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

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Chapter 5. Section 1
Page 1 of 3
C. Bellomo, revised 12-Nov-07
Section 5.1 โ€“ Systems of Equations in Two Variables
โ€ข 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
โ€ข 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
โ€ข This first section focuses on finding the solution to a system of two linear equations
โ€ข 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:
โ€ข This is not the best way, because you have to rely on your scale being exact, and your graphing
abilities being exact
โ€ข This does help you find an approximate solution, and help you determine if there is a solution at all
The Substitution Method:
โ€ข If you have one equation that is already solved for x or y, the substitution method is appropriate
โ€ข In general, for the system
11
2
ymxb
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
โ€ข Example. Solve 5, 4
x
yxyโˆ’=โˆ’ =โˆ’
Notice the second equation is solved for x
( 4 ) 5yyโˆ’โˆ’=โˆ’ (step 1, plug this into the first equation)
5 5, 1yyโˆ’=โˆ’ = (step 2, solve for y)
4(1) 4x=โˆ’ =โˆ’ (step 3, solve for x)
The solution is ( โ€“4, 1)
โ€ข Note that this method also works well when you have one equation that is 'almost' solved
โ€ข Example. Solve 21,27xy x yโˆ’= + =โˆ’
Notice the first equation is 'almost' solved for y, that is, 21yx
=
โˆ’ (so is the other, really)
Let's finish the stepsโ€ฆ
2(2 1) 7
427
55, 1
xx
xx
xx
+โˆ’=โˆ’
+โˆ’=โˆ’
=โˆ’ =โˆ’
And 2( 1) 1 3y=โˆ’โˆ’=โˆ’
Solution: (โ€“1, โ€“3)
Check: 2( 1) ( 3) 1, ( 1) 2( 3) 7โˆ’โˆ’โˆ’ = โˆ’+โˆ’ =โˆ’. Yes
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Section 5.1 โ€“ 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

  1. Plug the first equation into the second, m x 1 + b 1 (^) = m x 2 + 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 x โˆ’ y = โˆ’5, x = โˆ’ 4 y Notice the second equation is solved for x ( 4 )โˆ’ y โˆ’ y = โˆ’ 5 (step 1, plug this into the first equation) โˆ’ 5 y = โˆ’5, y = 1 (step 2, solve for y ) x = โˆ’4(1) = โˆ’ 4 (step 3, solve for x ) The solution is ( โ€“4, 1)
  • Note that this method also works well when you have one equation that is 'almost' solved
  • Example. Solve^2 x^ โˆ’^ y^ =^ 1,^ x^ +^2 y = โˆ’^7 Notice the first equation is 'almost' solved for y , that is, y = 2 x โˆ’ 1 (so is the other, really) Let's finish the stepsโ€ฆ 2(2 1) 7 4 2 7 5 5, 1

x x x x x x

And y = 2( 1)โˆ’ โˆ’ 1 = โˆ’ 3 Solution: (โ€“1, โ€“3) Check: 2( 1)โˆ’ โˆ’ โˆ’( 3) = 1, ( 1)โˆ’ + 2( 3)โˆ’ = โˆ’ 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

  1. You want to eliminate either the x or y term
  2. First, get the coefficients ( a 's or b 's) to have the same number and opposite signs
  3. Add the two equations, combine like terms
  4. You should have an equation with only one variable ( x or y ). Solve for that variable
  5. Plug that value into either one of the original equations and solve for the other variable
  • Example. Solve 3 x + 4 y = โˆ’2, โˆ’ 3 x โˆ’ 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) = โˆ’2, โˆ’ 3 x โˆ’ 5(1) = 1 โ‡’ x = โˆ’ 2 (Note that using both is a way to check) The solution is (โ€“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 โ€“2 to get โˆ’ 6 x โˆ’ 8 y = โˆ’ 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 โ‡’ x = โˆ’ 2 (Note that using both is a way to check) The solution is (โ€“2, 3)

What Can Go Wrong?

  • Example. Solve 4 x โˆ’ 2 y = 5, 6 x โˆ’ 3 y = โˆ’ 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 โ€“3 to get โˆ’ 12 x + 6 y = โˆ’ 15 And multiply the second equation by 2 to get 12 x โˆ’ 6 y = โˆ’ 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)