11.2: Basic Terminology
 Page ID
 14035
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Definitions:
Consecutive integers (specific numbers): 7, 8, 9.
Consecutive integers (general numbers in a box): \(x\), \(x+1\), \(x+2\).
Consecutive even integers (specific numbers): 8, 10, 12.
Consecutive even integers (general numbers in a box): \(x\), \(x+2\), \(x+4\).
Consecutive odd integers (specific numbers): 7, 9, 11
Consecutive integers (general numbers in a box): \(x\), \(x+2\), \(x+4\).
Examples
Some of the problems here are simple. The solution can be worked out fast by quick reasoning. The benefit of these problems is not to find the solution by reasoning, but by learning algebraic steps applicable to more challenging situations.
The general method is not to read a problem first and understand it. Many of my colleagues will disagree with me. I propose creating a preamble in which you write a symbol or set of symbols for each element in the problem. (Refer to exercise 10.2.9).
Look at the end of the problem where you usually find what to solve for. Let \(x\) be the number we are looking for. Write this as the first step in the preamble (list of symbols in nontrivial exercises whose solution we seek.)
Next start reading the problem from the beginning and develop the preamble step by step. Write the symbol(s) for each step on a new line.
Finish the preamble by converting all the steps in the problem into symbols. Now look at your preamble and gain an overview of the your problem.
It should be easier to obtain an equation using the symbols from the preamble. Then solve the equation by the method introduced this far for linear equations. (A linear equation has a variable to the first degree (exponent).)
Example \(\PageIndex{1}\)
The sum of three consecutive positive integers is \(705\). Find the integers.
Solution:
Preamble:
Let \(x\) be the smallest of the three consecutive integers (for example \(x=20\)).
Then \(x+1\) is the middle number \(20+1=21\).
And \(x+2\) is the largest number \(20+2=22\).
The sum of three integers is \(x+(x+1)+(x+2)=3x+3\).
Equation:
\[\begin{array}{rcl lll} \hbox{Sum of integers}&=&\hbox{sum of integers}\\[10pt] 3x+3&=&705\\[5pt] 3x+33&=&7053\\[5pt] 3x&=&702\\[5pt] \displaystyle \frac{3x}{3}&=&\displaystyle \frac{702}{3}\\[10pt] x&=&234\end{array}\]
Smallest integer: \(x=234\)
Middle integer: \(x+1=234+1=235\)
largest integer: \(x+2=234+2=236\)
Preamble:
Let \(x\) be the middle of the three consecutive integers (for example \(x=20\)).
[5pt] Then \(x1\) is the smallest number \(201=19\).
And \(x+1\) is the largest number \(20+1=21\).
The sum of three integers is \((x1)+(x+1)+(x)+(x+1)=3x\).
Equation: \[\begin{array}{rcl lll} \hbox{Sum of integers}&=&\hbox{sum of integers}\\[10pt] 3x&=&705\\[5pt] 3x&=&705\\[5pt] \displaystyle \frac{3x}{3}&=&\displaystyle \frac{705}{3}\\[10pt] x&=&235\end{array}\]
Smallest integer: \(x1=234\)
Middle integer: \(x=235\)
largest integer: \(x+1=234+2=236\)
Preamble:
Let \(x\) be the largest of the three consecutive integers (for example \(x=20\)).
[5pt] Then \(x1\) is the middle number \(20+1=19\).
And \(x2\) is the smallest number \(20+2=18\).
The sum of three integers is \((x2)+(x1)+x=3x3\).
Equation: \[\begin{array}{rcl lll} \hbox{Sum of integers}&=&\hbox{sum of integers}\\[10pt] 3x3&=&705\\[5pt] 3x+33&=&705+3\\[5pt] 3x&=&708\\[5pt] \displaystyle \frac{3x}{3}&=&\displaystyle \frac{708}{3}\\[10pt] x&=&236\end{array}\]
Smallest integer: \(x=234\)
Middle integer: \(x+1=234+1=235\)
largest integer: \(x+2=234+2=236\)
Example \(\PageIndex{3}\)
Find two integers whose sum is \(82\).
\(11\) more than three times the smaller number is the same as \(18\) less than twice the larger number.
Find the numbers.
Solution:
Preamble:
Let \(x\) be the smaller number (for example \(x=20\)).
Then the larger number is \(82x\) \(8220\).
\(11\) more than \(3\times\) the smaller number: \(3x+11\) \(3(20)+11\).
\(18\) less than \(2\times\)
the larger number: \(\begin{array}{lll lll} &2(82x)18\\ =&1642x18\\ =&1462x \end{array}\) \(2(8220)18\)).
Equation:
\(\begin{array}{rcl lll} 11\hbox{ more than three times the smaller number}&=&\hbox{\)18\(less than twice the larger number}\\[6pt] 3x+11&=&1462x\\[6pt] 3x+2x+11&=&1462x+2x\\[6pt] 5x+11&=&146\\[6pt] 5x+1111&=&14611\\[6pt] 5x&=&135\\[10pt] \displaystyle \frac{5}{5}x&=&\displaystyle \frac{135}{5}\\[5pt] x&=&27 \end{array}\)
The smaller number is \(x=27\).
The larger number is \(8227=55\).
Preamble:
Let \(x\) be the larger number (for example \(x=20\)).
Then the smaller number is \(82x\) \(8220\).
[5pt] \(11\) more than \(3\times\) \(3(82x)+11\) \(3(8220)+11\).
smaller number: \(=246+113x\) .
\(=2573x\) .
\(18\) less than \(2\times\) the larger number: \(2(x)18\) \(2(20)18\)).
Equation: \[\begin{array}{rcl lll} 11 more than 3 \times smaller &=& 18 less than twice larger \\[6pt] 2573x&=&2x18\\[6pt] 2573x+3x&=&2x+3x18\\[6pt] 257&=&5x18\\[6pt] 257+18&=&5x18+18\\[6pt] 275&=&5x\\[6pt] x&=&\displaystyle \frac{275}{5}\\[10pt] x&=&55 \end{array}\] The smaller number is \(8255=27\).
The larger number is \(55\).
Example \(\PageIndex{3}\)
Find two consecutive odd integers such that 60 less than three times the larger number equals 73 more than twice the smaller number.
Solution
Preamble:
Let \(x\) be the smaller odd number (for example \(x=21\)).
Then larger number is \(x+2\) \(21+2\).
[5pt] \(60\) less than \(3\times \hspace{0.7in} 3(x+2)60\) \(3(21+2)60\).
the larger number: \(=3x+660\) .
\(=3x54\) .
\(73\) more than \(2\times\) smaller: \(2x+73\) \(2(21)+73\).
Equation: \[\begin{array}{rcl lll} \text{60 less than 3 times larger}&=&text{73 more than 2\times smaller number}\\[5pt] 3x54&=&2x+73\\[5pt] 3x54+54&=&2x+73+54\\[5pt] 3x&=&2x+127\\[5pt] 3x2x&=&2x2x+127\\[5pt] x&=&127\\[5pt] \end{array}\] The smaller number is \(x=127\).
The larger number is \(127+2=129\).
Preamble:
Let \(x\) be the larger odd number (for example \(x=21\)).
Then the smaller number is \(x2\) \(212\).
[5pt] \(60\) less than \(3\times\) the larger number: \(3x60\) \(3(21)60\) .
\(73\) more than \(2\times\) smaller: \(2(x2)+73\) \(2(212)+73\).
\(=2x4+73\) .
\(=2x+69\) .
Equation: \[\begin{array}{rcl lll} \text{60 less than 3 \times larger}&=&\text{73 more than 2\times smaller number}\\[5pt] 3x60&=&2x+69\\[5pt] 3x60+60&=&2x+69+60\\[5pt] 3x&=&2x+129\\[5pt] 3x2x&=&2x2x+129\\[5pt] x&=&129\\[5pt] \end{array}\] The smaller number is \(x=1292=127\).
The larger number is \(129\).
Exercises 11

