February 15, 2013

Chapter #4 Lecture Videos

System of Linear Equations

Systems of Linear Equations with Three Variables

Solving Systems of Linear Inequalities

3 thoughts on “Chapter #4 Lecture Videos

  • In the first example of substitution method of solving system of linear equation with 2 variables:
    5x+8y = -1
    3x+y=7 the solution is ( 3, -2) because -57/-19 is equal to 3 not -3

  • Hi Eric- In video 2, you asked us to share the steps we would take to solve this particular type of system. Im confused, because you ended up walking us through each step in the video. Therefore, I am not sure what you are looking for? We combined 2 equations to eliminate the unknown values, set up a new system of linear equations to solve for the unknown then used the substitution method to solve for y, then z. From then on its a matter of multiple substitutions until the problem is solved. Please let me know if this is what you wanted.
    Thanks- Liz

  • Share the steps to solve this system: Holy Crap there are a lot of steps.

    1.OK, you have 3 equation, we will call them A,B,and C.
    2.First you added equations B and C to get rid of a variable.
    3.Then you multiplied equations A and B to get rid of a variable.
    4.Next you got y alone and used the Substitution Method.
    5.With that in place you where able to solve for z.
    6.Once you solved for z, you went back to the new system of linear equations and substituted for z and solved for y.
    7.Now that you had solved for y and z, you can substitute those and solve for x.
    8.Once you had x,y, and,z you went back to your original three equations A, B, and C.
    9.You chose one of the three equations and substituted for x,y, and z to make sure the statement checked out.

    Okay, maybe not that many steps, it seemed much longer in the video, no offence.

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