Lesson 2.3: Models and Applications

New Concept

This lesson teaches you to translate real-world scenarios into linear equations. You'll learn to build mathematical models to solve for unknowns in problems involving costs, distances, perimeters, and more, turning word problems into solvable equations.

What’s next

Now, let’s see this in action. You'll engage with interactive examples, a short video breakdown, and then test your skills with practice problems.

Property

To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable.
Then, we begin to interpret the words as mathematical expressions using mathematical symbols.
For example, a variable cost can be written as $0.10x$, while a fixed cost is a constant added or subtracted, such as in $C = 0.10x + 50$.

To model a linear equation:

1. Identify known quantities.
2. Assign a variable to represent the unknown quantity.
3. If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.
4. Write an equation interpreting the words as mathematical operations.
5. Solve the equation. Be sure the solution can be explained in words, including the units of measure.

Examples

• One number is 10 more than another, and their sum is 52. Let the first number be $x$. The second is $x+10$. The equation is $x + (x+10) = 52$, so $2x=42$, and $x=21$. The numbers are 21 and 31.
• A taxi charges 3 dollars plus 2 dollars per mile. The total cost $C$ for a ride of $x$ miles is modeled by the equation $C = 2x + 3$. A 5-mile ride costs $C = 2(5) + 3 = 13$ dollars.
• Two streaming services have different plans. Plan A is 15 dollars a month. Plan B is 5 dollars a month plus 2 dollars per movie. To find when they cost the same for $m$ movies, set $15 = 5 + 2m$. Solving gives $10 = 2m$, so $m=5$ movies.

Property

Distance problems are solved using the formula $d = rt$, where distance ($d$) equals rate ($r$) multiplied by time ($t$). It is critical that units of measurement are consistent; for example, if rate is in mi/h, time must be in hours.

When a problem involves two trips, you can write two separate equations. If the distances are equal for both trips, set the expressions for distance equal to each other and solve for the unknown variable.
$d = r_1 t_1$
$d = r_2 t_2$
$r_1 t_1 = r_2 t_2$

Examples

• A cyclist's trip to a friend's house takes 2 hours.

The return trip takes 2.5 hours because she rides 4 mi/h slower.
Let the initial rate be $r$.
The distance is $d=2r$ and $d=2.5(r-4)$.
So, $2r = 2.5r - 10$, which means $r=20$ mi/h.
The distance is $2(20)=40$ miles.

• Two cars leave a station at the same time, heading in opposite directions.

One travels at 50 mi/h and the other at 65 mi/h.
They will be 460 miles apart after $t$ hours.
The equation is $50t + 65t = 460$, so $115t = 460$, and $t=4$ hours.

• Leo starts walking at 4 mi/h.

One hour later, Mia starts jogging on the same path at 6 mi/h.
Mia will catch Leo after $t$ hours. Leo's distance is $4(t+1)$ and Mia's is $6t$.
Setting them equal: $4t+4=6t$, so $2t=4$, and $t=2$ hours.

Property

The perimeter formula for a rectangle is $P = 2L + 2W$, where $P$ is perimeter, $L$ is length, and $W$ is width.
In application problems, there are two unknown quantities, length and width.
However, we can write one in terms of the other, such as $L = W + 3$.
Substitute the perimeter value and the expression for length into the formula to solve for the dimensions.

Examples

• The perimeter of a garden is 100 ft. The length is 10 ft greater than the width. Let width be $W$. Then $L=W+10$. The formula is $100 = 2(W+10) + 2W$. This simplifies to $100 = 4W + 20$, so $W=20$ ft and $L=30$ ft.
• A rectangle's perimeter is 72 cm and its length is twice its width. Let width be $W$. Then $L=2W$. The formula is $72 = 2(2W) + 2W$. This simplifies to $72 = 6W$, so $W=12$ cm and $L=24$ cm.
• A picture frame has a perimeter of 36 inches. The width is 4 inches less than the length. Let length be $L$. Then $W=L-4$. The formula is $36 = 2L + 2(L-4)$. This simplifies to $36 = 4L - 8$, so $L=11$ inches and $W=7$ inches.

Explanation

To find a rectangle's dimensions, use the formula $P = 2L + 2W$. If you know the perimeter and the relationship between length and width, you can create a single-variable equation to solve.

Property

The standard formula for area is $A = LW$.
However, to solve some problems, you may first need to use the perimeter formula, $P = 2L + 2W$.
We use the perimeter formula because we may know enough information about the perimeter to solve for one of the unknowns, length or width.
Once the dimensions are found, they can be used to find the area.

Examples

• The perimeter of a rug is 26 feet, and its length is 3 feet more than its width. Find the area. First, find dimensions: $26 = 2(W+3) + 2W$, so $W=5$ ft and $L=8$ ft. The area is $A = 8 \times 5 = 40$ ft$^2$.
• A rectangular field has a perimeter of 320 meters. The length is 20 meters greater than the width. Find the area. First, solve $320 = 2(W+20) + 2W$. $W=70$ m and $L=90$ m. The area is $A = 70 \times 90 = 6300$ m$^2$.
• A poster has a perimeter of 120 inches. Its length is 12 inches more than its width. Find the area. First, solve $120 = 2(W+12) + 2W$. $W=24$ in and $L=36$ in. The area is $A = 24 \times 36 = 864$ in$^2$.

Explanation

To find the area of a rectangle ($A = LW$), you first need its length and width. Sometimes, you'll use the perimeter formula ($P = 2L + 2W$) and other clues to find these dimensions first.

Property

The formula for the volume of a box is given as $V = LWH$, the product of length, width, and height.
In application problems, we are often given relationships between the dimensions, such as the length being twice the width ($L = 2W$).
We substitute the given volume and the expressions for the dimensions into the formula to solve for the unknowns.

Examples

• The volume of a shipping box is 2400 in$^3$. The length is three times the width, and the height is 8 inches. Let width be $W$. Then $L=3W$. So, $2400 = (3W)(W)(8)$. This simplifies to $2400 = 24W^2$, so $W^2=100$, and $W=10$ in. The length is $L=30$ in.
• A block of wood has a volume of 720 cm$^3$. Its height is 10 cm and its length is twice its width. Let width be $W$. Then $L=2W$. So, $720 = (2W)(W)(10)$. This gives $720 = 20W^2$, so $W^2=36$, and $W=6$ cm. The length is $L=12$ cm.
• The volume of a cereal box is 256 cubic inches. The height is 16 inches and the base is a square, so $L=W$. Let the side of the base be $x$. So, $256 = (x)(x)(16)$. This gives $16 = x^2$, so $x=4$ inches. The dimensions are $L=4$ in, $W=4$ in, and $H=16$ in.

Explanation

For the volume of a box, use $V = LWH$. If you know the volume and how the sides relate to each other, you can set up an equation to find the exact length, width, and height.