3 thoughts on “Algebra – 11

  1. First of all, thanks so much Sir for the post!!

    I have understood the solution of first question x+3y+8z=20 but still confused with the approach that is followed here. I have tried to solve it by taking x=0,1,2,3…and obtained 5 solution..How is the equation with different value of z alone gives all the 14 possible solutions? I hope my question is clear.

  2. When we are solving an equation with 3 variables it gets very hard to juggle all three. So what we are doing here is just saying “OK look. Let me just keep one variable fixed. Then it reduces to the two-variable case I already know how to solve”. But the key thing is, any variable would do. However, given that time is a constraint, we need to choose well.

    The equation is x + 3y + 8z = 20. If we start with x, then we will have to consider 21 cases (from x = 0 to x = 20) and check them all. They will give you the same 14 solutions:
    For x = 0 (y, z) = (4, 1)
    For x = 1 (y, z) = (1, 2)
    For x = 2 (y, z) = (6, 0)
    For x = 3 (y, z) = (3, 1)
    For x = 4 (y, z) = (0, 2)
    …and so on. Some cases (x = 7, 10, 13, 15, 16, 18, 19) will give no solution. Chances are, you missed out a few of the 14 cases in your attempt.

    Starting with z makes more sense as I only need to check 3 cases. Reducing a complex 3-variable equation to 3 simpler 2-variable equations is fine. Reducing it to 21 simpler equations is not time-efficient, plus there is more chance of missing somethng. Hence we prefer to start with z.

    regards
    J

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