1. Find the values of x and y for $(a,b)=(81,57)$ using the Extended Euclidean Algorithm. Step 1: Compute $\gcd(81,57)$ using the Euclidean Algorithm: $$81 = 57 \times 1 + 24$$ $$57 = 24 \times 2 + 9$$ $$24 = 9 \times 2 + 6$$ $$9 = 6 \times 1 + 3$$ $$6 = 3 \times 2 + 0$$ So, $\gcd(81,57) = 3$. Step 2: Express $3$ as a linear combination of $81$ and $57$: From: $$3 = 9 - 6 \times 1$$ Recall: $$6 = 24 - 9 \times 2$$ So: $$3 = 9 - (24 - 9 \times 2) = 9 \times 3 - 24$$ Recall: $$9 = 57 - 24 \times 2$$ So: $$3 = (57 - 24 \times 2) \times 3 - 24 = 57 \times 3 - 24 \times 7$$ Recall: $$24 = 81 - 57 \times 1$$ So: $$3 = 57 \times 3 - (81 - 57) \times 7 = 57 \times 3 - 81 \times 7 + 57 \times 7 = 57 \times 10 - 81 \times 7$$ Therefore: $$x = -7, \quad y = 10$$ 2. Find the GCD of 120 and 35 and express it as $120x + 35y = \gcd(120,35)$. Step 1: Compute $\gcd(120,35)$: $$120 = 35 \times 3 + 15$$ $$35 = 15 \times 2 + 5$$ $$15 = 5 \times 3 + 0$$ So, $\gcd(120,35) = 5$. Step 2: Express $5$ as a linear combination: $$5 = 35 - 15 \times 2$$ Recall: $$15 = 120 - 35 \times 3$$ So: $$5 = 35 - (120 - 35 \times 3) \times 2 = 35 - 120 \times 2 + 35 \times 6 = 35 \times 7 - 120 \times 2$$ Therefore: $$x = -2, \quad y = 7$$ 3. Find whole numbers $x$ and $y$ such that $39x + 14y = 1$ with $x > 5$. Step 1: Compute $\gcd(39,14)$, which should be 1 for a solution to exist. $$39 = 14 \times 2 + 11$$ $$14 = 11 \times 1 + 3$$ $$11 = 3 \times 3 + 2$$ $$3 = 2 \times 1 + 1$$ $$2 = 1 \times 2 + 0$$ So, $\gcd(39,14) = 1$. Step 2: Express 1 as a linear combination: $$1 = 3 - 2 \times 1$$ Recall: $$2 = 11 - 3 \times 3$$ So: $$1 = 3 - (11 - 3 \times 3) = 3 \times 4 - 11$$ Recall: $$3 = 14 - 11 \times 1$$ So: $$1 = (14 - 11) \times 4 - 11 = 14 \times 4 - 11 \times 5$$ Recall: $$11 = 39 - 14 \times 2$$ So: $$1 = 14 \times 4 - (39 - 14 \times 2) \times 5 = 14 \times 4 - 39 \times 5 + 14 \times 10 = 14 \times 14 - 39 \times 5$$ Thus, one solution is $x = -5$, $y = 14$. Step 3: Find $x > 5$ by adjusting solutions: general solution is $$x = -5 + 14t, \quad y = 14 - 39t$$ Choose $t=1$ to make $x > 5$: $$x = -5 + 14 = 9, \quad y = 14 - 39 = -25$$ Therefore, one solution with $x > 5$ is: $$x = 9, \quad y = -25$$