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.

0.2 50d t Motorist:

60d tPolice:

d= the distance from the intersectiont= 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.

0.2 50d t

60d t

0.2 50d t

60d t

• 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.

60 0.2 50t t

• Solve this new equation for t.60 50 0.2 50 50t t t t

0.02t hrs

10 0.2t

0.2 50d t 60d t

0.02 hrst

So at t=0.02 hours the police and motorist will be the same distance from the intersection.

0.2 50 0.02

1.2 mi

d

d

60 0.02

1.2 mi

d

d

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?

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

Sample Data I

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.

90x y If x represents the amount of 5% solution and y represents the amount of 20% solution then

0.05 0.2 0.1(90)

0.05 0.2 9

x y

x y

Thinking about the salt in each solution gives another equation

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.

90

90

x y

y x

0.05 0.2 9

0.05 0.2(90 ) 9

x y

x x

0.05 18 0.2 9

0.15 18 9

0.15 9

60

x x

x

x

x

90

90 60

30

y x

y

y

Saw limitations to solving a problem graphically

Learned how to solve a system of equations using substitution.