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Mayerlin is the middle of three siblings whose ages are consecutive even integers. If the sum of their ages is 66, find Mayerlin's age. Answer:

Mayerlin is the middle of three siblings whose ages are consecutive even integers. If the sum of their ages is 6666, find Mayerlin's age.\newlineAnswer:

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Full solution

Q. Mayerlin is the middle of three siblings whose ages are consecutive even integers. If the sum of their ages is 6666, find Mayerlin's age.\newlineAnswer:
  1. Set Variables: Let xx be the age of the youngest sibling. Since the siblings' ages are consecutive even integers, Mayerlin's age would be x+2x + 2 and the oldest sibling's age would be x+4x + 4.
  2. Write Equation: The equation representing the sum of their ages is 6666 can be written as: x+(x+2)+(x+4)=66x + (x + 2) + (x + 4) = 66
  3. Simplify Equation: Simplify the left side of the equation:\newlinex+x+2+x+4=66x + x + 2 + x + 4 = 66\newlineCombine like terms:\newline3x+6=663x + 6 = 66
  4. Isolate Variable: Isolate the variable term:\newline3x+6−6=66−63x + 6 - 6 = 66 - 6\newline3x=603x = 60
  5. Solve for x: Solve for x:\newline3x=603x = 60\newlinex=603x = \frac{60}{3}\newlinex=20x = 20
  6. Calculate Mayerlin's Age: Since xx is the age of the youngest sibling, Mayerlin's age, being the middle sibling, is x+2x + 2: Mayerlin's age = 20+2=2220 + 2 = 22

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