The sum of three consecutive positive integers is \(1,128\). Find the integers.

Find two consecutive even integers such that 150 less than three times the smaller number equals 148 more than twice the larger number.

Find two integers whose sum is 425.
10 more than six times the smaller number is the same twice the larger number.
Find the numbers.

The sum of three consecutive positive integers is \(1,128\). Find the integers.
Solution:
Preamble:
Let \(x\) be the smallest of the three consecutive integers (for example \(x=20\)).
[5pt] Then \(x+1\) is the middle number \(20+1=21\).
And \(x+2\) is the largest number \(20+2=22\).
The sum of three integers is \(x+(x+1)+(x+2)=3x+3\).
Equation: \[\begin{array}{rcl lll} \hbox{Sum of integers}&=&\hbox{sum of integers}\\[10pt] 3x+3&=&1,128\\[5pt] 3x+33&=&1,1283\\[5pt] 3x&=&1,125\\[5pt] \displaystyle \frac{3x}{3}&=&\displaystyle \frac{1,125}{3}\\[10pt] x&=&375\end{array}\]Smallest integer: \(x=375\)
Middle integer: \(x+1=375+1=376\)
largest integer: \(x+2=375+2=377\) 
Find two consecutive even integers such that \(150\) less than three times the smaller number equals \(148\) more than twice the larger number.
Solution:
Preamble:
Let \(x\) be the smaller number (for example \(x=21\)).
Then the larger even number is \(x+2\) \(21+2\).
\(150\) less than \(3\times\) the smaller number: \(3x150\) \(3(21)150\).
\(148\) more than \(2\times\)
the larger number: \(\begin{array}{lll lll} &2(x+2)+148\\ =&2x+4+148\\ =&2x+152 \end{array}\) \(2(21+2)+148\)).Equation:
\(\begin{array}{rcl lll} 150\hbox{ less than\)3\(smaller number}&=&\hbox{\)148\(more than\)2\(larger number}\\[6pt] 3x150&=&2x+152\\[6pt] 3x2x150&=&2x2x+152\\[6pt] x150&=&152\\[6pt] x150+150&=&152+150\\[6pt] x&=&302\\[10pt] \end{array}\)The smaller number is \(x=302\).
The larger number is \(302+2=304\). 
Find two integers whose sum is \(425\).
\(10\) more than six times the smaller number is the same as twice the larger number.
Find the numbers.Solution:
Preamble:
Let \(x\) be the smaller number (for example \(x=20\)).
Then the larger number is \(425x\) \(42520\).
[5pt] \(10\) more than \(6\times\) smaller number: \(6x+10\) \(6(20)+10\).
\(10\) more than \(6\times\) smaller: \(6x+10\) \(6(20)+10\).Equation: \[\begin{array}{rcl lll} \text{10 more than 6 \times smaller}&=&\text{73 more than 2\times larger number}\\[5pt] 6x+10&=&8502x\\[5pt] 6x+2x+10&=&8502x+2x\\[5pt] 8x+10&=&850\\[5pt] 8x+1010&=&85010\\[5pt] 8x&=&840\\[5pt] \displaystyle \frac{8}{8}x&=&\displaystyle \frac{840}{8}\\[10pt] x&=&105 \end{array}\] The smaller number is \(x=105\).
The larger number is \(425105=320\).