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  1. #1
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    math question help

    Choose two different positive numbers. Then construct a system of two linear equations with two variables, such that its solution consists of exactly the two chosen numbers. Explain your method of creating the system.
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  2. #2
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    Hello, fancyface!

    Choose two different positive numbers.
    Then construct a system of two linear equations with two variables,
    such that its solution consists of exactly the two chosen numbers.
    Explain your method of creating the system.

    As a teacher, I've done this many many times.

    First, I make a "skeleton" of the problem.

    . . \begin{array}{cc}2\boxed{X} + \boxed{Y} \;=\;\boxed{A} \\<br />
4\boxed{X} - \boxed{Y} \;=\;\boxed{B}\end{array}

    Then I select values for X and Y . . . say, X = 2,\:Y = 1

    Plug them into the skeleton and determine A and B.
    . . \begin{array}{cc}2\boxed{2} + \boxed{1}\:=\:\boxed{5} \\ 4\boxed{2} - \boxed{1}\:=\:\boxed{7}\end{array}

    Then the system: . \begin{Bmatrix}2x + y \:=\:5 \\ 4x - y \:=\:7\end{Bmatrix} has the solution: x = 2,\:y = 1

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    We can make the answers as hard as we like: . x = \frac{1}{2},\:y = \text{-}\frac{1}{3}

    Using the skeleton: . \begin{array}{cc}4\boxed{X} + 3\boxed{Y}\:=\:\boxed{A} \\ 2\boxed{X} - 9\boxed{Y}\:=\:\boxed{B}\end{array}

    Then: . \begin{array}{cc}4\left(\frac{1}{2}\right) + 3\left(\text{-}\frac{1}{3}\right) \:=\:\boxed{1} \\ 2\left(\frac{1}{2}\right) - 9\left(\text{-}\frac{1}{3}\right)\:=\:\boxed{4} \end{array}

    Therefore: . \begin{Bmatrix}4x + 3y \:=\:1\\ 2x - 9y\:=\:4\end{Bmatrix} has the solution: x = \frac{1}{2},\:y = \text{-}\frac{1}{3}

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