Download Systems of Linear Equations and Their Solutions

ISHIK UNIVERSITY
FACULTY OF EDUCATION
Mathematics Education Department
MATH-203 Linear Algebra-I
System of Linear Equations and Solutions
Orhan TUĞ (PhDc)
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Systems of Linear Equations and Their Solutions
We have seen that all equations in the form Ax + By = C are straight lines
when graphed. Two such equations, such as those listed below, are called a
system of linear equations. A solution to a system of linear equations is an
ordered pair that satisfies all equations in the system. For example, (3, 4)
satisfies the system
x+y=7
(3 + 4 is, indeed, 7.)
x – y = -1
(3 – 4 is indeed, -1.)
Thus, (3, 4) satisfies both equations and is a solution of the system. The
solution can be described by saying that x = 3 and y = 4. The solution can also
be described using set notation. The solution set to the system is {(3, 4)} - that
is, the set consisting of the ordered pair (3, 4).
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Example:
Determining Whether an Ordered
Pair Is a Solution of a System
Determine whether (4, -1) is a solution of the system
x + 2y = 2
x – 2y = 6.
Solution Because 4 is the x-coordinate and -1 is the y-coordinate of (4, -1),
we replace x by 4 and y by -1.
x + 2y = 2
x – 2y = 6
?
?
4 + 2(-1) = 2
4 – 2(-1) = 6
?
?
4 + (-2) = 2
4 – (-2) = 6
?
2 = 2 true
4+2=6
6 = 6 true
The pair (4, -1) satisfies both equations: It makes each equation true. Thus, the
pair is a solution of the system. The solution set to the system is {(4, -1)}.
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The Number of Solutions to a System of Two
Linear Equations
The number of solutions to a system of two linear equations in two variables
is given by one of the following.
Number of Solutions
Exactly one ordered-pair solution
No solution
Infinitely many solutions
y
y
x
Exactly one solution
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What This Means Graphically
The two lines intersect at one point.
The two lines are parallel.
The two lines are identical.
y
x
No Solution (parallel lines)
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x
Infinitely many solutions
(lines coincide)
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Determining Types of solutions
One way to determine the type of solution you expect to get is by
looking at the coefficients of each variable in the two equations.
 A1 x  B1 y  C1
Consider the general systems:

 A2 x  B2 y  C2
Compare the corresponding coefficients:
A1 B1 C1
Same line:


A2
Parallel lines:
Unique Solution:
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B2
C2
A1 B1 C1


A2 B2 C2
A1 B1

A2 B2
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Solving Linear Systems by Substitution
1.
2.
3.
4.
5.
Solve either of the equations for one variable in terms of the other. (If
one of the equations is already in this form, you can skip this step.)
Substitute the expression found in step 1 into the other equation. This
will result in an equation in one variable.
Solve the equation obtained in step 2.
Back-substitute the value found in step 3 into the equation from step 1.
Simplify and find the value of the remaining variable.
Check the proposed solution in both of the system's given equations.
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Example: Solving a System by Substitution
Solve by the substitution method:
5x – 4y = 9
x – 2y = -3.
Solution
Step 1 Solve either of the equations for one variable in terms of the other.
We begin by isolating one of the variables in either of the equations. By
solving for x in the second equation, which has a coefficient of 1, we can avoid
fractions.
x - 2y = -3
This is the second equation in the given system.
x = 2y - 3
Solve for x by adding 2y to both sides.
Step 2 Substitute the expression from step 1 into the other equation. We
substitute 2y - 3 for x in the first equation.
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x = 2y – 3
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5 x – 4y = 9
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Example: Solving a System by Substitution
Solve by the substitution method:
5x – 4y = 9
x – 2y = -3.
Solution
This gives us an equation in one variable, namely
5(2y - 3) - 4y = 9.
The variable x has been eliminated.
Step 3 Solve the resulting equation containing one variable.
5(2y – 3) – 4y = 9
This is the equation containing one variable.
10y – 15 – 4y = 9
Apply the distributive property.
6y – 15 = 9
Combine like terms.
6y = 24
Add 15 to both sides.
y=4
Divide both sides by 6.
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mor
Example: Solving a System by Substitution
Solve by the substitution method:
5x – 4y = 9
x – 2y = -3.
Solution
Step 4 Back-substitute the obtained value into the equation from step 1.
Now that we have the y-coordinate of the solution, we back-substitute 4 for y
in the equation x = 2y – 3.
x = 2y – 3
Use the equation obtained in step 1.
x = 2 (4) – 3
Substitute 4 for y.
x=8–3
Multiply.
x=5
Subtract.
With x = 5 and y = 4, the proposed solution is (5, 4).
Step 5 Check. Take a moment to show that (5, 4) satisfies both given
equations. The solution set is {(5, 4)}.
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Solving Linear Systems by Addition
1.
2.
3.
4.
5.
6.
If necessary, rewrite both equations in the form Ax + By = C.
If necessary, multiply either equation or both equations by appropriate
nonzero numbers so that the sum of the x-coefficients or the sum of the
y-coefficients is 0.
Add the equations in step 2. The sum is an equation in one variable.
Solve the equation from step 3.
Back-substitute the value obtained in step 4 into either of the given
equations and solve for the other variable.
Check the solution in both of the original equations.
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Example: Solving a System by the Addition
Method
Solve by the addition method:
2x = 7y - 17
5y = 17 - 3x.
Solution
Step 1 Rewrite both equations in the form Ax + By = C. We first arrange
the system so that variable terms appear on the left and constants appear on
the right. We obtain
2x - 7y = -17
3x + 5y = 17
Step 2 If necessary, multiply either equation or both equations by
appropriate numbers so that the sum of the x-coefficients or the sum of
the y-coefficients is 0. We can eliminate x or y. Let's eliminate x by
multiplying the first equation by 3 and the second equation by -2.
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Solution
2x – 7y = -17
3x + 5y = 17
Steps 3 and 4
Multiply by 3.
Multiply by -2.
3•2x
-2•3x
– 3•7y = 3(-17)
+ (-2)5y = -2(17)
6x – 21y = -51
-6x + 10y = -34
Add the equations and solve for the remaining variable.
6x – 21y = -51
-6x – 10y = -34
-31y = -85
Add:
-31y = -85
-31
-31
y
= 85/31
Divide both sides by -31.
Simplify.
Step 5 Back-substitute and find the value for the other variable. Backsubstitution of 85/31 for y into either of the given equations results in
cumbersome arithmetic. Instead, let's use the addition method on the given
system in the form Ax + By = C to find the value for x. Thus, we eliminate y
by multiplying the first equation by 5 and the second equation by 7.
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Solution
2x – 7y = -17
3x + 5y = 17
Multiply by 5.
Multiply by 7.
5•2x – 5•7y = 5(-17)
7•3x + 7•5y = 7(17)
10x – 35y = -85
21x + 35y = 119
Add: 31x
x
= 34
= 34/31
Step 6 Check. For this system, a calculator is helpful in showing the
solution (34/31, 85/31) satisfies both equations. Consequently, the solution set
is {(34/31, 85/31)}.
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Examples
Determine the type of solution, then solve.
3x  2 y  1
1. 
x  y  3
4 x  6 y  12
2. 
6 x  9 y  36
4 x  5 y  3
3. 
2 x  3 y  7
2 x  y  3
4. 
4 x  2 y  5
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