Documents
Resources
Learning Center
Upload
Plans & pricing Sign in
Sign Out

Solving Systems of Linear Equations By Elimination

VIEWS: 6 PAGES: 8

									Solving Systems of Linear
Equations By Elimination
        What is Elimination?

   To eliminate means to get rid of or
    remove.

   You solve equations by eliminating one
    of the variables (x or y) using addition
    and or subtraction.
               Example 1
Solve the following system of linear
equations by elimination.
                  2x – 3y = 15 (1)
                  5x + 3y = 27 (2)
Add equation (1)
to equation (2)   7x + 0y = 42
                       7x = 42  By eliminating y, we
                                  can now solve for x
                        x=6
                  Example 1
Substitute x= 6 into               Check your solution x = 6
equation (1) to solve for y        and y = -1 in equation (2)
   2x – 3y = 15                         5x + 3y = 27
 2(6) – 3y = 15                     5(6) + 3(-1) = 27
   12 – 3y = 15                          30 – 3 = 27
      – 3y = 15 – 12                         27 = 27
      – 3y = 3                               LS = RS
         y = -1
                       Therefore, the solution set = {(6,-1)}
                Example 2
                5x + 4y = -28 (1)
               3x + 10y = -13 (2)

 If we were to add these equations we would obtain
8x + 14y = -41

 Even though we have only one equation now, we still
have 2 variables.

 We need to multiply the equations by terms that will
allow us to eliminate either x or y.
                 Example 2
              5x + 4y = -28 (1)
             3x + 10y = -13 (2)
  If we multiply equation (1) by 5 and equation (2) by -2,
 we be able to eliminate y.
 (1) x 5   25x + 20y = -140 (3)     When you change the
 (2) x -2   -6x – 20y = 26 (4)     equations you need to
Add (3) & (4)  19x = -114         renumber them.

                    x = -6
                    Example 2
Substitute x = -6 into equation (1)
    5x + 4y = -28         Check your answer x = -6 and
 5(-6) + 4y = -28         y = ½ into equation (2)

   -30 + 4y = -28            3(-6) + 10(½) = -13
         4y = -28 +30                 -18 + 5 = -13
         4y = 2                          -13 = -13
              2
           y                             LS = RS
              4
              1
           y        Therefore, the solution set = {(-6, ½)}
              2
             Questions?

   Any Questions?

   Homework: 1.4 # 5 – 12

   Complete Homework for Monday!

           Have a Great Weekend 

								
To top