answer:**1. Graphing the Linear Equation:** The line x + 2y = 2 can be rearranged to y = -frac{1}{2}x + 1. This is a straight line with y-intercept at (0, 1) and x-intercept at (2, 0). **2. Analyzing the Absolute Value Equation:** The equation left| |x| - 2|y| right| = 1 implies symmetry across both axes. In the first quadrant, where x geq 0 and y geq 0, this reduces to two possible equations: |x| - 2|y| = 1 or 2|y| - |x| = 1. These simplify to: - x - 2y = 1 (in the first quadrant) - 2y - x = 1 (in the first quadrant) The reflections of these lines across both axes fill out the other quadrants. **3. Finding Intersection Points:** Solve for intersections: - Intersection of y = -frac{1}{2}x + 1 and x - 2y = 1: [ x - 2(-frac{1}{2}x + 1) = 1 implies x + x - 2 = 1 implies 2x = 3 implies x = frac{3}{2}, quad y = -frac{1}{2}(frac{3}{2}) + 1 = frac{1}{4}. ] - Intersection of y = -frac{1}{2}x + 1 and 2y - x = 1: [ 2(-frac{1}{2}x + 1) - x = 1 implies -x + 2 - x = 1 implies -2x = -1 implies x = frac{1}{2}, quad y = -frac{1}{2}(frac{1}{2}) + 1 = frac{3}{4}. ] Check if these points satisfy the absolute value condition: - For (frac{3}{2}, frac{1}{4}), left| |frac{3}{2}| - 2|frac{1}{4}| right| = left| frac{3}{2} - frac{1}{2} right| = 1. - For (frac{1}{2}, frac{3}{4}), left| |frac{1}{2}| - 2|frac{3}{4}| right| = left| frac{1}{2} - frac{3}{2} right| = 1. **Conclusion:** The system has boxed{2} solutions.

answer:If there are 80 passengers in total and 20 of them are children, then the number of adults (men and women) on the airplane is 80 - 20 = 60. Since the number of men and women is equal, we can divide the number of adults by 2 to find the number of men. So, the number of men on the airplane is 60 / 2 = boxed{30} .

answer:1. **Introduction and Recall of Known Facts:** We aim to show that: [ sum_{d mid n} frac{mu(d)}{d} = frac{phi(n)}{n}. ] We will employ M"obius inversion and properties of arithmetic functions to establish this equality. 2. **M"obius Inversion Application:** Recall that by M"obius inversion, for two arithmetic functions ( f ) and ( g ), we have: [ f(n) = sum_{d mid n} g(d) muleft(frac{n}{d}right). ] Here, we use the specific functions ( f(k) = phi(k) ) and ( g(k) = k ). For these functions: [ phi(n) = sum_{d mid n} d muleft(frac{n}{d}right). ] 3. **Rewriting the Summation:** Let's rewrite the summation in terms of ( d' = frac{n}{d} ), where ( d' mid n ): [ phi(n) = sum_{d' mid n} frac{n}{d'} mu(d'). ] Notice that ( d = frac{n}{d'} ), and consequently ( sum_{d' mid n} ) sums over the same divisors of ( n ). 4. **Simplifying the Expression:** To isolate ( frac{phi(n)}{n} ), divide both sides of the equation by ( n ): [ frac{phi(n)}{n} = sum_{d' mid n} frac{mu(d')}{d'}. ] Thus, we reach the desired result. 5. **Conclusion:** The steps above show that: [ sum_{d mid n} frac{mu(d)}{d} = frac{phi(n)}{n}, ] which completes the proof. [ boxed{ sum_{d mid n} frac{mu(d)}{d} = frac{phi(n)}{n} } ]

answer:Let us consider a fraction (frac{a}{b}) that is irreducible, meaning (a) and (b) have no common divisors other than 1 (i.e., (gcd(a, b) = 1)). 1. **Formulation of the Complementary Fraction**: The fraction that complements (frac{a}{b}) to 1 is given by: [ 1 - frac{a}{b} = frac{b}{b} - frac{a}{b} = frac{b - a}{b} ] 2. **Analyze the Common Divisors**: Since (gcd(a, b) = 1), (a) and (b) are coprime. We need to verify whether the fraction (frac{b-a}{b}) is irreducible. 3. **Check the Numerator**: Consider the numerator (b - a). We need to show that (b - a) is coprime with (b), meaning (gcd(b - a, b) = 1). 4. **GCD Property**: Let's use the properties of gcd to check this: [ gcd(b-a, b) = gcd(b-a, a) ] This holds because (b - a) and (b) have the same common divisors as (b-a) and (a). 5. **Coprimeness Confirmation**: Since (gcd(a, b) = 1), it implies: [ gcd(b-a, a) = gcd(-a, a) = gcd(a, a) = gcd(a, b) = 1 ] Thus, (b-a) and (b) are coprime. 6. **Conclusion**: Because the numerator (b - a) and the denominator (b) of the fraction (frac{b-a}{b}) are coprime, this fraction is irreducible. Therefore, the fraction (frac{b-a}{b}), which complements the irreducible fraction (frac{a}{b}) to 1, is always irreducible. (blacksquare)