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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.

Feeling shaky? That's completely okay — it just means you're about to fill in a gap that will make the whole module easier. Read the plain-language explanation, follow the worked example step by step, then try one yourself. The answers are hidden so you can think first — click Show answer whenever you're ready. You can do this.

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Skills to Build First

Five small building blocks. Master these and Module 2 will feel like the next natural step, not a wall.

Skill 01

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 \((3,\,2)\)

  1. Start at the origin \((0,\,0)\) — the center where the two axes cross.
  2. The first number is \(3\). Move 3 units right along the x-axis (right because it's positive).
  3. The second number is \(2\). From there, move 2 units up (up because it's positive).
  4. 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."

Try it

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.

Show answer
1. The \(x\)-value is \(-2\), so move 2 units left first (left because it's negative). Then the \(y\)-value is \(4\), so move 4 units up. The dot lands in the upper-left region.

2. Left means negative \(x\): \(-5\). Down means negative \(y\): \(-3\). So the point is \((-5,\,-3)\).
Skill 02

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 \((4,\,8)\)

  1. Find the rise (vertical change): subtract the \(y\)-values. \(8 - 2 = 6\). The line goes up 6.
  2. Find the run (horizontal change): subtract the \(x\)-values in the same order. \(4 - 1 = 3\). The line goes right 3.
  3. Make the ratio: \(\text{slope} = \dfrac{\text{rise}}{\text{run}} = \dfrac{6}{3}\).
  4. 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.

Try it

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?

Show answer
1. Rise \(= 7 - 1 = 6\). Run \(= 2 - 0 = 2\). Slope \(= \dfrac{6}{2} = 3\). The line climbs 3 up for every 1 right.

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.
Skill 03

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 \(2x + 3 = 11\)

  1. Goal: get \(x\) alone. First undo the \(+3\) by subtracting 3 from both sides: \(2x + 3 - 3 = 11 - 3\).
  2. Simplify: \(2x = 8\).
  3. Now undo the \(\times 2\) by dividing both sides by 2: \(\dfrac{2x}{2} = \dfrac{8}{2}\).
  4. 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.

Try it

1. Solve the one-step equation \(x - 5 = 9\).

2. Solve the two-step equation \(3x - 4 = 14\).

Show answer
1. Undo the \(-5\) by adding 5 to both sides: \(x - 5 + 5 = 9 + 5\), so \(x = 14\). Check: \(14 - 5 = 9\). True.

2. Add 4 to both sides: \(3x = 18\). Divide both sides by 3: \(x = 6\). Check: \(3(6) - 4 = 18 - 4 = 14\). True.
Skill 04

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 \(x = 4\)

  1. Rewrite the expression and substitute the value: replace \(x\) with \(4\), using parentheses: \(3(4) + 5\).
  2. Do multiplication first (order of operations): \(3 \times 4 = 12\), giving \(12 + 5\).
  3. Then add: \(12 + 5 = 17\).
  4. 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.

Try it

1. Evaluate \(2x - 1\) when \(x = 6\).

2. Evaluate \(-2x + 7\) when \(x = -3\).

Show answer
1. Substitute: \(2(6) - 1\). Multiply: \(12 - 1\). Subtract: \(11\). So the value is \(11\).

2. Substitute: \(-2(-3) + 7\). A negative times a negative is positive: \(-2 \times -3 = 6\), giving \(6 + 7 = 13\). So the value is \(13\).
Skill 05

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 \(y = kx\)

  1. Find the constant of proportionality \(k\): the cost per apple is \(\$2\), so \(k = 2\).
  2. Write the relationship with \(y\) = total cost and \(x\) = number of apples: \(y = 2x\).
  3. Use it — how much for 5 apples? Substitute \(x = 5\): \(y = 2(5)\).
  4. 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 \(k = \dfrac{16}{8} = 2\).

Try it

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 \(y = kx\).

Show answer
1. The constant rate is \(60\), so \(k = 60\) and \(y = 60x\). After 4 hours: \(y = 60(4) = 240\) miles.

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.

  1. Plot — in your head or on paper — the point \((-3,\,1)\). Which direction do you move first?
    Show answer
    Move 3 units left first (negative \(x\)), then 1 unit up (positive \(y\)).
  2. Find the slope of the line through \((2,\,3)\) and \((6,\,11)\).
    Show answer
    Rise \(= 11 - 3 = 8\); run \(= 6 - 2 = 4\); slope \(= \dfrac{8}{4} = 2\).
  3. Solve \(5x + 2 = 17\).
    Show answer
    Subtract 2: \(5x = 15\). Divide by 5: \(x = 3\).
  4. Solve \(x + 9 = 4\).
    Show answer
    Subtract 9 from both sides: \(x = 4 - 9 = -5\).
  5. Evaluate \(4x - 3\) when \(x = -2\).
    Show answer
    \(4(-2) - 3 = -8 - 3 = -11\).
  6. If \(y = kx\) and \(y = 21\) when \(x = 3\), find \(k\), then find \(y\) when \(x = 10\).
    Show answer
    \(k = \dfrac{21}{3} = 7\), 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.


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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 \(y = kx\)


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When You're Ready →

No rush. When the skills above feel solid, step into Module 2.

Module 2 Overview

The full picture of Exploring Constant Change — vocabulary, standards, and what you'll learn.

Go to Module 2 →

The Visual Lab

Drag the sliders and see slope and intercepts come to life in the interactive Linear Function Lab.

Open the Visual Lab

Course Syllabus

Policies, the studio learning environment, and the full Algebra I itinerary by grading period.

Read the Syllabus

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.