Transcript Lesson 5.2 - James Rahn

Solving Systems of Equations
Using Substitution





Graphing systems and comparing
table values are good ways to see
solutions.
It is not always easy to find a good
graphing window or the right
settings for the table to view the
solution.
Often these solutions are only
approximations.
To find the exact solutions, you’ll
need to work algebraically with the
equations.
Let’s look at the substitution
method.

On a rural highway a police officer sees a
motorist run a red light at 50 mph and begins
pursuit. At the instant the police officer
passes through the intersection at 60 mph,
the motorist is 0.2 mi down the road. When
and where will the office catch up to the
motorist?
◦ Write an equation in two variables to model this
situation.
Motorist:
Police:
d  0.2  50t
d  60t
d= the distance from the
intersection
t= time traveled

On a rural highway a police officer sees a
motorist run a red light at 50 mph and begins
pursuit. At the instant the police officer
passes through the intersection at 60 mph,
the motorist is 0.2 mi down the road. When
and where will the office catch up to the
motorist?
◦ Solve this system by the substitution method and
check the solution.
d  0.2  50t
d  60t
d  0.2  50t
•
When the officer catches up to the
motorist they will the same distance
from the intersection (so both
equations will have the same d value.
•
Replace the d value in one equation
with the d value from the other
equation.
•
Solve this new equation for t.
d  60t
60t  0.2  50t
60t  50t  0.2  50t  50t
10t  0.2
t  0.02 hrs
t  0.02 hrs
So at t=0.02 hours the police and motorist will be the same distance from the
intersection.
d  0.2  50t
d  60t
d  0.2  50  0.02 
d  60  0.02 
d  1.2 mi
d  1.2 mi
At t=0.02 hours both the police and motorist will be 1.2 miles from the
intersection. This is the only ordered pair that works in both equations.

Materials needed
◦
◦
◦
◦
Two ropes of different thickness about 1 m long
Measuring tape
One 9 meter long thin rope (optional)
One 10 meter long thick rope (optional)

Step 1 Measure the length of
the thinner rope without any
knots.
◦ Then tie a knot and measure the
length of the rope again. Continue
tying knots until no more can be
tied. Knots should be of the same
kind, size, and tightness. Record
the data for number of knots and
length of rope in a table.

Step 2 Define variables and write an equation
in intercept form to model the data you
collected in Step 1.
◦ What are the slope and y-intercept, and how do
they relate to the rope?

Step 3 Repeat Steps 1 and 2 for the thicker
rope.

Step 4 Suppose you have a 9-meter-long thin
rope and a 10-meter-long thick rope.
◦ Write a system of equations that gives the length of
each rope depending on the number of knots tied.




Step 5 Solve this system of equations using
the substitution method.
Step 6 Select an appropriate window setting
and graph this system of equations. Estimate
coordinates for the point of intersection to
check your solution. Compare this solution
with the one from Step 5.
Step 7 Explain the real-world meaning of the
solution to the system of equations.
Step 8 What happens to the graph of the
system if the two ropes have the same
thickness? The same length?
Sample Data I
Number of Knots
Length (cm)
0
100
1
89.7
2
78.7
3
68.6
4
57.4
5
47.8
6
38.1
Number of Knots
Length (cm)
0
90
1
83.1
2
76
3
68.8
4
61.9
5
75
6
67.8
Number of Knots
Length (cm)
0
100
1
94
2
88
3
81.3
4
75.7
5
69.9
6
63.5
Number of Knots
Length (cm)
0
90
1
86
2
81.9
3
77.3
4
73
5
68.9
6
64.8
Sample Data II


So far you have seen equations written in
intercept form.
These equations make it easy to use the
substitution method since they are already
both solved for y.
◦ y=900 -6x
◦ y=1000-10.3x

Sometimes it is necessary to place equations
in intercept form before using substitution.

A pharmacist has 5% saline (salt) solution and
20% saline solution. How much of each
solution should be combined to create a
bottle of 90 ml of 10% solution.
◦ Write a system of equations that models this
situation.
If x represents the amount of 5% solution and
y represents the amount of 20% solution then
Thinking about the salt in each
solution gives another equation
x  y  90
0.05x  0.2y  0.1(90)
0.05x  0.2y  9

A pharmacist has 5% saline (salt) solution and
20% saline solution. How much of each
solution should be combined to create a
bottle of 90 ml of 10% solution.
◦ Solve the one equation for x or y and substitute it
into the other equation. Find a solution.
x  y  90
y  90  x
y  90  x
y  90  60
y  30
0.05x  0.2y  9
0.05x  0.2(90  x )  9
0.05x  18  0.2 x  9
0.15x  18  9
0.15x  9
x  60


Saw limitations to solving a problem
graphically
Learned how to solve a system of equations
using substitution.