Question

solve.
ann, betty and claire are sisters. ann is the youngest, and betty is in high school. betty’s age is ann’s age squared. claire’s age is one less than three times ann’s age. the sum of their ages is 31. ann’s age and claire’s age combined is 1 less than betty’s age.
let ( x = ) ann’s age. find all 3 girls’ ages.
ages
ann: ____ betty: __ claire: ____

Explanation:

Step1: Define variables and expressions

Let \( x \) be Ann's age. Then Betty's age is \( x^2 \) (since Betty’s age is Ann’s age squared), and Claire’s age is \( 3x - 1 \) (since Claire’s age is one less than three times Ann’s age).

Step2: Set up the equation from the sum of ages

The sum of their ages is 31, so we have the equation:
\[
x + x^2 + (3x - 1) = 31
\]
Simplify the left - hand side:
\[
x^2 + 4x - 1 = 31
\]
Subtract 31 from both sides to get a quadratic equation:
\[
x^2 + 4x - 32 = 0
\]

Step3: Solve the quadratic equation

We can factor the quadratic equation \( x^2 + 4x - 32 = 0 \). We need two numbers that multiply to - 32 and add up to 4. The numbers are 8 and - 4. So,
\[
(x + 8)(x - 4)=0
\]
Setting each factor equal to zero gives \( x=-8 \) or \( x = 4 \). Since age cannot be negative, we take \( x = 4 \).

Step4: Verify with the other condition

We also know that Ann’s age and Claire’s age combined is 1 less than Betty’s age. Let's check:
Ann's age \( x = 4 \), Betty's age \( x^2=16 \), Claire's age \( 3x - 1=3\times4 - 1 = 11 \)
Ann's age + Claire's age \(=4 + 11=15\)
Betty's age - 1 \(=16 - 1 = 15\). The condition is satisfied.

Answer:

Ann: 4, Betty: 16, Claire: 11