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11x + 15y = -23 ---- (1) ////// 7x - 2y = 20 ---- (2). Take equation (2). Thus, x = 20 + 2y / 7// ---- (A). Equating (A) to (1), we get :- 11 (20 + 2y / 7) + 15y = -23 => 220 + 22y / 7 + 15y = -23 => 220 + 22y + 105y = -161 => 127y = -161 - 220 => 127y = -381 => y = -3// Therefore, x = 20 + 2y / 7 = 20 - 6 / 7 = 14 / 7 = 2
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Preeti Dabral 4 years, 4 months ago
According to Euclid’s Division Lemma if we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition a = bq + r where 0 ≤ r ≤ b.
HCF is the largest number which exactly divides two or more positive integers.
Since 12576 > 4052
12576 = (4052 × 3) + 420
420 is a reminder which is not equal to zero (420 ≠ 0).
4052 = (420 × 9) + 272
271 is a reminder which is not equal to zero (272 ≠ 0).
Now consider the new divisor 272 and the new remainder 148.
272 = (148 × 1) + 124
Now consider the new divisor 148 and the new remainder 124.
148 = (124 × 1) + 24
Now consider the new divisor 124 and the new remainder 24.
124 = (24 × 5) + 4
Now consider the new divisor 24 and the new remainder 4.
24 = (4 × 6) + 0
Reminder = 0
Divisor = 4
HCF of 12576 and 4052 = 4.
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Adithya Dev A 4 years, 6 months ago
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