Math

Question Write and solve an inequality for the sum of 2 consecutive integers greater than 73. Solve the inequality and find the pair with the least sum.
Let x be the first integer and x+1 be the second integer.\text{Let } x \text{ be the first integer and } x+1 \text{ be the second integer.} x+(x+1)>73x + (x+1) > 73 2x+1>732x + 1 > 73 2x>722x > 72 x>36x > 36 The pair of integers with the least sum is 37 and 38.

Studdy Solution

STEP 1

Assumptions
1. We are dealing with consecutive integers, which means they follow one after the other without any gaps.
2. The sum of these two consecutive integers is greater than 73.
3. We need to find the smallest pair of such integers.

STEP 2

Let's denote the first consecutive integer as n n . Since the integers are consecutive, the next integer is n+1 n + 1 .

STEP 3

Write an inequality to represent the situation where the sum of these two consecutive integers is greater than 73.
n+(n+1)>73 n + (n + 1) > 73

STEP 4

Simplify the inequality by combining like terms.
2n+1>73 2n + 1 > 73

STEP 5

Subtract 1 from both sides of the inequality to isolate the term with the variable n n .
2n>72 2n > 72

STEP 6

Divide both sides of the inequality by 2 to solve for n n .
n>36 n > 36

STEP 7

Since n n must be greater than 36 and we are dealing with integers, the smallest integer n n can be is 37.

STEP 8

Now, find the next consecutive integer by adding 1 to n n .
n+1=37+1=38 n + 1 = 37 + 1 = 38

STEP 9

The smallest pair of consecutive integers whose sum is greater than 73 is 37 and 38.
The inequality is n>36 n > 36 and the smallest pair of integers with the least sum that satisfies this inequality is (37, 38).

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