Writing Equations and Constraints from Situations

From words to an equation

A gym charges a $50$-dollar sign-up fee and $35$ dollars a month. Maya has $400$ dollars. How many months can she pay for?

Four steps turn that into mathematics and back.

1. Name the unknown, with units. Let $m$ be the number of months.
2. Write what the situation says. The fee plus the monthly charges equals the total: $50 + 35m = 400$.
3. Solve. $35m = 350$, so $m = 10$.
4. Answer the question asked, in words. Maya can pay for $10$ months.

Step 1 carries more weight than it looks. “Let $m$ be the months” makes every later expression mean something, and the units say what the answer will be measured in.

The same steps handle other kinds of equations.

Situation	Equation	Kind
a garden $3$ feet longer than wide, area $88$ square feet	$w(w + 3) = 88$	quadratic
two pipes fill a pool in $6$ and $3$ hours alone	$\tfrac{1}{6} + \tfrac{1}{3} = \tfrac{1}{t}$	rational
$2{,}000$ dollars growing $5\%$ a year, until it reaches $3{,}000$	$2000(1.05)^t = 3000$	exponential

Why an equation’s solution may not be an answer

Solve the garden equation.

$$w(w + 3) = 88 \quad\Rightarrow\quad w^2 + 3w - 88 = 0 \quad\Rightarrow\quad (w + 11)(w - 8) = 0$$

The equation has two solutions, $w = 8$ and $w = -11$. The garden has one width.

The equation is not wrong. It records only what the words said about the numbers: one number times a number $3$ larger is $88$. It knows nothing about gardens, and $-11$ times $-8$ really is $88$.

The situation adds a condition the equation never saw: a width cannot be negative. So $w = -11$ is not viable, and the garden is $8$ feet by $11$ feet.

Every model carries unwritten conditions like this one:

Situation	Unwritten condition
lengths, times, ages	not negative
people, buses, tickets	whole numbers
a budget	at most the budget
a recipe or a requirement	at least the minimum

Checking each solution against them is the last step of every modeling problem, not an optional one.

Inequalities: at most and at least

Many situations set a limit rather than an exact amount.

Words	Symbol
at most, no more than, up to	$\le$
at least, no less than, a minimum of	$\ge$

Two adult tickets cost $12$ dollars each and student tickets cost $8$. With at most $96$ dollars, how many student tickets $s$ can be bought?

$$2(12) + 8s \le 96 \quad\Rightarrow\quad 8s \le 72 \quad\Rightarrow\quad s \le 9$$

Tickets come whole, so the viable answers are $0$ through $9$.

Several constraints at once

A snack mix uses peanuts and raisins. Peanuts have $7$ grams of protein an ounce and cost $30$ cents an ounce. Raisins have $1$ gram an ounce and cost $20$ cents an ounce. The mix needs at least $14$ grams of protein and must cost at most $120$ cents.

Let $p$ be ounces of peanuts and $r$ ounces of raisins. Each sentence becomes an inequality:

$$\begin{cases} 7p + r \ge 14 & \text{protein} \\[2pt] 30p + 20r \le 120 & \text{cost} \\[2pt] p \ge 0, \; r \ge 0 & \text{amounts} \end{cases}$$

An option is viable only when every inequality is true.

Option $(p, r)$	Protein	Cost	Viable?
$(2, 1)$	$15$ ✓	$80$ ✓	yes
$(4, 0)$	$28$ ✓	$120$ ✓	yes, right at the budget
$(1, 3)$	$10$ ✗	$90$ ✓	no: too little protein
$(3, 2)$	$23$ ✓	$130$ ✗	no: over budget

Graphing the system shows every viable mix at once. The shaded triangle is where both constraints hold, with $p$ across and $r$ up.

• protein: 7p + r = 14
• cost: 30p + 20r = 120

Points inside or on the edge of the triangle are viable. The two points outside it each break exactly one constraint, and the graph shows which: $(1, 3)$ is on the low-protein side of the steep line, and $(3, 2)$ is above the budget line.

Worked examples

Common mistakes

Practice problems