James is four years younger than Austin. If three times James' age is increased by the square of Austin's age, the result is 28. Use an algebraic solution to find the ages of James and Austin.

James is years old, and Austin is years old.

Let's use J for James's age and A for Austin's age. The equations are:

J = A - 4
3J + A² = 28

Just plug (A - 4) in the place of J in the second equation. This gives you:

3(A - 4) + A² = 28
-->
A² + 3A - 12 = 28
-->
A² + 3A - 40 = 0
-->
(A - 5)(A + 8) = 0
-->
A = 5 or -8

-8 is nonsense, so Austin is 5 years old. Therefore, James is 1 year old.

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James is four years younger than Austin. If three times James' age is increased by the square of Austin's age, the result is 28. Use an algebraic solution to find the ages of James and Austin. James is years old, and Austin is years old.

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James is four years younger than Austin. If three times James' age is increased by the square of Austin's age, the result is 28. Use an algebraic solution to find the ages of James and Austin.

James is years old, and Austin is years old.