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x, y ∈ Z+, s.t. 1/x + 1/y = 1/6.

Find the value of x and y

Sagot :

Nzmidn

PROBLEM:

[tex]\sf Find \: all \: solutions \: (x,y) \: s.t. \: \frac{1}{x} +\frac{1}{y} = \frac{1}{6}, \: x, y \in \mathbb{Z}^{+}[/tex]

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SOLUTION:

[tex]\sf \frac{1}{x} + \frac{1}{y} = \frac{1}{6}[/tex]

[tex]\sf \frac{1}{y} = \frac{1}{6} - \frac{1}{x}[/tex]

[tex]\sf \frac{1}{y} = \frac{x}{6x} - \frac{6}{6x}[/tex]

[tex]\sf \frac{1}{y} = \frac{x-6}{6x}[/tex]

[tex]\sf y = \frac{6x}{x-6}[/tex]

[tex]\sf y = \frac{6x-36+36}{x-6}[/tex]

[tex]\sf y = \frac{6x-36}{x-6} + \frac{36}{x-6}[/tex]

[tex]\sf y = \frac{6\cancel{(x-6)}}{\cancel{x-6}} + \frac{36}{x-6}[/tex]

[tex]\sf y= 6 + \frac{36}{x-6} \implies x - 6 \neq 0[/tex]

[tex]\textsf{This implies that} \: \sf x - 6 \: \big\vert \: 36 \: as \: x,y \in Z^{+}[/tex]

[tex]\textsf{It follows, } \sf x - 6 = 1, 2, 3, 4, 6, 9, 12, 18, 36[/tex]

[tex]\sf Thus,[/tex]

[tex]\begin{cases}\sf x - 6=1 \implies x = 7\\\sf x - 6 = 2 \implies x = 8\\\sf x - 6 = 3 \implies x = 9\\\sf x - 6 = 4 \implies x = 10\\\sf x - 6 = 6 \implies x = 12\\\sf x - 6 = 9 \implies x = 15\\\sf x - 6 = 12 \implies x = 18\\\sf x - 6 = 18 \implies x = 24\\\sf x - 6=36 \implies x= 42\end{cases}[/tex]

[tex]\textsf{Solving for y,}[/tex]

[tex]\begin{cases}\sf y = 6 + \frac{36}{7-6} \implies y = 42 \\\sf y = 6 + \frac{36}{8-6} \implies y = 24\\\sf y = 6 + \frac{36}{9-6} \implies y = 18\\\sf y = 6 + \frac{36}{10-6} \implies y = 15\\\sf y = 6 +\frac{36}{12-6} \implies y = 12\\\sf y= 6 + \frac{36}{18-6} \implies y = 9\\\sf y = 6 + \frac{36}{24-6} \implies y = 8\\\sf y = 6 + \frac{36}{42-6} \implies y = 7\end{cases}[/tex]

[tex]\textsf{Therefore we have 9 possible integer} \sf \: solutions, \: nam\textsf{ely:}[/tex]

[tex]\boxed{\sf (7,42), \: (8,24), \: (9,18), \: (10,15), \: (12,22), \: (15,10), \: (18,9),\: (24,8),\:(42,7)}[/tex]

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ANSWER:

[tex]\sf (7,42), \: (8,24), \: (9,18), \: (10,15), \: (12,22), \: (15,10), \: (18,9),\: (24,8),\:(42,7)[/tex]

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