SOLVING SPECIAL SYSTEMS ALGEBRAICALLY
About "Solving special systems algebraically"
Solving special systems algebraically :
When we solve a system of equations with no solution, we will get a false statement, and when we solve a system with infinitely many solutions,we will get a true statement.
Solving special systems algebraically - Examples
Example 1 :
Solve the system of linear equations by substitution.
x - y = - 2
- x + y = 4
Solution :
Step 1 :
Solve the equation - x + y = 4 for "y".
- x + y = 4
Add "x" to both sides.
(-x + y) + x = (4) + x
-x + y + x = 4 + x
Simplify.
y = 4 + x
Step 2 :
Substitute y = 4 + x in the other equation.
x - y = - 2
x - (4+x) = -2
x - 4 - x = -2
Simplify.
-4 = -2
Step 3 :
Interpret the solution.
The result (-4 = -2) is the false statement. So there is no solution for the given system.
Example 2 :
Solve the system of linear equations by elimination.
2x + y = - 2
4x + 2y = - 4
Solution :
Step 1 :
2x + y = - 2 ------- (1)
4x + 2y = - 4 ------- (2)
In the given system, to eliminate one of the variables, the coefficients of that variable must be same and signs must be opposite.
The coefficients of the variable "x" are not same in both the equations and also the variable "y".
To make the coefficients of "y" same and signs opposite, multiply the first equation by -2.
(1) ⋅ 2 ------ > (-2)(2x + y) = (-2)(-2)
-4x - 2y = 4 ------- (3)
Step 2 :
Now, we can add equations (2) and (3) to eliminate the variable 2y.
Step 3 :
Interpret the solution.
The result is the statement 0 = 0, which is always true.
So the system has infinitely many solutions.
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