant 60 miles per hour. Write the proportional equation for distance \(y\) after \(x\) hours, then find the distance after 4 hours.
2. A table shows that \(3\) notebooks cost \(\$12\). Find the constant of proportionality \(k\) and write \(y = kx\).
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2. \(k = \dfrac{y}{x} = \dfrac{12}{3} = 4\) (each notebook is \(\$4\)). So \(y = 4x\).
Quick Readiness Check
Six short questions across all five skills. Try each one first, then peek at the answer.
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Plot — in your head or on paper — the point \((-3,\,1)\). Which direction do you move first?
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Move 3 units left first (negative \(x\)), then 1 unit up (positive \(y\)). -
Find the slope of the line through \((2,\,3)\) and \((6,\,11)\).
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Rise \(= 11 - 3 = 8\); run \(= 6 - 2 = 4\); slope \(= \dfrac{8}{4} = 2\). -
Solve \(5x + 2 = 17\).
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Subtract 2: \(5x = 15\). Divide by 5: \(x = 3\). -
Solve \(x + 9 = 4\).
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Subtract 9 from both sides: \(x = 4 - 9 = -5\). -
Evaluate \(4x - 3\) when \(x = -2\).
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\(4(-2) - 3 = -8 - 3 = -11\). -
If \(y = kx\) and \(y = 21\) when \(x = 3\), find \(k\), then find \(y\) when \(x = 10\).
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\(k = \dfrac{21}{3} = 7\), so \(y = 7x\). When \(x = 10\): \(y = 7(10) = 70\).
If these feel comfortable, you're ready for the module. If a couple felt tricky, that's fine — revisit that skill card above and drill a little with the free links below. No shame, just steps.
Practice More — Free
Want extra reps? These free sites let you drill each skill until it clicks. (Links open in a new tab.)
Plotting points
Slope (rise over run)
One- & two-step equations
Evaluating expressions
Proportional relationships \(y = kx\)
When You're Ready →
No rush. When the skills above feel solid, step into Module 2.
Instructor: Dr. Goodluck Ijezie-Desbois, PharmD · Beta Academy · Room: TBA
Foundations support for the TEA Bluebonnet Learning — Secondary Mathematics open curriculum,
licensed CC BY-NC 4.0. Non-commercial classroom use.