Algebra & Manipulation

Why it is true: a quadratic with roots $r$ and $s$ and leading coefficient $1$ factors as $(x - r)(x - s) = x^2 - (r + s)x + rs$. Matching coefficients with $x^2 + \tfrac ba x + \tfrac ca$ gives $r + s = -\tfrac ba$ and $rs = \tfrac ca$. Before Vieta, though, comes the skill every AMC 8 algebra problem needs: turning words into one equation and solving it by undoing operations in reverse order.

Step 1 — multiply both sides by $4$, the common denominator, so no fractions remain: $x + 3 = 2(x - 1)$.

Step 2 — expand: $x + 3 = 2x - 2$.

Step 3 — collect $x$ on one side: $x = 5$. Check: $\tfrac84 = 2$ and $\tfrac42 = 2$.

Step 1 — name the MIDDLE one $m$; the five are $m - 4, m - 2, m, m + 2, m + 4$.

Step 2 — the $\pm$ terms cancel, so the sum is $5m = 135$ and $m = 27$.

Step 3 — the numbers are $23, 25, 27, 29, 31$. Naming the middle term instead of the smallest turns the equation into one division.

Step 1 — Vieta: $r + s = 5$ and $rs = 3$.

Step 2 — rewrite the target in terms of those: $r^2 + s^2 = (r + s)^2 - 2rs$.

Step 3 — $25 - 6 = 19$. The roots themselves, $\tfrac{5 \pm \sqrt{13}}{2}$, never appear.

Why the "$k$" works: a ratio $3 : 5$ says the amounts are made of equal-sized parts, $3$ of one and $5$ of the other. The size of one part is the unknown $k$; the total $40$ is $8$ parts, so one part is $5$. Drawing the parts as a bar makes the equation visible before it is written.

Step 1 — flour per pancake is constant, so the ratio flour : pancakes stays fixed: $\tfrac38 = \tfrac{x}{20}$.

Step 2 — cross-multiply: $8x = 60$, so $x = 7.5$ cups. Check by a friendlier route: $20$ pancakes is $2.5$ batches, and $2.5\cdot3 = 7.5$.

Step 1 — originally $3k$ boys and $4k$ girls.

Step 2 — the girls did not change, so after the change $3k + 6 = 4k$, giving $k = 6$.

Step 3 — originally $18$ boys and $24$ girls: $42$ students. The trick is to write the ratio with $k$ BEFORE the change and apply the change to the amounts, not to the ratio.

Step 1 — Simon's trick: $xy - 3x - 2y + 6 = (x - 2)(y - 3)$, so add $6$ to both sides: $(x - 2)(y - 3) = 16$.

Step 2 — $x - 2 \ge -1$ and $y - 3 \ge -2$, so a negative factor pair would need a factor below $-2$ on one side; none works.

Step 3 — positive pairs of $16$: $(1, 16), (2, 8), (4, 4), (8, 2), (16, 1)$, giving $(x, y) = (3, 19), (4, 11), (6, 7), (10, 5), (18, 4)$. There are $5$ pairs.

Most lost points in this section are translation errors, not algebra errors: "5 less than twice $x$" is $2x - 5$, not $5 - 2x$; "the ratio of boys to girls is $3:2$" means $b = 3k$, $g = 2k$, not $b = 3$, $g = 2$. Writing the units next to each quantity catches a surprising share of these before any solving happens.

A trip at $60$ km/h out and $40$ km/h back averages $\dfrac{2\cdot60\cdot40}{60+40} = 48$ km/h, NOT $50$ — the harmonic mean, because the two legs take different times. Work problems pair with it: someone finishing in $6$ hours has rate $\tfrac16$ of the job per hour, and rates add when people work together, so with a $3$-hour worker it is $\tfrac16 + \tfrac13 = \tfrac12$, i.e. $2$ hours.

Step 1 — do NOT average $12$ and $24$.

Step 2 — total distance $= 24$ km.

Step 3 — total time $= \tfrac{12}{12} + \tfrac{12}{24} = 1 + 0.5 = 1.5$ hours.

Step 4 — average speed $= \tfrac{24}{1.5} = 16$ km/h. Average speed is always total distance over total time; the slow leg counts for more because it lasts longer.

Step 1 — the gap between them shrinks by $12 + 18 = 30$ km every hour (their speeds add because they move toward each other).

Step 2 — time $=$ distance $\div$ closing speed $= 60/30 = 2$ hours. Check: in $2$ hours they ride $24$ km and $36$ km, which together make $60$.

Step 1 — one unknown: brother $b$, Kim $3b$.

Step 2 — in $6$ years: $3b + 6 = 2(b + 6)$.

Step 3 — $3b + 6 = 2b + 12$, so $b = 6$: the brother is $6$ and Kim is $18$. Check: in $6$ years, $24 = 2\cdot12$.