Question

10. The larger of two positive integers is three more than the smaller. If twice the

Answer 1

Answer:

Correct answer: n₁ = 8 and n₂ = 5

Step-by-step explanation:

If i understand correctly you claim that the first number by three is greater than the second.

We will form a system of equations according to the given data

Let be first positive integer n₁ and second n₂

n₁ = n₂ + 3

2 n₂² + 3 n₁ = 74

If we replace the first number n₁ = n₂ + 3  in the second equation, we get:

2 n₂² + 3 (n₂ + 3) = 74  ⇒ 2 n₂² + 3 n₂ + 9 = 74  ⇒

2 n₂² + 3 n₂ - 65 = 0

n₁₂ = (-b ±√b² - 4 a c) / 2 a

n₁₂ = (-3 ±√3² - 4 · 2 · (-65)) / 2 · 2 = (-3 ±√9 + 520) / 4

n₁₂ = (-3 ±√529) / 4 = (-3 ± 23) / 4

We accept the positive solution

n₂ = (-3 + 23) / 4 = 20 / 4 = 5

n₂ = 5

n₁ = 5 + 3 = 8

n₁ = 8

God is with you!!!

 

Answer 2

Answer: The integers are 8 and 5

Step-by-step explanation:

An integer is a whole number.

Let x represent the larger positive integer.

Let y represent the smaller positive integer.

The larger of two positive integers is three more than the smaller. It means that

x = y + 3

If twice the square of the smaller integer is increased by three times the larger integer, the result is 74. It means that

2y²+ 3x = 74- - - - - - - - - - -1

Substituting x = y + 3 into equation 1, it becomes

2y² + 3(y + 3) = 74

2y² + 3y + 9 = 74

2y² + 3y + 9 - 74 = 0

2y² + 3y - 65 = 0

2y² + 13y - 10y - 65 = 0

y(2y + 13) - 5(2y + 13) = 0

y - 5 = 0 or 2y - 13 = 0

y = 5 or y = 13/2

Since an integer is a whole number, then

y = 5

x = y + 3 = 5 + 3

x = 8