Math Core

On the AMC, a system of equations is rarely meant to be solved by grinding through substitution. The equations are usually symmetric or patterned, and one well-chosen move (adding everything, multiplying everything, or switching to new variables) cracks the whole thing. This module collects the moves that come up most.

Add everything

When each equation leaves out a different variable, or the variables appear in a cyclic pattern, add all the equations. Each variable shows up the same number of times, so you get the total $x + y + z$ in one step. Then subtract each original equation to peel off one variable at a time.

Worked example: Pairwise sums

Suppose $x + y = 10$ , $y + z = 14$ and $x + z = 16$ . Find $xyz$ .

Adding all three: $2(x + y + z) = 40$ , so $x + y + z = 20$ . Subtract each equation from this total: $z = 20 - 10 = 10$ , $x = 20 - 14 = 6$ , $y = 20 - 16 = 4$ . So $xyz = 6 \cdot 4 \cdot 10 = 240$ .

Multiply everything

When the equations give products of variables, multiply them all together. For $xy$ , $yz$ and $xz$ , the product is $(xyz)^2$ .

Worked example: Pairwise products

Positive numbers $x, y, z$ satisfy $xy = 6$ , $yz = 15$ and $xz = 10$ . Find $x + y + z$ .

Multiplying gives $(xyz)^2 = 6 \cdot 15 \cdot 10 = 900$ , so $xyz = 30$ (positive). Then $z = \dfrac{xyz}{xy} = 5$ , $x = \dfrac{30}{15} = 2$ and $y = \dfrac{30}{10} = 3$ . So $x + y + z = 10$ .

Switch to sum and product

For a symmetric system in two variables, let

s = x + y, \qquad p = xy.

Every symmetric expression can be written in $s$ and $p$ (the same identities as Vieta's formulas): $x^2 + y^2 = s^2 - 2p$ and $x^3 + y^3 = s^3 - 3ps$ . Often you never need $x$ and $y$ individually. If you do, they are the roots of $t^2 - st + p = 0$ , and they are real exactly when $s^2 \ge 4p$ .

Look for the structure first

Before substituting, ask:

Is the system symmetric or cyclic? Add the equations.
Are the equations products? Multiply them.
Is it symmetric in two variables? Use $s = x + y$ and $p = xy$ .
Is there an $xy$ term with $x$ and $y$ terms and you want integers? Factor by adding a constant.

Worked example: Sum and product

Real numbers satisfy $x + y = 5$ and $xy = 3$ . Find $x^2 + y^2$ and $x^3 + y^3$ .

With $s = 5$ and $p = 3$ : $x^2 + y^2 = 25 - 6 = 19$ and $x^3 + y^3 = 125 - 3 \cdot 3 \cdot 5 = 80$ .

Factoring with an added constant

An equation like $xy + 2x + 3y = 30$ looks unfactorable, but it is almost $(x + 3)(y + 2) = xy + 2x + 3y + 6$ . Add $6$ to both sides:

(x + 3)(y + 2) = 36.

For integer solutions, now just list factor pairs. This trick (often called Simon's Favorite Factoring Trick) turns $axy + bx + cy$ into a product.

Worked example: Counting integer solutions

How many ordered pairs of positive integers $(x, y)$ satisfy $xy + 2x + 3y = 30$ ?

From above, $(x + 3)(y + 2) = 36$ with $x + 3 \ge 4$ and $y + 2 \ge 3$ . The factor pairs $(x + 3, y + 2)$ of $36$ that meet those bounds are $(4, 9), (6, 6), (9, 4), (12, 3)$ . These give $(x, y) = (1, 7), (3, 4), (6, 2), (9, 1)$ : 4 pairs. Check $(3, 4)$ : $12 + 6 + 12 = 30$ .

Common mistake

After a multiply-everything step, you take a square root. Decide the sign from the problem ("positive numbers") or keep both signs. Also, in factoring problems, remember negative factor pairs, then check which ones break the bounds on the variables.