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.
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.
. . $\displaystyle \begin{array}{cc}2\boxed{X} + \boxed{Y} \;=\;\boxed{A} \\
4\boxed{X} - \boxed{Y} \;=\;\boxed{B}\end{array}$
Then I select values for $\displaystyle X$ and $\displaystyle Y$ . . . say, $\displaystyle X = 2,\:Y = 1$
Plug them into the skeleton and determine $\displaystyle A$ and $\displaystyle B$.
. . $\displaystyle \begin{array}{cc}2\boxed{2} + \boxed{1}\:=\:\boxed{5} \\ 4\boxed{2} - \boxed{1}\:=\:\boxed{7}\end{array}$
Then the system: .$\displaystyle \begin{Bmatrix}2x + y \:=\:5 \\ 4x - y \:=\:7\end{Bmatrix}$ has the solution: $\displaystyle x = 2,\:y = 1$
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We can make the answers as hard as we like: .$\displaystyle x = \frac{1}{2},\:y = \text{-}\frac{1}{3}$
Using the skeleton: .$\displaystyle \begin{array}{cc}4\boxed{X} + 3\boxed{Y}\:=\:\boxed{A} \\ 2\boxed{X} - 9\boxed{Y}\:=\:\boxed{B}\end{array}$
Then: .$\displaystyle \begin{array}{cc}4\left(\frac{1}{2}\right) + 3\left(\text{-}\frac{1}{3}\right) \:=\:\boxed{1} \\ 2\left(\frac{1}{2}\right) - 9\left(\text{-}\frac{1}{3}\right)\:=\:\boxed{4} \end{array}$
Therefore: .$\displaystyle \begin{Bmatrix}4x + 3y \:=\:1\\ 2x - 9y\:=\:4\end{Bmatrix}$ has the solution: $\displaystyle x = \frac{1}{2},\:y = \text{-}\frac{1}{3}$