Ask Experts Questions for FREE Help !
Ask
    990123e's Avatar
    990123e Posts: 3, Reputation: 1
    New Member
    		  
    #1

    May 14, 2009, 02:06 PM
    solving using the elimination method
    I don't understand solving using for the elimination method. Algebra is like chinese to me can some help me... here is an example of a problem...

    4x-2y=-2
    4x+3y=-12
    :mad::mad::mad:
    :confused::confused::confused:
    Perito's Avatar
    Perito Posts: 3,139, Reputation: 150
    Ultra Member
    		  
    #2

    May 14, 2009, 02:11 PM
    4x-2y=-2
    4x+3y=-12




    Subtract the second equation from the first



    Viola! X is eliminated!



    Substitute this back in to either of the equations to solve for x.

    If the coefficients of the variable you're going to eliminate aren't the same, multiply the entire equation -- or both equations, with constants so that when added, one of the variables will be "eliminated".
    990123e's Avatar
    990123e Posts: 3, Reputation: 1
    New Member
    		  
    #3

    May 14, 2009, 02:14 PM
    Quote Originally Posted by Perito View Post



    Subtract the second equation from the first



    Viola! x is eliminated!

    Substitute this back in to either of the equations to solve for x.

    If the coefficients of the variable you're going to eliminate aren't the same, multiply the entire equation -- or both equations, with constants so that when added, one of the variables will be "eliminated".
    so it's that easy because how do i no to subtract sometimes there are questions like
    8x-y=20
    5x+y=-8
    Perito's Avatar
    Perito Posts: 3,139, Reputation: 150
    Ultra Member
    		  
    #4

    May 14, 2009, 02:19 PM
    8x-y=20
    5x+y=-8
    multiply 8x-y=20 by 5
    multiply 5x+y=-8 by -8

    40x -5y = 100
    -40x -8y = 64

    add them

    -13y = 164
    y = -164/13 ≈ 12.616

    You could also just add them directly since the y-coefficients are the same (in absolute value). This would eliminate y

    8x-y=20
    5x+y=-8

    13x = 12
    x = 12/13


    You can select constants to multiply each equation with. It'll still be an equation.
    990123e's Avatar
    990123e Posts: 3, Reputation: 1
    New Member
    		  
    #5

    May 14, 2009, 02:21 PM
    Quote Originally Posted by Perito View Post
    multiply 8x-y=20 by 5
    multiply 5x+y=-8 by -8

    40x -5y = 100
    -40x -8y = 64

    add them

    -13y = 164
    y = -164/13 ≈ 12.616

    You can select constants to multiply each equation with. It'll still be an equation.
    how do I still find x and for the first equation I plugged in x and got 8. is that right? (8,-10)
    Perito's Avatar
    Perito Posts: 3,139, Reputation: 150
    Ultra Member
    		  
    #6

    May 14, 2009, 02:30 PM

    I made a sign error in my first post. :mad: :rolleyes: Y is -2 in the first problem. X will turn out to be -3/2

    Yes, you just plug that into either equation and solve for x. To check it, insert the values for x and y into the other equation and it has to be an equality. That's how I figured out that I had made a mistake.
Not your question? Ask your question View similar questions
 

Question Tools Search this Question
Search this Question:

Advanced Search


Check out some similar questions!

Solving using elimination. [ 5 Answers ]

Doe anyone know some of these answers? Please help!! :confused: x + y = 10 x - y = 4 4x - 3y = -2 -2x + 3y = 4

Solving systems with exponents using elimination? [ 1 Answers ]

My teacher didn't go over solving systems with exponents very well. How would I solve a problem like 3x^2-y^2=11 & x^2+4y^2=8

Solving using elimination [ 3 Answers ]

I need the answer for x+y=2 2x-y=4

Elimination method in solving an equation [ 1 Answers ]

I have two problems that I need help with I. 3x-2y=8 -12x+8y=32 II. -4x+3y=5 12x=9y=15

Solving Systems By Elimination.Algebra 1 [ 1 Answers ]

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

View more questions Search