EXPRESS theHCF/GCD of 48 and 18 as a linear combination

1.	EXPRESS theHCF/GCD of 48 and 18 as a linear combination
Answer» Let us find HCF of 48 and 18{tex}48 = 18 \\times 2 + 12{/tex}{tex}18 = 12 \\times 1 + 6{/tex}{tex}12 = 6 \\times 2 + 0{/tex}Hence HCF (48, 18) = 6Now, 6 = 18 - 12\xa0{tex}\\times{/tex}\xa016 = 18 - (48 - 18\xa0{tex}\\times{/tex}\xa02)6 = 18 - 48\xa0{tex}\\times{/tex}\xa01 + 18{tex}\\times{/tex}26 = 18\xa0{tex}\\times{/tex}\xa03 - 48\xa0{tex}\\times{/tex}\xa016 = 18\xa0{tex}\\times{/tex}\xa03 + 48\xa0{tex}\\times{/tex}(-1)i.e., 6 = 18x + 48y ........ (1)where x= 3, y = -1{tex}\\therefore{/tex}\xa06 = 18\xa0{tex}\\times{/tex}\xa03 + 48\xa0{tex}\\times{/tex}\xa0(-1)= 18\xa0{tex}\\times{/tex}\xa03 + 48\xa0{tex}\\times{/tex}\xa0(-1) + 18\xa0{tex}\\times{/tex}\xa048 - 18\xa0{tex}\\times{/tex}\xa048= 18(3 + 48) + 48(-1 - 18)= 18\xa0{tex}\\times{/tex}\xa051 + 48\xa0{tex}\\times{/tex}\xa0(-19)6 = 18x + 48y ...... (2)where x = 51, y = -19Hence, x\xa0and y are not unique.

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