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Get Ready: Module 3 — Modeling Linear Equations & Inequalities

Everyone starts somewhere. Before we put up the building, we pour the floor — the few skills that make the whole module click into place. Take it one card at a time, at your own pace. There is no clock here.

How to use this page: work through the four skill cards below in order. Each one tells you what it is, shows you a fully worked example, then gives you a problem or two to try. The answers are hidden — open them after you try. If a card already feels easy, skim it and move on. When the cards feel comfortable, you are ready for the module.
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Skills to build first

Four building blocks. Master these and Module 3 stops feeling like a wall and starts feeling like a staircase.

Skill 1

Solving multi-step linear equations

What it is: finding the one value of \(x\) that makes an equation true, when it takes more than one move — distributing, combining, then undoing addition and multiplication. Why you need it for this module: a system of equations is just two of these at once. To find where two lines cross, you set them equal and solve a multi-step equation. If solving for \(x\) is shaky, the intersection point will feel like magic instead of math.

Worked example

Solve \(\;3(x - 2) + 5 = 2x + 7\;\)

  1. Distribute the 3 across the parentheses: \(\;3x - 6 + 5 = 2x + 7\)
  2. Combine like terms on the left (\(-6 + 5 = -1\)): \(\;3x - 1 = 2x + 7\)
  3. Get the variables together. Subtract \(2x\) from both sides: \(\;x - 1 = 7\)
  4. Undo the subtraction. Add \(1\) to both sides: \(\;x = 8\)
  5. Check: \(3(8-2)+5 = 18+5 = 23\) and \(2(8)+7 = 23\). Both sides match. ✓

Try it

(a) Solve \(\;2(x + 4) = 3x - 1\)

(b) Solve \(\;5x - 3 = 2x + 12\)

Show answer

(a) Distribute: \(2x + 8 = 3x - 1\). Subtract \(2x\): \(8 = x - 1\). Add \(1\): \(x = 9\). Check: \(2(9+4)=26\) and \(3(9)-1=26\). ✓

(b) Subtract \(2x\): \(3x - 3 = 12\). Add \(3\): \(3x = 15\). Divide by \(3\): \(x = 5\). Check: \(5(5)-3=22\) and \(2(5)+12=22\). ✓

Skill 2

Graphing a line from slope-intercept form

What it is: reading a line written as \(y = mx + b\) and drawing it — \(b\) is where it crosses the \(y\)-axis (the starting point), and \(m\) is the slope (rise over run, how steep and which way it tilts). Why you need it for this module: the whole lab is about where two lines meet. You cannot find the meeting point if you cannot place each line. Plotting from \(y=mx+b\) is the single most-used move in Module 3.

Worked example

Graph \(\;y = \tfrac{2}{3}x + 1\)

  1. Plot the \(y\)-intercept. Here \(b = 1\), so start at the point \((0, 1)\).
  2. Read the slope. \(m = \tfrac{2}{3}\) means rise \(2\), run \(3\).
  3. Step off the slope. From \((0,1)\) go up \(2\) and right \(3\) to land on \((3, 3)\).
  4. Repeat or go backward. From \((0,1)\) you can also go down \(2\), left \(3\) to \((-3,-1)\).
  5. Draw the line straight through those points and extend it both ways.

Try it

(a) For \(\;y = -2x + 4\), name the \(y\)-intercept and the slope, then give one more point on the line.

(b) A line has slope \(\tfrac{1}{2}\) and passes through \((0,-3)\). Write its equation and find the point you reach by stepping the slope once.

Show answer

(a) \(y\)-intercept \((0, 4)\); slope \(m = -2 = \tfrac{-2}{1}\) (down \(2\), right \(1\)). Stepping once from \((0,4)\) gives \((1, 2)\).

(b) \(y = \tfrac{1}{2}x - 3\). From \((0,-3)\), up \(1\) and right \(2\) lands on \((2, -2)\).

Skill 3

Combining like terms & the distributive property

What it is: tidying an expression. Like terms have the same variable part, so \(3x\) and \(5x\) add to \(8x\) (but \(3x\) and \(5\) cannot combine). Distributing means multiplying what is outside the parentheses by everything inside: \(a(b+c) = ab + ac\). Why you need it for this module: real-world equations almost always show up messy. Before you can solve or graph, you have to clean them up — and that is exactly these two moves.

