Chapter 0 wrap-up

Readiness: the algebra this course leans on

Closing thought

None of this chapter was new arithmetic. Each section checked a tool the rest of the course will pick up without warning. An equation is linear when each unknown appears only to the first power, multiplied by a number and never by another unknown. Sigma notation packs a long sum of look-alike terms into one line. A function is a rule from a domain to a codomain. A proof is a chain of steps where each one follows from the last.

So you can now handle the raw materials with confidence. Given $2x + y = 7$ and $x - y = 2$ , add the equations to get $3x = 9$ . That gives $x = 3$ and $y = 1$ , the one point where two lines cross. You can also expand $\sum_{i=1}^{3} i^2$ into $1 + 4 + 9 = 14$ . Chapter 1 opens the course proper. It takes a list of numbers you can also draw as an arrow, and shows how adding and stretching such lists builds almost everything that follows.

Recipes in this chapter

Step-by-step procedures for the problem types this chapter teaches.

Find a triangle's angle from its three side lengths Angles and cosine — the trig this course actually uses Write an explicit sum in sigma notation Sigma notation and index bookkeeping Find the domain of a function Functions: domain, codomain, and composition Write a direct proof of a universal claim Reading and writing a proof Prove a claim by contradiction Reading and writing a proof

Chapter review

One practice set per topic this chapter taught — same fixed, ungraded ladders as the practice panels inside each section.

Practice: Chapter review: linear equations and systems (5 problems · doesn’t affect your progress)

Which equation is NOT linear? (For reference, the choices given are: $-4x + y = 0$ ; $x + y + z = 9$ ; $2x - \frac{1}{3}y = 8$ ; $4^x + y = 6$ .)

Correct! 🎉 Not quite.

Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.

This one IS linear — both variables appear alone, to the first power. Look again for the equation with an unusual exponent or variable placement.

This one IS linear — three variables, each alone to the first power, all added together.

Fractional coefficients like $-\frac{1}{3}$ are still just constants — this equation is linear.

Why

In $4^x + y = 6$ , the variable $x$ sits in an exponent, not as something multiplied by a constant — exponential terms are never linear.

$4^x + y = 6$

How many of these are linear equations? $8x + y = 3$ , $x^3 = 2$ , $-y = 5x$ , $xy + 1 = 0$ (Work through it specifically for $8x + y = 3$ and $x^3 = 2$ and $-y = 5x$ .)

Correct! 🎉 Not quite.

Why

$8x + y = 3$ and $-y = 5x$ (rewritten $5x + y = 0$ ) are linear. $x^3 = 2$ has an exponent above $1$ ; $xy + 1 = 0$ multiplies two variables together.

Correct answer

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Why

Solving $x - 3y = 4$ for $x$ gives what expression?

Correct! 🎉 Not quite.

Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.

Why

Add $3y$ to both sides: $x = 3y + 4$ . Since $x$ already had coefficient $1$ , this costs no fractions — the signal to substitute for $x$ rather than $y$ .

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Why

Isolate $x$ by moving the $-3y$ term to the other side of the equation.

3*y + 4

For the system $y = 3x - 5$ and $2x + y = 10$ , which move avoids the most arithmetic? (For reference, the choices given are: Substitute $3x - 5$ for $y$ directly into the second equation; Scale both equations to make the $x$ -coefficients opposites, then add; Scale both equations to make the $y$ -coefficients opposites, then add; Graph both lines and read the crossing point off the grid.)

Correct! 🎉 Not quite.

That works eventually, but the first equation is already solved for $y$ — using that head start with substitution is far shorter than scaling both equations.

Same issue: elimination is a fine method in general, but it's extra setup here when $y$ is already isolated for you.

A graph only gives an exact answer when the crossing point happens to fall on a grid line — algebra gives the exact solution every time.

Why

The first equation already reads $y = 3x - 5$ — that's substitution's ideal case, a variable handed to you pre-isolated. Reaching for elimination here means undoing that head start and redoing the work.

Look for an equation that's already solved for one variable -- substituting that in avoids extra algebra.

Substitute $3x - 5$ for $y$ directly into the second equation

A store sells notebooks and pens. Three notebooks and two pens cost $\$54$ ; two notebooks and five pens cost $\$80$ . What is the price of one notebook, in dollars? (Work through it specifically for $\$ and$ ; two notebooks and five pens cost $and$ 80$.)

