ALGEBRA/EQUATION

I believe I have the answers already. Just need confirmation please!
Solve the following system of equations.

x + 2y = 5
x = 2 - 2 y;

For the first equation, I canceled out the x and divided by 2 on both sides. So y = 5 - x/ y.
For the second equations, this is what I got. x = 2 -2 y / Y = 2 - x/ 2. Thank you so much.

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asked by Amy
  1. When we have a system of linear equations, it can fall into one of three cases:
    1. no solution,
    2. a unique solution, and
    3. an infinite number of solutions.

    I am not sure which level of linear algebra you are working on, but I will explain it in the easiest terms:

    1. happens when the equations are inconsistent, or contradictory, for example:
    x + 2y = 5
    x + 2y = 2
    The left hand sides are identical, and the right hand sides are different, which is impossible. So we say that there is no solution.

    2. happens most of the time when there is a unique solution.
    For example:
    2x+y=3
    x+2y=3
    If you solve the system of equations, x=1, y=1. No other values will satisfy both equations.

    3. happens when there are more unknowns than equations, so the solution is based on some "assumed values" of arbitrary variables of our choice, called free variables. This way, there is an infinite number of solutions, depending on the choice of the values of the free variables.
    Example:
    2x+y=6
    If we take y as the free variable, then we have:
    x=(6-T)/2
    y=T
    where T can take on any finite value, and the equation will be satisfied.

    Decide on which case the given equations fall, and post your answer for checking if you wish.

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  2. I would say it falls under the category of number 2. I'm trying. What do you think?

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    posted by Amy
  3. Have you tried to rearrange the equations to the form where variables are on the left of the equal sign, and the constant terms on the right? Perhaps it will help you see better.

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