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Bao is older than Jeriel. Their ages are consecutive integers. Find Bao's age if the sum of Bao's age and 2 times Jeriel's age is 40.
Answer:

Bao is older than Jeriel. Their ages are consecutive integers. Find Bao's age if the sum of Bao's age and 22 times Jeriel's age is 4040.\newlineAnswer:

Full solution

Q. Bao is older than Jeriel. Their ages are consecutive integers. Find Bao's age if the sum of Bao's age and 22 times Jeriel's age is 4040.\newlineAnswer:
  1. Set Up Equations: Let's denote Bao's age as BB and Jeriel's age as JJ. Since their ages are consecutive integers, we can express Bao's age as B=J+1B = J + 1. The problem states that the sum of Bao's age and 22 times Jeriel's age is 4040. We can write this as an equation: B+2J=40B + 2J = 40 Substituting BB with J+1J + 1, we get: (J+1)+2J=40(J + 1) + 2J = 40
  2. Simplify Equation: Now, let's simplify the equation:\newlineJ+1+2J=40J + 1 + 2J = 40\newlineCombine like terms:\newline3J+1=403J + 1 = 40
  3. Isolate Jeriel's Age: Next, we will isolate JJ by subtracting 11 from both sides of the equation:\newline3J+11=4013J + 1 - 1 = 40 - 1\newline3J=393J = 39
  4. Solve for Jeriel's Age: Now, we will divide both sides by 33 to solve for JJ: \newline3J3=393\frac{3J}{3} = \frac{39}{3}\newlineJ=13J = 13\newlineSince JJ represents Jeriel's age, we have found that Jeriel is 1313 years old.
  5. Find Bao's Age: Finally, since Bao is older than Jeriel by 11 year, we can find Bao's age:\newlineB=J+1B = J + 1\newlineB=13+1B = 13 + 1\newlineB=14B = 14\newlineTherefore, Bao is 1414 years old.

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