Worked example

Simplify \(\;4(2x + 3) - (x - 5)\)

  1. Distribute the 4: \(4 \cdot 2x + 4 \cdot 3 = 8x + 12\).
  2. Distribute the subtraction (a hidden \(-1\)) across \((x - 5)\): \(-x + 5\).
  3. Write it all together: \(8x + 12 - x + 5\).
  4. Combine like terms: the \(x\)'s: \(8x - x = 7x\); the numbers: \(12 + 5 = 17\).
  5. Result: \(7x + 17\).

Try it

(a) Simplify \(\;2(3x - 1) + 4x\)

(b) Simplify \(\;5 - 2(x + 4)\)

Show answer

(a) Distribute: \(6x - 2 + 4x\). Combine the \(x\)'s: \(6x + 4x = 10x\). Result: \(10x - 2\).

(b) Distribute the \(-2\): \(5 - 2x - 8\). Combine the numbers: \(5 - 8 = -3\). Result: \(-2x - 3\).

Skill 4

Inequalities and number lines

What it is: solving statements like \(x > 3\) that describe a range of answers, not one number, and showing them on a number line (open circle for \(<\) or \(>\), filled circle for \(\le\) or \(\ge\)). The one trap: flip the inequality sign when you multiply or divide by a negative. Why you need it for this module: Module 3 turns single-variable inequalities into shaded regions on the plane. Getting the direction and the circle right on a number line is the warm-up for shading the correct side of a line later.

Worked example

Solve and graph \(\;-2x + 1 \le 7\)

  1. Undo the \(+1\). Subtract \(1\) from both sides: \(-2x \le 6\).
  2. Divide by \(-2\) — and because it is negative, flip the sign: \(x \ge -3\).
  3. Graph it: a filled circle at \(-3\) (because \(\ge\) includes \(-3\)), shaded to the right toward larger numbers.
  4. Sanity check: try \(x = 0\): \(-2(0)+1 = 1\), and \(1 \le 7\) is true, and \(0 \ge -3\) is true. ✓

Try it

(a) Solve \(\;3x - 4 < 5\) and describe its number-line graph.

(b) Solve \(\;-x + 2 \ge 6\) and describe its number-line graph.

Show answer

(a) Add \(4\): \(3x < 9\). Divide by \(3\) (positive, no flip): \(x < 3\). Graph: open circle at \(3\), shaded left.

(b) Subtract \(2\): \(-x \ge 4\). Divide by \(-1\) and flip: \(x \le -4\). Graph: filled circle at \(-4\), shaded left.

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Quick readiness check

Five short questions across all four skills. Try each in your head or on paper, then open the answer. If these feel comfortable, you're ready for the module.

  1. Solve: \(\;4x + 3 = 2x + 11\).
    Show answer

    Subtract \(2x\): \(2x + 3 = 11\). Subtract \(3\): \(2x = 8\). Divide: \(\mathbf{x = 4}\).

  2. For \(\;y = -3x + 2\), what is the \(y\)-intercept and the slope?
    Show answer

    \(y\)-intercept \((0, 2)\); slope \(\mathbf{m = -3}\) (down \(3\), right \(1\)).

  3. Simplify: \(\;3(2x - 1) + 4\).
    Show answer

    Distribute: \(6x - 3 + 4\). Combine: \(\mathbf{6x + 1}\).

  4. Solve and describe the graph: \(\;-2x \ge 10\).
    Show answer

    Divide by \(-2\) and flip: \(\mathbf{x \le -5}\). Filled circle at \(-5\), shaded left.

  5. Two lines: \(y = x + 1\) and \(y = -x + 5\). At the point where they meet, both \(y\)-values are equal — set them equal and solve for \(x\).
    Show answer

    \(x + 1 = -x + 5\). Add \(x\): \(2x + 1 = 5\). Subtract \(1\): \(2x = 4\), so \(\mathbf{x = 2}\) (and \(y = 3\)). They meet at \((2, 3)\) — a preview of Module 3!

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Practice more (free)

Want to drill a skill until it feels automatic? These free sites have endless practice. Pick the skill you want to strengthen.

Solving multi-step equations

Khan Academy → IXL →

Graphing slope-intercept lines

Khan Academy → IXL →

Like terms & distributive property

Khan Academy → IXL →

Inequalities & number lines

Khan Academy → IXL →

When you're ready →

No rush. When the cards above feel comfortable, step into the module.

Module 3 Overview Open the Visual Lab Course Syllabus