127+96+73 class 6 chapter name is whol numbers
Answers
Step-by-step explanation:
Topic
Diophantine equations.
-Simon's favorite factorization.
It refers to the method of factorization by isolating unknowns and constant terms.
Solution
To solve this Diophantine equation, let's isolate unknowns by unknowns, constants by constants.
\implies xy-x-y=3⟹xy−x−y=3
Adding 11 on both sides,
\implies (x-1)(y-1)=4⟹(x−1)(y−1)=4
This is called Simon's favorite factorization.
Now as the left-hand side is factorized, we can think of factorizing the right-hand side as well.
\implies (x-1)(y-1)=4\cdot 1⟹(x−1)(y−1)=4⋅1
\implies \boxed{x=5\ \text{and}\ y=2}⟹
x=5 and y=2
\implies (x-1)(y-1)=2\cdot 2⟹(x−1)(y−1)=2⋅2
\implies \boxed{x=3\ \text{and}\ y=3}⟹
x=3 and y=3
\implies (x-1)(y-1)=1\cdot 4⟹(x−1)(y−1)=1⋅4
\implies \boxed{x=2\ \text{and}\ y=5}⟹
x=2 and y=5
Of course, the factors can be negative integers.
\implies (x-1)(y-1)=(-4)\cdot (-1)⟹(x−1)(y−1)=(−4)⋅(−1)
\implies \boxed{x=-3\ \text{and}\ y=0}⟹
x=−3 and y=0
\implies (x-1)(y-1)=(-2)\cdot (-2)⟹(x−1)(y−1)=(−2)⋅(−2)
\implies \boxed{x=-1\ \text{and}\ y=-1}⟹
x=−1 and y=−1
\implies (x-1)(y-1)=(-1)\cdot (-4)⟹(x−1)(y−1)=(−1)⋅(−4)
\implies \boxed{x=0\ \text{and}\ y=-3}⟹
x=0 and y=−3
Advanced problems
Question: Solve 3x+5y=903x+5y=90 over natural numbers.
Answer: (x,y)=(5,15), (10,12), (15,9), (20,6), (25,3)(x,y)=(5,15),(10,12),(15,9),(20,6),(25,3)
Answer key:
Solution: From 3x=90-5y=5(18-y)3x=90−5y=5(18−y) , and since 3 and 5 are co-prime, xx is a multiple of 5.
Let x=5kx=5k . Then 15k+5y=9015k+5y=90 .
Divide both sides by 5 to obtain 3k+y=183k+y=18 . Then, similarily y=18-3k=3(6-k)y=18−3k=3(6−k) , so yy is a multiple of 3.
Let y=3ly=3l . Then 15k+15l=9015k+15l=90 . Divide both sides by 15 to obtain k+l=6k+l=6 .
Then, solution pairs are (k,l)=(1,5),(2,4),(3,3),(4,2),(5,1)(k,l)=(1,5),(2,4),(3,3),(4,2),(5,1) . And we know that x=5kx=5k and y=3ly=3l . So the solution pairs to this equation are \boxed{(x,y)=(5,15), (10,12), (15,9), (20,6), (25,3)}
(x,y)=(5,15),(10,12),(15,9),(20,6),(25,3)
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