Question

If d is the HCF of 56 and 72, find x,y satisfying $d=56x+72y$ . Also, show that x and y are not unique.

Answer 1

Hint:We will first find the HCF of 56 and 72 normally. Then we will try to find x and y by inspection method. Then if we can find more than one point $\left( x,y \right)$ which satisfies the equation, then we can say that x and y are not unique.

Complete step-by-step answer:
First, let us find HCF of 56 and 72. Let us find the factors of 56 and 72 and if any of the factors are common, then these common factors consist of the HCF.
Now, 56 can be written as $2\times 2\times 2\times 7$ .
Thus, $56={{2}^{3}}\times 7$ .
Also, in the same way, 72 can be written as $2\times 2\times 2\times 3\times 3$ .
Thus $72={{2}^{3}}\times {{3}^{2}}$ .
As you can see, ${{2}^{3}}$ is common in both the numbers. Thus the HCF of 56 and 72 is ${{2}^{3}}$ which is 8.
Now, we move to the equation that needs to be satisfied.
$d=56x+72y,$ Where d is the HCF of 56 and 72.
Thus $8=56x+72y$.
Taking out 8 as a common factor from both numbers in RHS, we get,
$8=8\times \left( 7x+9y \right)$ .
Cancelling 8 on both sides, we get,
$1=7x+9y$ .
Now, if we put x=0, we get,
$\begin{aligned} & 1=7\left( 0 \right)+9y \\\ & \Rightarrow 1=9y \\\ \end{aligned}$
Cross multiplying 9, we get,
$y=\dfrac{1}{9}$ .
Thus one of the $\left( x,y \right)$ pairs is $\left( 0,\dfrac{1}{9} \right)$ .
Now, let us put y=0.
Therefore,
$\begin{aligned} & 1=7x+9\left( 0 \right) \\\ & \Rightarrow 1=7x \\\ \end{aligned}$
Cross multiplying 7, we get,
$x=\dfrac{1}{7}$ .
Thus one more pair is $\left( x,y \right)=\left( \dfrac{1}{7},0 \right)$ .
Thus we got two pairs of $\left( x,y \right)$ from the inspection method.
Thus, it is proved that x and y are not unique.

Note: You can also observe that the equation $7x+9y=1$ is a straight line. It is satisfied by infinitely many $\left( x,y \right)$ points. Thus, as we said before, if we can find more than one $\left( x,y \right)$ pair, we can say that x,y are not unique. But here we found that there are infinitely many points $\left( x,y \right)$ that satisfy the equation. Thus x and y are not unique.