Equation word problems describe real-life situations using words. The goal is to translate the
story into one or more equations, solve the equations, and interpret the answer in context.
1. The basic strategy
Most equation word problems can be solved using four steps:
\[
\text{define} \rightarrow \text{translate} \rightarrow \text{solve} \rightarrow \text{check}.
\]
The most important step is defining the unknown. For example, if a problem asks for a speed,
you might write:
\[
x=\text{speed of the boat in still water}.
\]
2. Translating words into equations
Common word phrases often translate directly into algebra.
3. Age problems
Age problems often compare ages now and later. If B’s current age is \(x\), and A is
\(m\) times as old as B, then:
\[
B=x,\qquad A=mx.
\]
If the comparison happens \(t\) years from now, then both ages increase by \(t\):
\[
A+t=n(B+t).
\]
Substituting \(A=mx\) and \(B=x\) gives:
\[
mx+t=n(x+t).
\]
4. Mixture problems
Mixture problems usually use the idea:
\[
\text{pure amount}=\text{concentration}\times\text{total amount}.
\]
If \(x\) liters of a new solution are added, the equation often has the form:
\[
p_1A+p_2x=p_t(A+x),
\]
where \(p_1\), \(p_2\), and \(p_t\) are written as decimals.
5. Distance, rate, and time problems
Distance problems are based on:
\[
d=rt.
\]
Therefore:
\[
r=\frac{d}{t}.
\]
For boat-current problems:
\[
\text{upstream speed}=b-c,
\]
\[
\text{downstream speed}=b+c,
\]
where \(b\) is the speed of the boat in still water and \(c\) is the speed of the current.
6. Work-rate problems
If a person can complete one job in \(a\) hours, the work rate is:
\[
\frac{1}{a}\text{ job per hour}.
\]
If two people work together, their rates add:
\[
\frac{1}{a}+\frac{1}{b}=\frac{1}{t}.
\]
Solving for \(t\) gives the time needed together.
7. Number problems
If the sum of two numbers is \(S\) and their difference is \(D\), then:
\[
x+y=S,
\]
\[
x-y=D.
\]
Adding the equations gives:
\[
2x=S+D.
\]
So:
\[
x=\frac{S+D}{2}.
\]
8. Rectangle problems
Rectangle problems often use perimeter and area:
\[
P=2L+2W,
\]
\[
A=LW.
\]
If the length is \(d\) more than the width, then:
\[
L=W+d.
\]
Substituting into the perimeter formula gives a linear equation:
\[
2(W+d)+2W=P.
\]
9. Quadratic application problems
Some word problems lead to quadratic equations. For a rectangle with known perimeter and area,
first use:
\[
L+W=\frac{P}{2}.
\]
If one side is \(x\), then the other side is:
\[
\frac{P}{2}-x.
\]
The area equation becomes:
\[
x\left(\frac{P}{2}-x\right)=A.
\]
After expansion, this becomes a quadratic equation.
10. Checking the answer
A numerical answer is not complete until it is checked in the original context.
For example, if a rectangle has perimeter \(50\), length \(15\), and width \(10\), check:
\[
2(15)+2(10)=50.
\]
Also check that the answer makes sense. Ages, distances, times, and lengths should usually be positive.
11. Formula summary
12. Common mistakes
- Using a number before defining what the variable means.
- Confusing “more than” with “less than.”
- Using percent values without converting them to decimals in mixture equations.
- Adding times instead of rates in work problems.
- Forgetting that upstream speed is boat speed minus current speed.
- Writing a correct equation but giving an answer without checking the story.
Key idea: a word problem becomes easier when you clearly define the unknown, write the equation from the story, solve carefully, and check the result in context.