Correct! 🎉 Not quite.

Why

Let $n$ = notebook price, $p$ = pen price: $3n+2p=54$ and $2n+5p=80$ . Scale by $5$ and $2$ : $15n+10p=270$ and $4n+10p=160$ . Subtracting gives $11n=110$ , so $n=10$ . (Then $p=12$ .)

Correct answer

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Why

Let $n$ be the price of a notebook and $p$ the price of a pen, translate each sentence into an equation, then solve the resulting system.

Practice: Chapter review: coordinates, distance, and angles (6 problems · doesn’t affect your progress)

A student finds the distance between $(3, 5)$ and $(9, 1)$ as $\sqrt{(3-9)^2+(5-1)^2}$ instead of $\sqrt{(9-3)^2+(1-5)^2}$ . What happens to the answer? (For reference, the choices given are: It comes out wrong; It comes out the same, since each difference just gets squared; It comes out negative; It can't be computed that way.)

Correct! 🎉 Not quite.

Squaring a difference erases its sign — try expanding both versions and compare.

A squared real number is never negative, and a square root returns the non-negative root, so distance can never be negative.

Subtracting in either order is perfectly valid — the formula only cares about the squared differences.

Why

(3-9)^2 = (-6)^2 = 36 and (9-3)^2 = 6^2 = 36 — same value either way. Swapping which point is 'first' flips the sign of each difference but not its square, so distance never changes.

Compare what happens when you square each difference -- squaring removes the sign, so check whether swapping the order changes anything once you square.

It comes out the same, since each difference just gets squared

Find the distance between $(4, 6)$ and $(-2, -2)$ , moving left and down.

Correct! 🎉 Not quite.

Why

Legs: Δx = 4-(-2) = 6, Δy = 6-(-2) = 8. Distance = sqrt(6^2 + 8^2) = sqrt(36+64) = sqrt(100) = 10.

Correct answer

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Why

Find the horizontal and vertical differences between the points before squaring and adding them.

Write an expression for the distance between $(a, 0)$ and $(0, b)$ , in terms of $a$ and $b$ . (Work through it specifically for $(a, 0)$ and $(0, b)$ and $a$ .)

Correct! 🎉 Not quite.

Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.

Why

Legs are Δx = a - 0 = a and Δy = 0 - b = -b. Distance = sqrt(a^2 + (-b)^2) = sqrt(a^2 + b^2).

Correct answer

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Why

Set up the distance formula with $(a,0)$ and $(0,b)$ as the two points, then simplify the differences before squaring.

sqrt(a2 + b2)

Write $300°$ in radians as an exact multiple of $\pi$ (e.g. enter as pi/4 style). (Work through it specifically for $300°$ and $\pi$ .)

Correct! 🎉 Not quite.

Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.

Why

Multiply by $\frac{\pi}{180°}$ : $300°\cdot\frac{\pi}{180°} = \frac{300\pi}{180} = \frac{5\pi}{3}$ .

Correct answer

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Why

5*pi/3

Which of these numbers could NOT be a value of $\cos\theta$ for any angle $\theta$ ? (For reference, the choices given are: 0.5; -0.9; 1; 1.2.)

Correct! 🎉 Not quite.

$0.5$ is achievable — for instance $\cos 60° = 0.5$ exactly.

Cosine can absolutely be negative — it's negative throughout Quadrants II and III, so $-0.9$ is a value it actually takes.

$1$ is achievable at the extreme: $\cos 0° = 1$ , when the terminal point sits right at $(1,0)$ .

Why

Every point on the unit circle has $x$ -coordinate between $-1$ and $1$ , so $\cos\theta$ can never exceed that range in either direction — $1.2$ is outside the circle entirely.

1.2

Find $\cos(-60°)$ . (Work through it specifically for $\cos(-60°)$ .)

Correct! 🎉 Not quite.

Why

Cosine is an even function: $\cos(-\theta) = \cos\theta$ , so $\cos(-60°) = \cos 60° = 0.5$ — swinging $60°$ below the axis instead of above reaches a point with the same $x$ -coordinate.

