Answer
- 103 and 104
First, the variables have to be defined. Let \(x\) be the first integer; therefore, \(x + 1\) is the second integer. This is a two-step problem.
The sum of three times the first and two less than the second is translated into the following expression: \(3x + (x + 1 – 2)\). This expression is set equal to 411 to obtain \(3x + (x + 1 – 2) = 411\).
The left-hand side is simplified to obtain \(4x – 1 = 411\). To solve for \(x\), first add 1 to both sides and then divide both sides by 4 to obtain \(x = 103\). The next consecutive integer is 104.
Two consecutive integers exist such that the sum of three times the first and two less than the second is equal to 411. What are those integers?
- 103 and 104
- 104 and 105
- 102 and 103
- 100 and 101
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