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.)

The two consecutive even integers = (4, 6)

Step-by-step 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)

0.85% : )

step-by-step explanation:

unless you made a typo, the answers are both a and c.

step-by-step explanation:

1.5/3 is .5 whereas 3/9 is .3333333 not proportional

22/33 is .66666 and 4/6 is also .66666 proportional

2/4 is .5 whereas 142/213 is .66666 not proportional

14/7 is 2 and 7/3.5 is also 2 proportional

you could've also checked it by doing this simple division:

if the larger fraction divides with the smaller fraction to equal 1, it is proportional. so let me show you what i mean:

when i divide 22/33 by 4/6 i get 1, so we know it is proportional, and it is the same with 14/7 and 7/3.5. now if i divide 3/9 by 1.5/3 i get .66666 so therefore we know it is not proportional.

there are many ways to do this so i wanted to show you some possible ways.