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Solving Systems of Linear Equations in Two Variables: Methods and Examples, Study notes of Algebra

University of Nevada - Las Vegas (UNLV)Algebra

An introduction to systems of linear equations in two variables, explaining the concepts of inconsistent and non-unique solutions. It presents two methods for solving such systems: the substitution method and the elimination method. Examples and explanations for each method.

Typology: Study notes

2009/2010

Uploaded on 02/24/2010

koofers-user-xz4 🇺🇸

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Chapter 5. Section 1

Page 1 of 3

C. Bellomo, revised 12-Nov-07

Section 5.1 – Systems of Equations in Two Variables

• When you have two or more equations that need to be solved at the same time, we refer to them as a

system of equations

• Count the number of equations and count the number of variables. In general, the system can be

solved uniquely if there are the same number of each

• This first section focuses on finding the solution to a system of two linear equations

• Is it possible there is no solution (inconsistent system), or there are many solutions (non-unique

solutions)?

Yes. When finding the intersection point of two lines, it is possible the lines will not intersect.

How can this happen?

There can also be many solutions if the lines are the same (one is on top of the other)

Solving A System of Linear Equations Graphically:

• This is not the best way, because you have to rely on your scale being exact, and your graphing

abilities being exact

• This does help you find an approximate solution, and help you determine if there is a solution at all

The Substitution Method:

• If you have one equation that is already solved for x or y, the substitution method is appropriate

• In general, for the system

11

2

ymxb

=+

1. Plug the first equation into the second, 112

mx b mx b

+

=+

2. You should have an equation with only one variable (x or y). Solve for that variable

3. Plug that value into either one of the original equations and solve for the other variable

• Example. Solve 5, 4

x

yxy−=− =−

Notice the second equation is solved for x

( 4 ) 5yy−−=− (step 1, plug this into the first equation)

5 5, 1yy−=− = (step 2, solve for y)

4(1) 4x=− =− (step 3, solve for x)

The solution is ( –4, 1)

• Note that this method also works well when you have one equation that is 'almost' solved

• Example. Solve 21,27xy x y−= + =−

Notice the first equation is 'almost' solved for y, that is, 21yx

=

− (so is the other, really)

Let's finish the steps…

2(2 1) 7

427

55, 1

xx

+−=−

=− =−

And 2( 1) 1 3y=−−=−

Solution: (–1, –3)

Check: 2( 1) ( 3) 1, ( 1) 2( 3) 7−−− = −+− =−. Yes

Partial preview of the text

Download Solving Systems of Linear Equations in Two Variables: Methods and Examples and more Study notes Algebra in PDF only on Docsity!

Page 1 of 3

Section 5.1 – Systems of Equations in Two Variables

When you have two or more equations that need to be solved at the same time, we refer to them as a system of equations
Count the number of equations and count the number of variables. In general, the system can be solved uniquely if there are the same number of each
This first section focuses on finding the solution to a system of two linear equations
Is it possible there is no solution (inconsistent system), or there are many solutions (non-unique solutions)? Yes. When finding the intersection point of two lines, it is possible the lines will not intersect. How can this happen? There can also be many solutions if the lines are the same (one is on top of the other)

Solving A System of Linear Equations Graphically :

This is not the best way, because you have to rely on your scale being exact, and your graphing abilities being exact
This does help you find an approximate solution, and help you determine if there is a solution at all

The Substitution Method :

If you have one equation that is already solved for x or y , the substitution method is appropriate
In general, for the system 1 1 2

y m x b y m x b

Plug the first equation into the second, m x 1 + b 1 (^) = m x 2 + b
You should have an equation with only one variable ( x or y ). Solve for that variable
Plug that value into either one of the original equations and solve for the other variable

Example. Solve x − y = −5, x = − 4 y Notice the second equation is solved for x ( 4 )− y − y = − 5 (step 1, plug this into the first equation) − 5 y = −5, y = 1 (step 2, solve for y ) x = −4(1) = − 4 (step 3, solve for x ) The solution is ( –4, 1)
Note that this method also works well when you have one equation that is 'almost' solved
Example. Solve^2 x^ −^ y^ =^ 1,^ x^ +^2 y = −^7 Notice the first equation is 'almost' solved for y , that is, y = 2 x − 1 (so is the other, really) Let's finish the steps… 2(2 1) 7 4 2 7 5 5, 1

x x x x x x

And y = 2( 1)− − 1 = − 3 Solution: (–1, –3) Check: 2( 1)− − −( 3) = 1, ( 1)− + 2( 3)− = − 7_._ Yes

Page 2 of 3

The Elimination Method :

The elimination method works well when both the equations have coefficients in front of the x and y terms
In general, for the system 1 1 1 2 2 2

a y b x c a y b x c

You want to eliminate either the x or y term
First, get the coefficients ( a 's or b 's) to have the same number and opposite signs
Add the two equations, combine like terms
You should have an equation with only one variable ( x or y ). Solve for that variable
Plug that value into either one of the original equations and solve for the other variable

Example. Solve 3 x + 4 y = −2, − 3 x − 5 y = 1 Notice it would not be 'easy' to solve for either variable in either equation It is already set up that the x 's coefficients have the same number and opposite signs Add these equations together 3 4 2 3 5 1 1, 1

x y x y y y

3 x + 4(1) = −2, − 3 x − 5(1) = 1 ⇒ x = − 2 (Note that using both is a way to check) The solution is (–2, 1)

Example. Solve 3 x + 4 y = 6, 2 x + 3 y = 5 Notice it would not be 'easy' to solve for either variable in either equation Let's eliminate the x. The least common multiple of 3 and 2 is 6 Multiply the first equation by –2 to get − 6 x − 8 y = − 12 And multiply the second equation by 3 to get 6 x + 9 y = 15 Add these equations together 6 8 12 6 9 15 1 3

x y x y y

3 x + 4(3) = 6 or 2 x + 3(3) = 5 ⇒ x = − 2 (Note that using both is a way to check) The solution is (–2, 3)

What Can Go Wrong?

Example. Solve 4 x − 2 y = 5, 6 x − 3 y = − 10 Notice it would not be 'easy' to solve for either variable in either equation. Use elimination The least common multiple of 4 and 6 is 12 Multiply the first equation by –3 to get − 12 x + 6 y = − 15 And multiply the second equation by 2 to get 12 x − 6 y = − 20 Add these equations together 12 6 15 12 6 20 0 35

x y x y

No ( x , y ) satisfy the system (no solution)

Solving Systems of Linear Equations in Two Variables: Methods and Examples, Study notes of Algebra

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Partial preview of the text

Download Solving Systems of Linear Equations in Two Variables: Methods and Examples and more Study notes Algebra in PDF only on Docsity!

Section 5.1 – Systems of Equations in Two Variables