This document provides an overview of how to solve different types of applied problems using linear equations, including geometry problems, motion problems, mixture problems, and investment problems. It gives examples of each type of problem and shows the step-by-step process for setting up and solving the linear equation to find the unknown value. Problem-solving strategies include using diagrams, tables, variables to represent unknowns, and checking the answer. The document is intended to teach students how to model and solve word problems involving linear equations.

1. 1.2 Applications and Modeling
with Linear Equations
Chapter 1 Equations and Inequalities

2. Concepts & Objectives
⚫ Solving applied problems
⚫ Geometry problems
⚫ Motion problems
⚫ Mixture problems
⚫ Modeling with linear equations

3. Solving Applied Problems
Step 1 Read the problem carefully until you understand
what is given and what is to be found.
Step 2 Assign a variable to represent the unknown value,
using diagrams or tables as needed. Express any
other unknown values in terms of the variable.
Step 3 Write an equation using the variable expression(s).
Step 4 Solve the equation
Step 5 State the answer to the problem. Does it seem
reasonable?
Step 6 Check your answer to the original problem.

4. Geometry Problems
Example: If the length of each side of a square is increased
by 3 cm, the perimeter of the new square is 40 cm more
than twice the length of each side of the original square.
Find the dimensions of the original square.
⚫ Let x represent the original length of the side.
⚫ The new length would then be x + 3, and the new
perimeter would be 40 + 2x.
⚫ The equation would then be comparing the perimeters:
( )+ = +4 3 40 2x x

5. Geometry Problems (cont.)
⚫ Solve the equation:
⚫ The original length of a side is 14 cm.
⚫ To check: the new length is 14+3 = 17 cm. The new
perimeter would be 4(17) = 68 cm. 40 + 2(14) = 68 cm,
which matches.
( )4 3 40 2x x+ = +
4 12 40 2x x+ = +
2 28x =
14x =

6. Motion Problems
⚫ In a motion problem, the components distance, rate, and
time are denoted by d, r, and t, respectively. (The rate is
also called speed or velocity. In these problems, rate is
understood to be constant.) These variables are related
by the equation
.d rt=

7. Motion Problems (cont.)
Example: Maria and Eduardo are traveling to a business
conference. The trip takes 2 hours for Maria and 2.5 hours
for Eduardo, since he lives 40 miles farther away. Eduardo
travels 5 mph faster than Maria. Find their average rates.

8. Motion Problems (cont.)
Example: Maria and Eduardo are traveling to a business
conference. The trip takes 2 hours for Maria and 2.5 hours
for Eduardo, since he lives 40 miles farther away. Eduardo
travels 5 mph faster than Maria. Find their average rates.
We also know that Eduardo’s distance is 40 more than
Maria’s.
r (mph) t (hours) d = rt
Maria x 2 2x
Eduardo x + 5 2.5 2.5(x + 5)

9. Motion Problems (cont.)
Eduardo’s distance = 40 more than Maria’s
This means that Maria’s rate is 55 mph and Eduardo’s is
55 + 5 = 60 mph.
( )2.5 5 2 40x x+ = +
2.5 12.5 2 40x x+ = +
.5 27.5x =
55x =

10. Mixture Problems
⚫ Problems involving mixtures of two types of the same
substance, salt solution, candy, and so on, often involve
percent.
⚫ In mixture problems, the rate (percent) of concentration
is multiplied by the quantity to get the amount of pure
substance present. The concentration of the final mixture
must be between the concentrations of the two solutions
making up the mixture.

11. Mixture Problems (cont.)
Example: A chemist needs a 20% solution of alcohol. She
has a 15% solution on hand, as well as a 30% solution.
How many liters of the 15% solution should she add to 3 L
of the 30% solution to obtain her 20% solution?

12. Mixture Problems (cont.)
Example: A chemist needs a 20% solution of alcohol. She
has a 15% solution on hand, as well as a 30% solution.
How many liters of the 15% solution should she add to 3 L
of the 30% solution to obtain her 20% solution?
Strength Liters of Solution
Liters of Pure
Alcohol
15% x .15x
30% 3 .30(3)
20% x + 3 .20(x + 3)

13. Mixture Problems (cont.)
Since the number of liters of pure alcohol in the 15% + 30%
solution must equal the number of liters in the 20%
solution, we can set them equal to each other:
This means that the chemist needs 6 L of 15% solution to
combine with 3 L of 30% solution to produce a solution
which is 20% alcohol.
( ) ( ).15 .30 3 .20 3x x+ = +
.15 .90 .20 .60x x+ = +
.30 .05x=
6 x=

14. Investment Problems
⚫ In mixed investment problems, multiply each principal
by the interest rate to find the amount of interest
earned.
⚫ Remember that if a certain total amount is invested, say
$100,000, then if one quantity is x, the other quantity is
100000 – x.

15. Investment Problems (cont.)
Example: Last year, Owen earned a total of $1456 in
interest from two investments. He invested a total of
$28,000; part of it he invested at 4.8% and the rest at 5.5%.
How much did he invest at each rate?

16. Investment Problems (cont.)
Example: Last year, Owen earned a total of $1456 in
interest from two investments. He invested a total of
$28,000; part of it he invested at 4.8% and the rest at 5.5%.
How much did he invest at each rate?
Amount
Invested
Interest
Rate (%)
Time (in
years)
Interest Earned
$28,000 mixed 1 yr $1456
x 4.8 1 yr x(.048)(1)
28,000 – x 5.5 1 yr (28000–x)(.055)(1)

17. Investment Problems (cont.)
The partial interest amounts have to add up to the total
interest:
So, Owen invested $12,000 at 4.8% and $16,000 at 5.5%.
( )( ) ( )( )( ).048 1 28000 .055 1 1456x x+ − =
.048 1540 .055 1456x x+ − =
.007 84x− = −
12000x =

18. Classwork
⚫ 1.2 Assignment
⚫ Page 97: 10-16 (even); pg. 89: 30-46 (even);
pg. 71: 94-106 (even)
⚫ 1.2 Classwork Check (due 10/2)
⚫ Quiz 1.1 (due 10/2)