addition and multiplication principles

[tina4545]

I need to solve this by Friday 8-13-10

2/3-x/5<4/15 using the addition and multiplication principles

Any help would be greatly appreciated.
Thank You

 

[masters]

I need to solve this by Friday 8-13-10

2/3-x/5<4/15 using the addition and multiplication principles

Any help would be greatly appreciated.
Thank You

Hi tina4545,

$\displaystyle \frac{2}{3}-\frac{x}{5}<\frac{4}{15}$

First, multiply all the terms by the least common denominator which is 15.

$\displaystyle 15\left(\frac{2}{3}\right)-15\left(\frac{x}{5}\right)<15\left(\frac{4}{15}\right)$

 

[BigGlenntheHeavy]

$\displaystyle \frac{2}{3}-\frac{x}{5} \ < \ \frac{4}{15}, \ solve \ for \ x.$

$\displaystyle -\frac{x}{5} \ < \ \frac{4}{15}-\frac{2}{3}$

$\displaystyle -\frac{x}{5} \ < \ -\frac{2}{5}$

$\displaystyle -x \ < \ -2$

\(\displaystyle Hence, \ x \ > \ 2, \ note, \ when \ we \ divide \ or \ multiply \ by \ -1 \ (or \ any \ negative \ number), \ we \ change \\)

$\displaystyle the \ sense \ of \ the \ inequality, \ why?$

$\displaystyle I'll \ leave \ the \ check \ up \ to \ you.$

 

[BigGlenntheHeavy]

$\displaystyle To \ continue:$

$\displaystyle We \ have \ -x \ < \ -2 \ and \ we \ want \ to \ rid \ ourselves \ of \ that \ pesky \ negative \ sign \ on \ x.$

$\displaystyle What \ we \ really \ have \ is \ (-1)x \ < \ (-1)2, \ the \ one \ being \ understood.$

$\displaystyle Proposition \ One: \ Aha, \ I \ know, \ I'll \ multiply \ both \ sides \ by \ -1, \ I \ can \ do \ that.$

$\displaystyle Hence, \ (-1)(-1)x \ < \ (-1)(-1)2 \ \implies \ x \ < \ 2, \ problem \ solved, \ no \ more \ pesky \ - \ sign.$

$\displaystyle But \ wait \ a \ minute \ if \ x \ = \ 3, \ then \ -3 \ < \ -2, \ but \ 3 \ < \ 2, \ I \ don't \ think \ so.$

$\displaystyle Naughty, \ naughty, \ you \ are \ a \ bad \ person, \ get \ thee \ hence \ to \ bed \ with \ no \ supper.$

$\displaystyle Proposition \ Two: \ -x \ < \ -2, \ -x \ + \ 2 \ < \ -2 \ +2 \ \implies \ -x+2 \ < \ 0, \ and$

$\displaystyle x-x+2 \ < \ 0+x \ \implies \ 2 \ < \ x \ or \ x \ > \ 2.$

$\displaystyle Now, \ if \ x \ = \ 3, \ then \ -3 \ < \ -2 \ true \ and \ 3 \ > \ 2 \ true.$

$\displaystyle Hence, \ the \ second \ proposition \ is \ true \ and \ the \ first \ is \ false.$

$\displaystyle However, \ mathematicians, \ realizing \ this \ use \ a \ shorthand \ method, \ knowing \ that \ if \ you$

$\displaystyle divide \ or \ multiply \ an \ inequality \ by \ a \ negative \ number, \ you \ change \ the \ sense \ of \ the$

$\displaystyle inequality, \ thereby \ avoiding \ the \ unecessary \ (necessary \ to \ the \ novice) \ grunt \ work.$

$\displaystyle In \ other \ words, \ if \ you \ have \ -x \ < \ -2, \ multiply \ both \ sides \ by \ -1 \ and \ change \ the \ sense.$

$\displaystyle -x \ < \ -2, \ \implies \ x \ > \ 2 \ and \ -x \ < \ 2 \ \implies \ x \ > \ -2.$