Correct answer

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Why

0.5

Practice: Chapter review: sigma notation and functions (6 problems · doesn’t affect your progress)

Evaluate $\sum_{i=1}^{3} 3i$ , tripling each index before adding.

Correct! 🎉 Not quite.

Why

Substitute $i=1,2,3$ into $3i$ to get $3, 6, 9$ , then add: $3+6+9=18$ .

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Why

Which sum correctly represents $x_0 + x_1 + x_2 + x_3 + x_4$ ? (For reference, the choices given are: $\sum_{i=0}^{4} x_i$ ; $\sum_{i=1}^{4} x_i$ ; $\sum_{i=0}^{5} x_i$ ; $\sum_{i=1}^{5} x_i$ .)

Correct! 🎉 Not quite.

Starting at $i=1$ skips $x_0$ , the first term in the list — always check the first subscript actually written, not just assume sums start at 1.

Stopping at $i=5$ would include $x_5$ , a term that's not in the list — the ending value must match the last subscript shown, which is 4.

This shifts both bounds up by one and would produce $x_1$ through $x_5$ instead of $x_0$ through $x_4$ .

Why

The subscripts run 0 through 4, so the index must start at $i=0$ and end at $i=4$ : $\sum_{i=0}^{4} x_i$ . Sums don't have to start at 1.

Figure out what starting and ending value the index must take so that the subscripts run $0, 1, 2, 3, 4$ .

$\sum_{i=0}^{4} x_i$

If $u_1=2, u_2=-1, u_3=4$ and $v_1=3, v_2=5, v_3=-2$ , evaluate $\sum_{i=1}^{3} u_iv_i$ . (Work through it specifically for $u_1=2, u_2=-1, u_3=4$ and $v_1=3, v_2=5, v_3=-2$ and $\sum_{i=1}^{3} u_iv_i$ .)

Correct! 🎉 Not quite.

Why

$u_1v_1 + u_2v_2 + u_3v_3 = (2)(3) + (-1)(5) + (4)(-2) = 6 - 5 - 8 = -7$ .

Correct answer

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Why

Multiply each matching pair $u_i v_i$ separately first, then add the three products together.

-7

What is the domain of $g(x) = \sqrt{x-6}$ ? (For reference, the choices given are: $x \ge 6$ ; $x \le 6$ ; $x > 6$ ; All real numbers.)

Correct! 🎉 Not quite.

Flip the inequality — a square root needs its radicand to be at least zero, so $x-6\ge0$ gives $x\ge6$ , not $x\le6$ .

$x=6$ itself is allowed since $\sqrt{0}=0$ is defined — the boundary point belongs in the domain, so the inequality isn't strict.

A square root does restrict the domain — negative radicands aren't real numbers, so not every $x$ works.

Why

A square root is only real-valued when its radicand is non-negative. Solving $x-6\ge0$ gives $x\ge6$ .

$x \ge 6$

Using the same $f(x)=3x-2$ and $g(x)=x^2$ , find $(g\circ f)(2)$ . (Work through it specifically for $f(x)=3x-2$ and $g(x)=x^2$ and $(g\circ f)(2)$ .)

Correct! 🎉 Not quite.

Why

Apply $f$ first: $f(2)=3(2)-2=4$ . Then apply $g$ : $g(4)=4^2=16$ . Notice this is different from $(f\circ g)(2)=10$ from the previous question — composition order matters.

Correct answer

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Why

Work from the inside out: first find $f(2)$ , then plug that result into $g$ .

$f(x) = \log(x+4)$ requires which restriction on its domain? (For reference, the choices given are: $x > -4$ ; $x < -4$ ; $x \ge -4$ ; $x \ne -4$ .)

Correct! 🎉 Not quite.

Flip the inequality — a logarithm needs a strictly positive argument, so $x+4>0$ gives $x>-4$ , not $x<-4$ .

The inequality must be strict: $x=-4$ gives $\log(0)$ , which is undefined, not $x\ge-4$ .

This is a strict inequality over a whole interval, not just one excluded point like a denominator restriction would be.

Why

A logarithm's argument must be strictly positive: $x+4>0$ gives $x>-4$ . Unlike a square root, the boundary itself ( $x=-4$ , giving $\log(0)$ ) is excluded too.

