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2.1: Solution of Linear Systems by the Echelon Method
After completing this section, you will be able to do the following:


Use echelon method to evaluate linear systems.
Identify and interpret types of solutions:
o no solution,
o one solution, and
o infinite number of solutions.
Introduction to systems of equations:
System of equations: two or more equations are used to find a common solution
Solution of a system: an ordered pair (x, y) that makes all the involved equations true
Determine whether the given ordered pair is a solution of the system.
Ex.)
Check:
(4, 1)
3 = 3 True
8–3=5
5 = 5 True
Solution: (4, 1)
Ex.)
Check:
True
False
(-3, 3)  Not a solution
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How do we find the solution to the system?
Method 1: Graphing (This can be done on a graphing calculator to check work.)
Unique solution/One solution
Consistent: at least one solution
Independent
No solution
Inconsistent
Independent
Infinite solutions
Consistent
Dependent
Method 2: Substitution method
1.
2.
3.
4.
Solve for a variable.
Plug the expression into that variable of the second equation.
Find the value of the one variable left by solving.
Find the value of the other variable by substituting the value found in step 3 into the equation
from step 1.
5. Plug the ordered pair into both equations to check if it is a solution.
Ex.)
(
)
 No solution
2
Ex.)
Plug in: y = 1
Plug in:
Solve for a variable:
-6
-6
Solution: (4, 1)
Method 3: Echelon method/combination
1. Rewrite each equation in standard form.
2. Multiply each equation by a number that causes the coefficient of x or y to be opposites of
each other.
3. Add the equations together.
4. Solve for the solution to the left over variable.
5. Find the solution of the second variable by substituting the value found in step 4 into one of
the original equations.
6. Check to see if the ordered pair is a solution to the system.
What is allowed when transforming a system?
The following transformations can be applied to a system of equations to get an equivalent system:
1. Exchange any two equations
2. Multiplying both sides of an equation by any nonzero real number (if you want to
divide/multiply by a fraction)
Replacing any equation by nonzero multiple of that equation plus a non zero multiple of any
other equation
Ex.)
plug into an equation:
2*R2 2*(
)
-6
(divide by -2)
y = -2
+2
+2
(divide by -3)
Unique solution: ( , -2 )
3
Ex.)
-2*R1
-2*(
+
0 ≠ 6 Therefore, no solution
Ex.)
3*R1
3(
)
9
0 = 0  Infinite number of solutions
How do we express the solution?
We are going to write the answer using a parameter: rewriting the other equations in terms of this
variable. We will always use the rightmost variable.
1. Solve one equation for x:
+y
+y
(divide both sides by 3)
2. Write the answer as a point: (
Ex.)
= 31
*Always get rid of fractions first.
Do this by multiplying every term by
the common denominator.*
------------------------
5(
5(
= 31)
)
5*R1
5*R2
------------------------
4
-10* R1
-
-151y = -1505 (divide by -151)
y = 9.97
Plug back into the first equation.
= 31
= 31
31 (add 29.91 to both sides)
(multiply both sides by 5)
x = 5.45
Solution: (5.45, 9.97)
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