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Solving Systems of Equations: Applications and Examples, Schemes and Mind Maps of Linear Algebra

A lesson on solving systems of equations using substitution and elimination methods. It includes examples of real-life applications, such as finding the number of miles driven in the city and on the highway based on gasoline consumption, and determining the speed of a canoe and the current. The document also covers systems of equations with linear and quadratic functions.

What you will learn

  • How do you solve a system of equations using the substitution method?
  • What is the difference between the substitution and elimination methods for solving systems of equations?
  • How do you solve a system of equations using the elimination method?

Typology: Schemes and Mind Maps

2021/2022

Uploaded on 09/27/2022

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16-week Lesson 36 (8-week Lesson 30) Applications of Systems of Equations
1
The two methods we have used to solve systems of equations are
substitution and elimination. Either method is acceptable for solving the
systems of equations that we will be working with in this lesson.
Method of Substitution:
1. solve one equation for one variable (it doesnโ€™t matter which equation
you choose or which variable you choose).
2. substitute the solution from step 1 into the other equation.
3. solve the new equation from step 2.
4. back substitute to solve the equation from step 1.
Method of Elimination:
1. multiply at least one equation by a nonzero constant so the
coefficients for one variable will be opposites (same absolute value)
2. add the equations so the variable with the opposite coefficients will
be eliminated.
3. take the result from step 2 and solve for the remaining variable.
4. take the solution from step 3 and back substitute to any of the
equations to solve for the remaining variable.
Steps for solving applications:
1. assign variables to represent the unknown quantities
2. set-up equations using the variables from step 1
3. solve using substitution or elimination; it makes no difference which
method you use
Just as in the previous lesson, all of these application problems should
result in a system of equations with two equations and two variables:
{๐‘Ž๐‘ฅ+๐‘๐‘ฆ=๐‘
๐‘‘๐‘ฅ+๐‘’๐‘ฆ=๐‘“
The equations will usually be linear, but not always.
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The two methods we have used to solve systems of equations are

substitution and elimination. Either method is acceptable for solving the

systems of equations that we will be working with in this lesson.

Method of Substitution:

  1. solve one equation for one variable (it doesnโ€™t matter which equation

you choose or which variable you choose).

  1. substitute the solution from step 1 into the other equation.
  2. solve the new equation from step 2.
  3. back substitute to solve the equation from step 1.

Method of Elimination:

  1. multiply at least one equation by a nonzero constant so the

coefficients for one variable will be opposites (same absolute value)

  1. add the equations so the variable with the opposite coefficients will

be eliminated.

  1. take the result from step 2 and solve for the remaining variable.
  2. take the solution from step 3 and back substitute to any of the

equations to solve for the remaining variable.

Steps for solving applications:

  1. assign variables to represent the unknown quantities
  2. set-up equations using the variables from step 1
  3. solve using substitution or elimination; it makes no difference which

method you use

Just as in the previous lesson, all of these application problems should

result in a system of equations with two equations and two variables:

The equations will usually be linear, but not always.

Example 1 : Set-up a system of equations and solve using any method.

A salesperson purchased an automobile that was advertised as averaging

25 miles

gallon

of gasoline in the city and

40 miles

gallon

on the highway. A recent sales

trip that covered 1800 miles required 51 gallons of gasoline. Assuming

that the advertised mileage estimates were correct, how many miles were

driven in the city and how many miles were driven on the highway?

How many gallons of gasoline were used in the city?

How many gallons of gasoline were used on the highway?

Write an equation to represent the total number of gallons of gasoline

used on the trip.

How many miles were driven in the city, assuming the advertised mileage

of

25 miles

gallon

is correct?

25

miles

gallon

โˆ™ ๐‘ฅ gallons =

How many miles were driven on the highway, assuming the advertised

mileage of

40 miles

gallon

is correct?

40

miles

gallon

โˆ™ ๐‘ฆ gallons =

Write an equation to represent the total number of miles driven on the trip.

Example 2 : Set-up a system of equations and solve using any method.

For a particular linear function ๐‘“

= 11 and

= โˆ’ 9. Find the values of ๐‘š and ๐‘.

Example 3 : Set-up a system of equations and solve using any method.

For a particular quadratic function ๐‘“

2

and ๐‘“

= 7. Find the values of ๐‘ and ๐‘, given that ๐‘Ž = 5.

2

2

2

2

At this point I have two equations, but both equations have three

unknowns

๐‘Ž, ๐‘, and ๐‘

. Since weโ€™re told in the direction that ๐‘Ž = 5 , I

can replace ๐‘Ž with 5 in both equations, and this will leave me with two

equations with only two unknowns.

Since ๐‘ฅ represents how fast the people can move the boat, that means the

people can row the boat 7. 5 miles per hour. To find the value of ๐‘ฆ (the

speed of the current), Iโ€™ll back substitute to one of the prior equations by

replacing ๐‘ฅ with 7. 5.

Since ๐‘ฆ represents the speed of the current, that means the current is

moving at 4. 5 miles per hour.

Be sure to pay attention to units when working with distance, rate, and

time problems; if a rate is in terms of miles per hour, then time must be in

terms of hours in order to get a distance in terms of miles

miles

hour

โˆ™ hours = miles

This will be important on Example 5, because we will need to convert our

time units from minutes to hours.

Example 5 : Set-up a system of equations and solve using any method.

A short airplane trip between two cities took 30 minutes when traveling

with the wind. The return trip took 45 minutes when traveling against the

wind. If the speed of the plane with no wind is 320 mph, find the speed of

the wind (๐’˜) and the distance (๐’…) between the two cities (pay attention to

units).

miles =

miles

hour

โˆ™ hours

With Wind

Against Wind