A logarithm is only defined when its argument is strictly positive -- set up that inequality for $x+4$ .

$x > -4$

Practice: Chapter review: matrices and proofs (6 problems · doesn’t affect your progress)

Find $\det\begin{pmatrix} 4 & 6 \\ 6 & 9 \end{pmatrix}$ for this pair of rows.

Correct! 🎉 Not quite.

Why

$4(9) - 6(6) = 36 - 36 = 0$ . Row 2 is exactly $1.5\times$ row 1, so the two rows are dependent — the parallelogram they span is flattened to zero area.

Correct answer

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Why

$\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = 9$ . What is $\det\begin{pmatrix} 3a & 3b \\ c & d \end{pmatrix}$ ? (Work through it specifically for $\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = 9$ and $\det\begin{pmatrix} 3a & 3b \\ c & d \end{pmatrix}$ .)

Correct! 🎉 Not quite.

Why

Scaling one whole row by $k$ scales the determinant by that same $k$ . Row 1 was scaled by $3$ , so the new determinant is $3 \times 9 = 27$ .

Correct answer

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Why

Recall the row-operation rule for how scaling a single row by a constant affects the determinant.

In terms of $a$ , $b$ , $c$ , $d$ , what is $\det\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ ? (Work through it specifically for $a$ and $b$ and $c$ .)

Correct! 🎉 Not quite.

Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.

Why

The $2\times2$ determinant is the main-diagonal product minus the other diagonal's product: $ad - bc$ .

Correct answer

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Why

Recall the direct formula for a $2\times2$ determinant in terms of the entries $a,b,c,d$ .

ad - bc

The claim "for every real number x, |x − 4| = x − 4" is false. Give a value of x less than 4 that disproves it. (Answer for this specific instance.)

Correct! 🎉 Not quite.

Why

At x = 0, the left side is |0 − 4| = 4, while the right side is 0 − 4 = −4, and 4 ≠ −4. Any x < 4 works, since |x−4| = 4−x there, not x−4.

Correct answer

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Why

A proof reads: "Since x > 3, we have x^2 > 9. Since x^2 > 9, the equation y = x^2 − 9 has y > 0." What logical structure holds this together? (For reference, the choices given are: A chain of implications, each justified and following from the line before it.; A single unjustified leap from the hypothesis straight to the conclusion.; A proof by contradiction, since a false statement was assumed at the start.; Three unrelated observations that happen to be true simultaneously..)

Correct! 🎉 Not quite.

Each line does state its justification ("since...") and connects to the previous line — it's not an unjustified leap.

No assumption of the negation appears anywhere; this is a forward chain, not a contradiction argument.

The lines are explicitly linked by "since," each depending on the one before — they aren't independent observations.

Why

A valid direct proof is exactly this: a sequence of implications, each one justified and building on the line immediately before it, from hypothesis to conclusion.

Notice that each sentence's conclusion becomes the next sentence's starting premise -- what's that pattern of linked if-then steps called?

A chain of implications, each justified and following from the line before it.

In a proof by contradiction, after assuming ¬C and validly deriving a statement of the form R ∧ ¬R, what can be concluded? (For reference, the choices given are: C must be true, since the assumption ¬C is the only thing that could have caused the impossibility.; C must be false, since a contradiction was reached.; Nothing can be concluded — the proof needs to be restarted with a different assumption.; Both C and ¬C are true simultaneously..)

Correct! 🎉 Not quite.

This reverses the correct conclusion — read the logic again: the contradiction is blamed on the assumption ¬C, so ¬C is rejected, leaving C true, not false.

A valid derivation of R ∧ ¬R from ¬C is a complete, successful proof — nothing needs restarting.

The contradiction R ∧ ¬R is impossible and never actually true; it's the signal that the assumption feeding into it was false, not a genuine simultaneous truth.

Why

Since every step from ¬C to R ∧ ¬R was valid, and R ∧ ¬R is impossible, the only faulty ingredient is the assumption ¬C itself. Rejecting ¬C leaves C, which is the desired conclusion.

A statement of the form $R \wedge \neg R$ is always false -- think about what that forces you to conclude about the assumption that produced it.

C must be true, since the assumption ¬C is the only thing that could have caused the impossibility.

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