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Let a, b, c, d are distinct numbers to be chosen from the set {1, 2, 3, 4, 5}. If the least possible positive solution for $x$ to the system of equations
$\left. \begin{matrix}
   ax+by=1 \\
   cx+dy=2 \\
\end{matrix} \right\}$ can be represented in the form $\dfrac{p}{q}$ where p and q are relatively prime, then find the value of (p + q).

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: We must first solve these two linear equations simultaneously for $x$. Then, we must try to minimize this solution, using trial and error techniques separately for both numerator and denominator. We must, thus, find the minimum solution, and change it in rational form, if not already, which will give the values for p and q. Thus, we can add p and q to get our final answer.

Complete step by step answer:
Here, we are given two equations, $ax+by=1$ and $cx+dy=2$. Let us now solve these two equations for $x$.
To solve these two linear equations simultaneously, we can multiply the first equation with d, and the second equation with b, and finally, we can subtract the two resulting equations.
We have $ax+by=1$. On multiplying d on both sides, we get
$\left( ax+by \right)d=d$.
We can also write this as
$adx+bdy=d...\left( i \right)$
Similarly, we have $cx+dy=2$. On multiplication with b on both sides, we get
$\left( cx+dy \right)b=2b$.
This can also be written as
$bcx+bdy=2b...\left( ii \right)$
Now, on subtracting equation (i) and equation (ii), we get
$adx+bdy-bcx-bdy=d-2b$.
By simplification, we can write
$\left( ad-bc \right)x=d-2b$.
Hence, the solution for $x$ is $x=\dfrac{d-2b}{ad-bc}$.
We need to find the least possible positive solution, so, we can say that $x$ must be minimum.
We can, thus, say that $\left( d-2b \right)$ must be minimum, and $\left( ad-bc \right)$ must be maximum.
Now, we need $\left( d-2b \right)$ to be minimum, but greater than 0, because we need positive solution.
So, we can say that $d-2b=1$.
With trial-and-error technique, we can see that this equation is satisfied for two cases
Case (i) d = 5 and b = 2.
Case (ii) d = 3 and b = 1.
Substituting the value of $d-2b$, we get
$x=\dfrac{1}{ad-bc}$.
Now, we need to maximise $\left( ad-bc \right)$.
Case (i): d = 5 and b = 2.
We want $\left( ad-bc \right)$ to be maximum, and also know that a, b, c, d are all distinct numbers. So, we can say that a must be equal to 4, and c must be equal to 1.
Thus, $ad-bc=20-2=18$. And so, we have $x=\dfrac{1}{18}$.
Case (ii): d = 3 and b = 1.
We want $\left( ad-bc \right)$ to be maximum, and also know that a, b, c, d are all distinct numbers. So, we can say that a must be equal to 5, and c must be equal to 2.
Thus, $ad-bc=15-2=13$. And so, we have $x=\dfrac{1}{13}$.
We know very well that $\dfrac{1}{18}<\dfrac{1}{13}$.
Hence, the least possible positive solution is $x=\dfrac{1}{18}$.
Here, 1 and 18 are relatively prime, and so we can say that p = 1, and q = 18.
Thus, (p + q) = 1 +18 = 19.

Note: We must understand that for p and q to be relatively prime, p and q must have no common factors other than 1. Then, we can say that these numbers are prime with respect to one another. Also, we must note that all a, b, c, d are distinct numbers.