2.3 Simplifying algebraic expressions  (Page 2/3)

Consecutive integers have the property that if

${\displaystyle \begin{array}{ccc}\hfill n& =& \text{the smallest integer, then}\hfill \\ \hfill n+1& =& \text{the next integer, and}\hfill \\ \hfill n+2& =& \text{the next integer, and so on.}\hfill \end{array}}$

Consecutive odd or even integers have the property that if

${\displaystyle \begin{array}{ccc}\hfill n& =& \text{the smallest integer, then}\hfill \\ \hfill n+2& =& \text{the next odd or even integer (since odd or even numbers differ by 2), and}\hfill \\ \hfill n+4& =& \text{the next odd or even integer, and so on.}\hfill \end{array}}$

The sum of three consecutive odd integers is equal to one less than twice the first odd integer. Find the three integers.

1. ${\displaystyle \begin{array}{cccc}Let& \hfill n& =& \text{the first odd integer. Then,}\hfill \\ & \hfill n+2& =& \text{the second odd integer, and}\hfill \\ & \hfill n+4& =& \text{the third odd integer.}\hfill \end{array}}$
2. Translate the words to mathematical symbols and construct an equation. Read phrases.

   ${\displaystyle \begin{array}{cc}\text{The sum of:}\hfill & \text{add some numbers}\hfill \\ \text{three consecutive odd integers:}\hfill & n,n+\mathrm{2,}n+4\hfill \\ \text{is equal to:}\hfill & =\hfill \\ \text{one less than:}\hfill & \text{subtract 1 from}\hfill \\ \text{twice the first odd integer:}\hfill & 2n\hfill \end{array}\}n+(n+2)+(n+4)=2n-1}$

3. ${\displaystyle \begin{array}{cc}n+n+2+n+4=2n-1\hfill & \\ 3n+6=2n-1\hfill & \text{Subtract}2n\text{from}\mathit{\text{ both }}\text{sides.}\hfill \\ 3n+6-2n=2n-1-2n\hfill \\ n+6=-1\hfill & \text{Subtract 6 from}\mathit{\text{ both }}\text{sides.}\hfill \\ n+6-6=-1-6\hfill & \\ n=-7\hfill & \text{The first integer is -7.}\hfill \\ n+2=-7+2=-5\hfill & \text{The second integer is -5.}\hfill \\ n+4=-7+4=-3\hfill & \text{The third integer is -3.}\hfill \end{array}}$

4. Check this result.
   The sum of the three integers is

   ${\displaystyle \begin{array}{ccc}\hfill -7+(-5)+(-3)& =& -\text{12}+(-3)\hfill \\ & =& -\text{15}\hfill \end{array}}$

   One less than twice the first integer is ${\displaystyle 2(-7)-1=-\text{14}-1=-\text{15}}$ . Since these two results are equal, the solution checks.

5. The three odd integers are -7, -5, -3.
   

The perimeter (length around) of a rectangle is 20 meters. If the length is 4 meters longer than the width, find the length and width of the rectangle.

1. Let ${\displaystyle x=}$ the width of the rectangle. Then,
   ${\displaystyle x+4=}$ the length of the rectangle.
2. We can draw a picture.
   

   

   The length around the rectangle is
   ${\displaystyle \underset{\text{width}}{\underbrace{x}}+\underset{\text{length}}{\underbrace{\left(x+4\right)}}+\underset{\text{width}}{\underbrace{x}}+\underset{\text{length}}{\underbrace{\left(x+4\right)}}=20}$

3. ${\displaystyle \begin{array}{cc}x+x+4+x+x+4=\text{20}\hfill & \\ 4x+8=\text{20}\hfill & \text{Subtract 8 from}\mathit{\text{ both }}\text{sides.}\hfill \\ 4x=\text{12}\hfill & \text{Divide}\mathit{\text{ both }}\text{sides by 4.}\hfill \\ x=3\hfill & \text{Then,}\hfill \\ x+4=3+4=7\hfill & \end{array}}$

4. Check:
   

   
   

5. The length of the rectangle is 7 meters.
   The width of the rectangle is 3 meters.