177k views
4 votes
Find two consecutive even integers such that 5 times their sum is 26 more than their product. (Enter your answers as a comma-separated list.)

User Anuruddha
by
8.2k points

1 Answer

1 vote

Answer:

The two consecutive even integers = (4, 6)

Explanation:

We are asked in the question to find two consecutive even integers such that 5 times their sum is 26 more than their product.

Two consecutive even integers is represented as: x, x + 2

Five times their sum is 26 more than their products

5( x + x + 2) = [x × (x + 2)] + 26

5( 2x + 2) = (x² + 2x) + 26

10x + 10 = (x² + 2x ) + 26

x² + 2x + 26 - 10x - 10 = 0

x² - 8x + 16 = 0

We factorise

x² - 4x - 4x + 16 = 0

(x² - 4x) -(4x + 16) = 0

x(x - 4) -4(x - 4) = 0

(x - 4)(x - 4) = 0

(x - 4)²

Hence

x - 4 = 0

x = 4

Since, we know that two consecutive even integers = (x , x + 2)

First even integer = x = 4

Second even integer = x + 2 = 4 + 2

= 6

Therefore, the two consecutive even integers = (4, 6)

User Guillaume Poussel
by
7.7k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories