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

hope this helps you......

Answer 2

Answer:-

$\begin{lgathered}72 = 56 \times 1 = 16 \\ 56 = 16 \times 3 + 8 \\ 16 = 8 \times 2 + 10\end{lgathered}$

Now L. C. M =8

$\begin{lgathered}56 - (16) \times 3 = 18 \\ 72 - 56 = 16\end{lgathered}$

$\begin{lgathered}56 - (72 - 56) \times 3 = 8 \\ 8 = 56 + 72( - 3) + 56(3) \\ 8 = 56(4) + 72( - 3) \\ d = 56 x + 72y\end {lgathered}$

$\begin{lgathered}x = 4 \: and \: y = - 3\\(72)(56)-(72)(56)\end{lgathered}$

$\begin{lgathered}8 = 56(4) + 72( - 3) + (72)(56) - (72)(56) \\ 8 = 56(4 + 72) + 72( - 3 - 56) \\ 8 = 56(76) + 72( - 59) \\ d = 56 x + 72y\end{lgathered}$

$\begin{lgathered}therefore.. \\ x = 56 \: and \: y = \: 59 \\ x \: and \: ar \: not \: unique\end{lgathered}$