Question

Find two consecutive integers such that the sum of the greatest integer and twice the lesser integer is 40.​

Answer 1

Answer:

13 and 14.

Step-by-step explanation:

So we have two consecutive integers.

Let's call the first integer a.

Since the integers are consecutive, the other integer must be (a+1) (one more than the last one).

We know that the sum of the greatest integer (or a+1) and twice the lesser integer (a) is 40. Therefore, we can write the following equation:

$(a+1)+2(a)=40$

The first term represents the greatest integer. The second term represents 2 times the lesser integer. And together, they equal 40.

Solve for a. Combine like terms:

$a+1+2a=40\\3a+1=40$

Subtract 1 from both sides. The 1s on the left cancel:

$(3a+1)-1=(40)-1\\3a=39$

Divide both sides by 3:

$\frac{3a}{3}=\frac{39}{3}\\a=13$

Therefore, a or the first integer is 13.

And the second integer is 14.

And we can check:

14+2(13)=14+26=40

Answer 2

Let the two consecutive integers be x and x + 1.

According to the question,

★ Greatest integer = x + 1

★ Lesser integer = x

The sum of the greatest integer and twice the lesser integer is 40. [ Given ]

⇒ ( x + 1 ) + 2 ( x ) = 40

⇒ x + 1 + 2x = 40

⇒ 3x + 1 = 40

⇒ 3x = 40 - 1

⇒ 3x = 39

⇒ x = 39/3

⇒ x = 13

★ x + 1 = 13 + 1 = 14