Question

One positive integer is 1 less than twice another. The sum of their squares is 461. Find the integers.

Answer 1

Answer:

10 and 19

Step-by-step explanation:

• In Algebra, an even number = 2x, because anything multiplied by 2 is even.

1. There are two numbers in the question, I'll define the first one as 2x
2. the other number is 1 less than twice of the first, so I'm going to define it in algebra as 2(2x) - 1, which is just 4x - 1
3. the square of the first is $4x^2$
4. the square of the second is $16x^2-8x+1$
5. add them together: $20x^2-8x+1$
6. this equation = 461, so $20x^2-8x+1 = 461$
7. simplify to $20x^2-8x-460 = 0$
8. I wont explain how to solve this here, since it is very complicated, if you don't know search up solving quadratics.
9. the solutions are 5 or -4.6, it cannot be -4.6 since the question says positive, so the solution for x = 5
10. substitute x into our equations for the two numbers in steps 1 and 2
11. the first number = 10, the second = 19