SOLUTION: you gave me a site to go on and i can't get on it i am trying to see how to do these because i am out of school right now and hve no one to show me use linear combinations to

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Question 41929: you gave me a site to go on and i can't get on it i am trying to see how to do these because i am out of school right now and hve no one to show me

use linear combinations to solve the system linera equation

1. x + 2 =5
3x - 2y =7



2. x + y = 1
2x -3y = 12


3. x -y = -4
x + 2y = 5

I NEED SOME HELP
Found 2 solutions by AnlytcPhil, venugopalramana:
Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!
USE LINEAR COMBINATIONS TO SOLVE THE LINEAR EQUATIONS 

1.

   x + 2y = 5
  3x - 2y = 7 

Since the coefficients of y are already equal in
absolute value but opposite in sign, we can add
1 times the first to 1 times the second and the
y terms will cancel

 1(  x + 2y = 5)
 1( 3x - 2y = 7)
————————————————
    4x      = 12
          x = 3

Then substitute to find y:

     x + 2y = 5
   (3) + 2y = 5
         2y = 5 - 3
         2y = 2
          y = 1



2.    x +  y =  1
     2x - 3y = 12 

Multiply the 1st equation by 3
Multiply the second by 1

  3[  x +  y =  1]
  1[ 2x - 3y = 12]

     3x + 3y =  3
     2x - 3y = 12
    —————————————
     5x      = 15
           x = 3

Then substitute to find y

       x + y = 1
       3 + y = 1
           y = -2


3. x -  y = -4
   x + 2y = 5

Multiply the first equation by 2 and the second
by 1 to make the y's cancel out:

2[x -  y = -4]
1[x + 2y =  5]

  2x - 2y = -8
   x + 2y =  5
 —————————————
  3x      = -3
        x = -1 

Substitute to find y

  x + 2y = 5
 -1 + 2y = 5
      2y = 6
       y = 3

Edwin
AnlytcPhil@aol.com
Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!
you gave me a site to go on and i can't get on it i am trying to see how to do these because i am out of school right now and hve no one to show me
use linear combinations to solve the system linera equation
THESE ARE ALL SIMILAR PROBLEMS.I SHALL SHOW ONE AND YOU MAY TRY THE OTHERS.
1. x + 2 =5....................EQN.I
3x - 2y =7.....................EQN.II
1.THERE ARE 2 UNKNOWNS AND 2 EQNS.
2.IF THERE IS 1 UNKNOWN AND 1 EQN.WE CAN SOLVE IT
3.SO WE TRY TO REMOVE ONE UNKNOWN BY ELIMINATION.
EQN.I GIVES X=5-2=3
EQN.II ..3*3-2Y=7
2Y=9-7=2
Y=1
2. x + y = 1......................III
2x -3y = 12........................IV
EQN.I*2-EQN.II
2X+2Y-2X+3Y=2-12=-10
5Y=-10
Y=-2
PUT THIS IN EQN.III
X-2=1
X=3....ETC..
3. x -y = -4
x + 2y = 5
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