Unit 2: Systems of Linear Equations and Inequalities

Topic 5

Modeling Real World Problems with Systems

Many word problems translate to two linear equations in two variables. Typical contexts include:

Cost and revenue with a fixed fee plus a per unit rate.
Mixtures that combine amounts and concentrations.
Motion with rate, time, and distance.

Modeling steps:

Define variables with units.
Write equations from relationships in the text.
Solve the system by substitution or elimination.
State the answer with units and interpret the coordinates.
Check that values are realistic for the context.

Core Skills

Translate fixed fee and per unit into $C = F + r\cdot q$ .
For mixtures, write one equation for total amount and one for total of the substance.
For motion, use $d = rt$ and align times or distances across objects.
Decide substitution or elimination based on the simplest path.

Example 1: Cost and Revenue Break Even

A gym offers Plan A with a $40 sign up fee plus $15 per class, and Plan B with a $10 sign up fee plus $20 per class. For how many classes do they cost the same and what is that common cost? Variables

c

classes,

C

dollars. Plan A:

C = 40 + 15c

. Plan B:

C = 10 + 20c

. Set equal:

40 + 15c = 10 + 20c \Rightarrow 30 = 5c \Rightarrow c = 6

. Cost:

C = 40 + 15(6) = 130

. Answer

(c, C) = (6, 130)

. Same cost after 6 classes.

Example 2: Ticket Sales

At a fundraiser, student tickets cost $6 and adult tickets cost $10. In total 120 tickets were sold for $960. How many of each were sold? Variables

s

students,

a

adults. Amount:

s + a = 120

. Money:

6s + 10a = 960

. Eliminate

s

: multiply the first by 6 and subtract.

6s + 6a = 720

. Subtract from money:

(6s + 10a) - (6s + 6a) = 960 - 720

4a = 240 \Rightarrow a = 60

. Then

s = 60

. Answer 60 students and 60 adults.

Example 3: Mixture by Concentration

How many liters of a 30 percent acid solution must be mixed with a 10 percent acid solution to get 20 liters of a 25 percent solution? Variables

x

liters of 30 percent,

y

liters of 10 percent. Total volume:

x + y = 20

. Total acid:

0.30x + 0.10y = 0.25(20) = 5

. Substitute

y = 20 - x

0.30x + 0.10(20 - x) = 5

0.30x + 2 - 0.10x = 5 \Rightarrow 0.20x = 3 \Rightarrow x = 15

. Then

y = 5

. Answer 15 L of 30 percent with 5 L of 10 percent.

Example 4: Motion Toward Each Other

Two cyclists start 45 miles apart on a straight road and ride toward each other. One rides at 12 mph and the other at 15 mph. How long until they meet and how far did each travel? Variables

t

hours until meeting. Distances:

12t

and

15t

. Together they cover 45 miles.

12t + 15t = 45 \Rightarrow 27t = 45 \Rightarrow t = \tfrac{45}{27} = \tfrac{5}{3}

hours. Distances:

12 \cdot \tfrac{5}{3} = 20

miles and

15 \cdot \tfrac{5}{3} = 25

miles. Answer Meet after

1\tfrac{2}{3}

hours. Distances 20 miles and 25 miles.

Example 5: Two trains with different departure times

Train A leaves a station at 8:00 a.m. at 50 mph. Train B leaves the same station on the same track at 9:00 a.m. at 70 mph in the same direction. At what clock time does Train B catch Train A? Variables Let

t

be hours after 9:00 a.m. Distances from station at time

t

after 9:00 a.m.: Train A time is

t + 1

hours, distance

50(t + 1)

. Train B time is

t

hours, distance

70t

. Catch up when

70t = 50(t + 1)\Rightarrow 70t = 50t + 50 \Rightarrow 20t = 50 \Rightarrow t = 2.5

. Clock time

= 9{:}00 + 2.5\text{ h} = 11{:}30

a.m. Answer 11:30 a.m.

Key Takeaways

Define variables first and attach units.
Align equations to the structure of the context. For mixtures use amount and substance equations. For motion use distance equals rate times time.
Solve, interpret, and check that answers match the language of the question.