Diophantine EquationsPractise finding integer solutions to equations with more than one unknown. |
This is level 1: linear diophantine equations of the form \(ax+by=c\). You will be awarded a trophy if you get at least 9 answers correct and you do this activity online.
Find a set of positive whole number solutions to the following equations, though there may be more.
InstructionsTry your best to answer the questions above. Type your answers into the boxes provided leaving no spaces. As you work through the exercise regularly click the "check" button. If you have any wrong answers, do your best to do corrections but if there is anything you don't understand, please ask your teacher for help. When you have got all of the questions correct you may want to print out this page and paste it into your exercise book. If you keep your work in an ePortfolio you could take a screen shot of your answers and paste that into your Maths file. |
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❎Level 1 - Linear Diophantine equations of the form \(ax+by=c\)
Level 2 - Equations derived from real life situations
Level 3 - Similar to Level 1 but with larger numbers
Level 4 - Non linear Diophantine equations
More on this topic including lesson Starters, visual aids, investigations and self-marking exercises.
A classic example of a Diophantine equation is \(3x + 2y = 11\).
The goal is to find values for \(x\) and \(y\) that are whole numbers (integers) and satisfy this equation.
To solve this, we can start by trying different values for \(x\) and see if we can find a corresponding value for \(y\) that makes the equation true. For example, let's try \(x = 1\):
\(3(1) + 2y = 11\)
\(3 + 2y = 11\)
\(2y = 11 - 3\)
\(2y = 8\)
\(y = \frac{8}{2}\)
\(y = 4\)
So, when \(x = 1\), \(y = 4\). Therefore, one solution to the equation is \(x = 1\) and \(y = 4\).
We can check if there are other solutions by trying different values for \(x\). If we try \(x = 2\):
\(3(2) + 2y = 11\)
\(6 + 2y = 11\)
\(2y = 11 - 6\)
\(2y = 5\)
This doesn't give us an integer solution for \(y\), since 5 divided by 2 is 2.5, not a whole number. Therefore, \(x = 2\) doesn't work.
Similarly, if we try \(x = 3\):
\(3(3) + 2y = 11\)
\(9 + 2y = 11\)
\(2y = 11 - 9\)
\(2y = 2\)
\(y = \frac{2}{2}\)
\(y = 1\)
So, another solution is \(x = 3\) and \(y = 1\).
Therefore, the Diophantine equation \(3x + 2y = 11\) has two positive integer solutions:
\((x, y) = (1, 4)\) and \((x, y) = (3, 1)\).
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