Elimination Method:
 Obtain coefficients of x(or y) that differ only in sign by multiplying all terms of one or both equations by suitably chosen constants.
 Add the equations to eliminate one variable, and solve the resulting equation.
 Back substituting the value obtained in step 2 into either of the original equations and solve for the other variable.
 Check the solution in both of the original equations.
Intersecting Lines:
Example:
Solve the system of linear equations:
3x + 2y = 4  Equation 1
5x  2y = 8  Equation 2
Solution:
Adding both equations, to eliminate y, we get
3x + 2y = 4 equation 1
5x  2y = 8  equation 2.
Adding equations.
8x = 12
x = 12/8 = 3/2. By back substituting we can solve for y.
substituting the value of x in equation 1 we have
3x + 2y = 4
3(3/2) + 2y = 4
y = 1/4
The solution of the two equations is x = 3/2 and y = 1/4
Example:
Solve the system of linear equations:
2x  3y = 7  Equation 1
3x + y = 5  Equation 2
Solution:
Here we can obtain the coefficients that differ only in sign by multiplying equation 2 by 3.
2x  3y = 7 equation 1
3( 3x + y = 5)  equation 2.
The equations we get are:
2x  3y = 7 equation 1
9x + 3y = 15  equation 2. Adding both equations, to eliminate y, we get

11x = 22
x = 2. By back substituting we can solve for y.
substituting the value of x in equation 1 we have
2x  3y = 7 equation 1
2(2)  3y = 7
4  3y = 7
 3y = 7 + 4
 3y = 3
y = 1
The solution of the two equations is x = 2 and y = 1
Verification:
Substituting the value of x and y in one of the two equations:
2x  3y = 7 equation 1
2x  3y
2(2)  3(1)
4  3
 7
Substituting in the equation 2:
3x + y = 5  equation 2.
3(2) + 1
6 + 1
5
Hence the values of x and y are correct.
Example:
Solve the system of linear equations:
5x + 3y = 9  Equation 1
2x  4y = 14  Equation 2
Solution:
Here we can obtain the coefficients that differ only in sign by multiplying equation 1 by 4 and equation 2 by 3.
4(5x + 3y = 9) equation 1
3(2x  4y = 14)  equation 2.
The equations we get are:
20x + 12y = 36 equation 1
6x  12y = 42  equation 2. Adding both equations, to eliminate y, we get

26x = 78
x = 3. By back substituting we can solve for y.
substituting the value of x in equation 1 we have
5x + 3y = 9 equation 1
5(3) + 3y = 9
15 + 3y = 9
3y = 9  15
3y = 6
y = 2
The solution of the two equations is x = 3 and y = 2
Verification:
Substituting the value of x and y in one of the two equations:
5x + 3y = 9 equation 1
5(3) + 3(2)
15 + 6
9
Hence the values of x and y are correct.
Parallel Lines:
Method of Elimination: Nosolution Case
Example:
Solve the system of linear equations:
x  2y = 3  Equation 1
2x + 4y = 1  Equation 2
Solution:
Here we obtain the coefficients that differ only in sign by multiplying equation 1 by 2 .
2(x  2y = 3)  Equation 1
2x + 4y = 1  Equation 2
The equations we get are:
2x  4y = 6  Equation 1
2x + 4y = 1  Equation 2 Adding both equations, to eliminate y, we get

0 = 7  False statement
Since there are no values of x and y for which 0 = 7 we can conclude that the system is inconsistent and has no solution.
In other words the two lines do not intersect they are parallel lines
Two lines Coincide:
Method of Elimination: Manysolution Case
Example:
Solve the system of linear equations:
2x  y = 2  Equation 1
4x  2y = 2  Equation 2
Solution:
To obtain the coefficients that differ only in sign by multiplying equation 2 by 1/2 .
The equations we get are:
2x  y = 2  Equation 1
1/2( 4x 2y = 2)  Equation 2
2x + y = 1 Adding both equations, to eliminate y, we get

0 = 0
Since the two equations turn out to be equivalent or have the same solution set, the system has infinitely many solutions.
In other words the two lines coincide
Directions: Solve for the variables using the method of Elimination. Also write at least 5 examples of your own.
