Systems of Equations

What you need to know

• Elimination and substitution solve any linear system. A $2\times2$ linear system has a unique solution exactly when $a_1b_2-a_2b_1\ne0$; otherwise none (parallel lines) or infinitely many (the same line).
• Symmetric systems reduce to $s=x+y$ and $p=xy$; then $x,y$ are the roots of $t^2-st+p=0$. For three variables use $x+y+z$, $xy+yz+zx$, $xyz$.
• Key identities: $x^2+y^2=(x+y)^2-2xy$, $x^3+y^3=(x+y)^3-3xy(x+y)$, $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$.
• Adding all equations of a cyclic system gives the total; subtracting neighbors gives differences.
• For product systems $xy=a$, $yz=b$, $zx=c$, multiply: $(xyz)^2=abc$, then divide back out.

How AMC 10 tests it

• Problems 5–12: a word problem with two or three unknowns asking for a combination like $x+y+z$ that falls out of adding the equations.
• Problems 10–16: cyclic systems ($x+y=a$, $y+z=b$, $z+x=c$, or the same with products).
• Problems 12–18: "$x+y=5$ and $x^2+y^2=13$; find $x^3+y^3$" — the individual values are never needed.
• A system with a parameter: for which $k$ is there no solution, or infinitely many?
• Systems from a table, a Venn diagram, or a sequence of transactions, with one entry asked for.

Standard approaches

1. Before solving, check whether the requested quantity is a linear combination of the equations (add them all, subtract, or weight them).
2. For symmetric systems, introduce $s=x+y$, $p=xy$ and rewrite everything in those terms.
3. Multiply or divide equations when they are products; add or subtract when they are sums.
4. Only as a last resort substitute and solve for each variable; then back-substitute to check.
5. For "how many solutions", view the equations as curves and count intersections.

Worked example

Real numbers $x$ and $y$ satisfy $x+y=4$ and $x^2+y^2=14$. What is $x^5+y^5$?

(A) $676$ (B) $700$ (C) $724$ (D) $728$ (E) $752$

From $(x+y)^2=x^2+y^2+2xy$ we get $16=14+2xy$, so $xy=1$. Then
$$x^3+y^3=(x+y)^3-3xy(x+y)=64-12=52.$$
Multiplying the two known power sums,
$$(x^2+y^2)(x^3+y^3)=x^5+y^5+x^2y^2(x+y),$$
so $x^5+y^5=14\cdot52-1\cdot4=724$. The answer is $\boxed{\textbf{(C)}\ 724}$.

Pitfalls

• Solving for $x$ and $y$ individually (here $2\pm\sqrt3$) when only $x+y$ and $xy$ are needed.
• Sign slips in $x^3+y^3=(x+y)^3-3xy(x+y)$.
• Losing the second solution after a square root, or keeping an extraneous one after squaring.
• Reading "no solution" as $\det=0$ alone; $\det=0$ also allows infinitely many solutions.

Traps that recur

• Adding the three equations to get ab + bc + ca + a + b + c = 247 and answering 247, forgetting to subtract a + b + c = −29. (2024 AMC 10A #23)
• Assuming the bricks keep their orientation, so that the new width 2a should be 3a + 1, which is impossible and stalls the whole problem. (2024 AMC 10B #25)
• Counting only systems with a zero row or a repeated row (302) and missing the 36 systems whose three distinct rows satisfy u + v = w. (2022 AMC 10B #18)
• Writing all eight equations and solving blindly; instead pair lines that share a variable so the unknown cancels immediately. (2001 AMC 10 #22)