Systems of Equations

What you need to know

• Elimination and substitution solve any linear system. Elimination adds or subtracts equations to remove a variable. Substitution solves one equation for a variable and plugs it into another.
• When does a $2\times2$ linear system have exactly one solution? Take $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$. There is exactly one solution when $a_1b_2-a_2b_1\ne0$. If $a_1b_2-a_2b_1=0$, the two lines point the same way, so there is either no solution (parallel lines) or infinitely many (the same line). Example: $2x+3y=7$ and $4x+6y=10$ give $2\cdot6-4\cdot3=0$ and no solution, because $4x+6y$ would have to be both $10$ and $14$.
• Symmetric systems (swapping $x$ and $y$ changes nothing) are about $s=x+y$ and $p=xy$. Once you know $s$ and $p$, $x$ and $y$ are the two roots of $t^2-st+p=0$. Example: $x+y=5$ and $xy=6$ make $t^2-5t+6=0$, so $x$ and $y$ are $2$ and $3$. With three variables, use $x+y+z$, $xy+yz+zx$, and $xyz$.
• Identities to know: $x^2+y^2=(x+y)^2-2xy$ and $x^3+y^3=(x+y)^3-3xy(x+y)$. Example: with $x+y=5$ and $xy=6$, $x^2+y^2=25-12=13$ and $x^3+y^3=125-90=35$. Also $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$.
• Cyclic systems. Adding all the equations gives the total; subtracting neighbors gives differences. Example: $x+y=7$, $y+z=9$, $z+x=10$ add to $2(x+y+z)=26$, so $x+y+z=13$. Then $z=13-7=6$, $x=13-9=4$, and $y=13-10=3$.
• Product systems. If $xy=a$, $yz=b$, and $zx=c$, multiply all three: $(xyz)^2=abc$. Take the square root (watch the sign), then divide back out. Example: $xy=6$, $yz=15$, $zx=10$ give $(xyz)^2=900$. For positive numbers $xyz=30$, so $z=\frac{30}6=5$, $x=\frac{30}{15}=2$, and $y=\frac{30}{10}=3$.

How AMC 10 tests it

• Problems 5–12: a word problem with two or three unknowns that asks for something like $x+y+z$, which comes 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 $x^3+y^3$." You never need $x$ and $y$ themselves.
• A system with a letter $k$ in it: for which $k$ is there no solution, or infinitely many?
• Systems that come from a table, a Venn diagram, or a list of purchases, with one entry asked for.

Standard approaches

1. Before solving, check whether the quantity asked for is a combination of the equations: add them all, subtract, or multiply some by numbers first.
2. For symmetric systems, set $s=x+y$ and $p=xy$, and rewrite everything with $s$ and $p$.
3. Multiply or divide equations when they are products; add or subtract them when they are sums.
4. Only as a last resort, substitute and solve for each variable. Then plug back in to check.
5. For "how many solutions," think of the equations as graphs and count where they cross.

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) $752$

We never find $x$ and $y$. We build up from $x+y$ and $xy$.

Find $xy$. $(x+y)^2=x^2+y^2+2xy$, so $16=14+2xy$ and $xy=1$.

Find $x^3+y^3$.
$$x^3+y^3=(x+y)^3-3xy(x+y)=64-3\cdot1\cdot4=52.$$
Multiply the two sums we know. Expanding,
$$(x^2+y^2)(x^3+y^3)=x^5+x^2y^3+x^3y^2+y^5=x^5+y^5+x^2y^2(x+y).$$
The left side is $14\cdot52=728$. The extra piece is $x^2y^2(x+y)=1^2\cdot4=4$. So $x^5+y^5=728-4=724$. The answer is $\boxed{\textbf{(C)}\ 724}$.

Pitfalls

• Solving for $x$ and $y$ one at a time (here they are $2+\sqrt3$ and $2-\sqrt3$) when only $x+y$ and $xy$ are needed.
• Sign slips in $x^3+y^3=(x+y)^3-3xy(x+y)$: the $3xy(x+y)$ is subtracted.
• Losing the second solution after taking a square root ($x^2=9$ means $x=3$ or $x=-3$), or keeping a false one after squaring. Example: squaring $\sqrt{x+2}=x$ gives $x=2$ or $x=-1$, but $x=-1$ fails, since $\sqrt1\ne-1$.
• Thinking $a_1b_2-a_2b_1=0$ means "no solution." It can also mean infinitely many. Example: $2x+3y=7$ and $4x+6y=14$ are the same line.

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)