8x+2<=-14 and 4x-2>10
Simple. For the first inequality subtract two from each side, then divide by 8, you get x < -2, for the next one add two to each side, divide by four, and get x > 3. Then draw a number line from -2 to 3, make an open dot at the -2 and three point, draw a line inbetween them.
I'm confused htmlmaster...
We have: $\displaystyle 8x\,+\,2\,\le\,-\,14$
Subtract 2 from both sides: $\displaystyle 8x\,\le\,-\,16$
Divide both sides by 8: $\displaystyle x\,\le\,-\,2$
Next we have: $\displaystyle 4x\,-\,2\,>\,10$
Which goes to: $\displaystyle x\,>\,3$
Thus we get: $\displaystyle \boxed{x\,\le\,-\,2\;\;and\;\;x\,>\,3}$
How can I have a number greater than 3 and be less than or equal to -2? This seems like a no solution to me.
You didn't mention are those inequalites part of system of inequalities or are they independent?
My guess is that htmlmaster solve it believing that they are independent.
If they are system of inequalities then they have no solution since there is no x which satisfies both inequalities.
Hello, Ashiebug3!
I wish you would respond and clear up the mystery . . .
If these are two separate problems . . .Graph: .$\displaystyle 8x+2 \:\leq \:-14$ .and .$\displaystyle 4x-2\:>\:10$
The first is: .$\displaystyle 8x + 2 \:\leq\:-14\quad\Rightarrow\quad 8x \:\leq\:16\quad\Rightarrow\quad x \:\leq\:-2$
Its graph is:Code:======* - - + - - - - - - - - - - -2 0
The second is: .$\displaystyle 4x - 2 \:>\:10\quad\Rightarrow\quad4x\:>\:12\quad\Rightar row\quad x \:>\:3$
Its graph is:Code:- - - - - - - + - - - - o======= 0 3
Are you really having difficulty with the algebra?