Answer:
(x, y) = (1/2, -1)
Step-by-step explanation:
Subtracting twice the first equation from the second gives ...
(2/x +1/y) -2(1/x -5/y) = (3) -2(7)
11/y = -11 . . . . simplify
y = -1 . . . . . . . multiply by y/-11
Using the second equation, we can find x:
2/x +1/-1 = 3
2/x = 4 . . . . . . . add 1
x = 1/2 . . . . . . . multiply by x/4
The solution is (x, y) = (1/2, -1).
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<em>Additional comment</em>
If you clear fractions by multiplying each equation by xy, the problem becomes one of solving simultaneous 2nd-degree equations. It is much easier to consider this a system of linear equations, where the variable is 1/x or 1/y. Solving for the values of those gives you the values of x and y.
A graph of the original equations gives you an extraneous solution of (x, y) = (0, 0) along with the real solution (x, y) = (0.5, -1).
Yes
Example: some multiples of 8 are 8,16,24,32,40 and these are multiples of 2
Answer:
La afirmación es falsa, no todos los divisores de 100 son divisores de 50, ya que solo se toman en cuenta sus divisores comunes, los cuales son todos los divisores de 50. Expresamos a 100 en sus factores primos: 100 = 2 · 2 · 5 · 5 = 2² · 5² Divisores de 100: {1, 2, 4, 5, 10, 20, 25, 50, 100}
Step-by-step explanation: