Lesson 4.2: Solve Applications with Systems of Equations

New Concept

Learn to translate word problems into systems of linear equations. We will tackle applications in geometry and uniform motion, using two variables to model and solve complex scenarios.

What’s next

Soon, we'll dive into worked examples for each application type, followed by a sequence of interactive practice cards to sharpen your skills.

Property

To solve an application problem using a system of equations, follow these steps:

1. Read the problem to understand all words and ideas.
2. Identify what you are looking for.
3. Name the unknown quantities by choosing variables to represent them.
4. Translate the problem's details into a system of equations.
5. Solve the system of equations using an appropriate algebraic method (substitution or elimination).
6. Check that the solution makes sense in the context of the original problem.
7. Answer the question with a complete sentence.

Examples

• The sum of two numbers is 25 and their difference is 9. Let the numbers be $x$ and $y$. The system is $x + y = 25$ and $x - y = 9$. Adding the equations gives $2x = 34$, so $x = 17$. Then $17 + y = 25$, so $y = 8$. The numbers are 17 and 8.

• A company offers two jobs. Job A pays 30,000 dollars plus 20 dollars per sale. Job B pays 20,000 dollars plus 60 dollars per sale. Let $S$ be salary and $n$ be sales. The salaries are equal when $30000 + 20n = 20000 + 60n$. Solving gives $10000 = 40n$, so $n=250$ sales.

Property

• Two angles are complementary if the sum of the measures of their angles is 90 degrees.
• Two angles are supplementary if the sum of the measures of their angles is 180 degrees.

Examples

• The difference of two complementary angles is 36 degrees. Let the angles be $x$ and $y$. The system is $x + y = 90$ and $x - y = 36$. Adding them gives $2x = 126$, so $x=63$. The angles are 63 and 27 degrees.

• Two angles are supplementary. The larger angle is 30 degrees more than twice the smaller angle. Let the angles be $L$ and $S$. The system is $L + S = 180$ and $L = 2S + 30$. Substituting gives $(2S + 30) + S = 180$, so $3S = 150$. The angles are 50 and 130 degrees.

Property

Perimeter problems can be modeled with a system of equations. For a rectangle with length $L$ and width $W$, the perimeter is $P = 2L + 2W$. If a fence is needed for only three sides (e.g., one length and two widths), the fencing required is $F = L + 2W$. The system is formed by using the perimeter formula along with a second equation that describes the relationship between the length and width.

Examples

• A farmer has 150 feet of fencing for a rectangular pen adjacent to a barn (3 sides). The length is 30 feet more than the width. The system is $L + 2W = 150$ and $L = W + 30$. Substituting gives $(W+30) + 2W = 150$, so $3W = 120$. The width is 40 feet and the length is 70 feet.

• A picture frame has a perimeter of 120 inches. Its length is twice its width. The system is $2L + 2W = 120$ and $L = 2W$. Substituting gives $2(2W) + 2W = 120$, so $6W = 120$. The width is 20 inches and the length is 40 inches.

Property

The fundamental relationship for uniform motion is Distance = Rate × Time ($D=rt$). In catch-up problems, two objects travel the same distance, so their distances can be set equal: $D_1 = D_2$. This means $r_1 t_1 = r_2 t_2$. A second equation relates their travel times, often accounting for a delayed start, such as $t_2 = t_1 - h$, where $h$ is the head start time.

Examples

• Maria leaves driving 50 mph. One hour later, her brother follows at 70 mph. Let Maria's time be $t_M$ and her brother's be $t_B$. The system is $50t_M = 70t_B$ and $t_B = t_M - 1$. Substituting gives $50t_M = 70(t_M-1)$, so $50t_M = 70t_M - 70$. This gives $20t_M = 70$, so $t_M=3.5$ hours. The brother catches up in $2.5$ hours.

• A train leaves a station at 60 mph. A second train leaves 30 minutes ($\frac{1}{2}$ hour) later on a parallel track at 75 mph. The distances are equal when $60t = 75(t-0.5)$. Solving gives $60t = 75t - 37.5$, so $15t = 37.5$. The first train travels for $t=2.5$ hours.

Property

An object's speed is affected by the current of the medium it travels through (water or air). Let $v$ be the object's speed in still conditions and $c$ be the current's speed.

• Speed with the current (downstream or tailwind): The speeds add together. The effective rate is $v + c$.
• Speed against the current (upstream or headwind): The speeds subtract. The effective rate is $v - c$.

A system of equations is formed using $D = rt$ for both directions.

Examples

• A boat sails 45 miles downstream in 3 hours and returns upstream in 5 hours. The system is $3(b+c) = 45$ and $5(b-c) = 45$. This simplifies to $b+c=15$ and $b-c=9$. Adding the equations gives $2b=24$, so the boat's speed is 12 mph and the current's speed is 3 mph.

• A plane flies 1200 miles in 4 hours with a tailwind. The return trip against the same wind takes 5 hours. The system is $4(p+w) = 1200$ and $5(p-w) = 1200$. This simplifies to $p+w=300$ and $p-w=240$. Adding gives $2p=540$, so the plane's speed is 270 mph and the wind speed is 30 mph.