Math Core

Many real questions involve two unknown quantities tied together by two separate facts: how many tickets of each type, how much of each solution to mix, how fast a boat goes and how fast the river flows. Each fact becomes an equation, and the system you've learned to solve delivers both answers at once.

A plan for every word problem

Solving a word problem with a system

Define two variables. Say exactly what each stands for, including units.
Write two equations, one for each independent fact in the problem.
Solve the system by whichever method fits best.
Answer the question that was asked, in words and with units.
Check your answer against the original wording, not just your equations.

Step 2 is where the thinking happens. In most problems, the two facts are of two different kinds. One equation usually counts things (how many items, how many liters, how many hours), and the other measures their total value (cost, amount of pure substance, distance). Organizing the information in a table makes both equations easy to read off.

Counting and value problems

Worked example: Ticket sales

A school play sold $300$ tickets. Adult tickets cost $8 and student tickets cost $5. Ticket sales brought in $2,070. How many of each type were sold?

Let $a$ be the number of adult tickets and $s$ the number of student tickets.

	number	price (dollars)	money (dollars)
adult	$a$	$8$	$8a$
student	$s$	$5$	$5s$
total	$300$		$2070$

The "number" column gives one equation and the "money" column gives the other:

\begin{cases} a + s = 300 \\ 8a + 5s = 2070 \end{cases}

Substitution is natural: $s = 300 - a$ . Then

\begin{aligned} 8a + 5(300 - a) &= 2070 \\ 8a + 1500 - 5a &= 2070 \\ 3a &= 570 \\ a &= 190 \end{aligned}

and $s = 300 - 190 = 110$ .

Answer: $190$ adult tickets and $110$ student tickets.

Check: $190 + 110 = 300$ tickets, and $190 \cdot 8 + 110 \cdot 5 = 1520 + 550 = 2070$ dollars.

Coin problems, mixtures and investments all follow the same pattern: a count equation plus a value equation.

Worked example: Mixing solutions

A chemist has a $10\%$ acid solution and a $30\%$ acid solution. How many liters of each should she mix to make $40$ liters of a $25\%$ acid solution?

Let $x$ be the liters of $10\%$ solution and $y$ the liters of $30\%$ solution. The "value" here is the amount of pure acid: liters times the percent, written as a decimal.

	liters	percent acid	liters of acid
$10\%$ solution	$x$	$0.10$	$0.10x$
$30\%$ solution	$y$	$0.30$	$0.30y$
mixture	$40$	$0.25$	$0.25(40) = 10$

\begin{cases} x + y = 40 \\ 0.10x + 0.30y = 10 \end{cases}

Multiply the second equation by $10$ to clear decimals: $x + 3y = 100$ . Now subtract the first equation from it:

(x + 3y) - (x + y) = 100 - 40 \quad\Longrightarrow\quad 2y = 60 \quad\Longrightarrow\quad y = 30.

Then $x = 40 - 30 = 10$ .

Answer: $10$ liters of the $10\%$ solution and $30$ liters of the $30\%$ solution.

Check: $0.10(10) + 0.30(30) = 1 + 9 = 10$ liters of acid, which is $25\%$ of $40$ liters.

Tip

Estimate before you solve. The target, $25\%$ , is much closer to $30\%$ than to $10\%$ , so the mixture should be mostly the $30\%$ solution. An answer of $30$ liters of it passes that test.

Rate problems with a current or wind

A boat moving with a river's current goes faster than in still water; against the current it goes slower. If the boat's still-water speed is $b$ and the current's speed is $c$ :

\text{downstream speed} = b + c, \qquad \text{upstream speed} = b - c.

The same idea applies to an airplane with or against the wind. Combine it with $\text{distance} = \text{rate} \times \text{time}$ .

Worked example: Upstream and downstream

A boat travels $36$ miles downstream in $2$ hours. The return trip upstream takes $3$ hours. Find the speed of the boat in still water and the speed of the current.

Let $b$ be the boat's speed in still water and $c$ the current's speed, both in miles per hour.

Downstream, $36 = (b + c) \cdot 2$ , so $b + c = 18$ . Upstream, $36 = (b - c) \cdot 3$ , so $b - c = 12$ .

\begin{cases} b + c = 18 \\ b - c = 12 \end{cases}

Add the equations: $2b = 30$ , so $b = 15$ . Then $c = 18 - 15 = 3$ .

Answer: the boat goes $15$ mph in still water, and the current flows at $3$ mph.

Check: downstream $18 \text{ mph} \times 2 \text{ h} = 36$ miles; upstream $12 \text{ mph} \times 3 \text{ h} = 36$ miles.

Comparing two options

When two plans each have a starting cost and a rate, the system asks: at what point do they cost the same? The solution is the break-even point, and a graph shows which plan is cheaper on each side of it.

Worked example: Two gym memberships

Gym A charges a $40 sign-up fee plus $25 per month. Gym B charges $10 to sign up plus $30 per month. After how many months is the total cost the same? Which gym is cheaper for a full year?

Let $m$ be the number of months and $C$ the total cost in dollars.

\begin{cases} C = 25m + 40 \\ C = 30m + 10 \end{cases}

Set the costs equal: $25m + 40 = 30m + 10$ , so $30 = 5m$ and $m = 6$ . Then $C = 25(6) + 40 = 190$ .

Gym B (steeper line) starts cheaper, but Gym A is cheaper after 6 months.Open in grapher →

Answer: after $6$ months both gyms have cost $190. Gym B starts cheaper, but its line is steeper, so after month $6$ Gym A is cheaper. For $12$ months, Gym A costs $25(12) + 40 = 340$ dollars and Gym B costs $30(12) + 10 = 370$ dollars, so Gym A is cheaper.

Common mistake

Two common traps:

Answering the wrong question. If the problem asks for the current's speed, solving for the boat's speed and stopping is only half the job. Reread the question before you write the answer.
Using percents as whole numbers. In a mixture or interest equation, $25\%$ is $0.25$ , not $25$ . Mixing the two in one equation gives nonsense.