I need help with this equation :)

[fivetwoone]

8m-(3-m)<*-{-[-(3-9m)}] *it is supposed to be less than or equal to - i could not find the button for that symbol

 

[fivetwoone]

sorry, i wrote the equation wrong, it is supposed to be:

8m-(3-m)<*-{-[-(3-9m)]}

 

[soroban]

Hello, fivetwoone!

What a strange-looking problem . . .

\(\displaystyle 8m-(3-m)\; \le \;-\big\{-\left[-(3-9m)\,\right]\,\big\} \)

We have: .\(\displaystyle 8m - 3 + m \;\le \;-(3-9m)\)

. . . . . . . . . . n\(\displaystyle 9m - 3 \;\le \; -3 + 9m\)

. . . . . . . . . . . . m. . \(\displaystyle 0 \;\le\;0\)

This is true for all values of \(\displaystyle m.\)


Therefore: .\(\displaystyle -\infty \:\le \:m \;\le\: \infty\)

 

[mmm4444bot]

Do not type an asterisk to denote "equal to" because an asterisk is already used as a multiplication sign.

Simply type <= to show "less than or equal to"

Your first step should be to simplify the expression on each side of the inequality symbol.

Combine like-terms on the left-hand side, and use the Order of Operations and the Distributive Property to simplify the right-hand side.

 

[mmm4444bot]

Or, just wait for soroban to do your homework for you.

 

[fivetwoone]

for soroban:

wouldn't the 'answer' come out as 3<=3? i know that the final answer is all solutions but i don't know how you got 0<=0...

 

[fivetwoone]

Soroban subtracted 9m - 3 from both sides of the inequality and got 0 <= 0. That is true no matter what m is.

If you subtract 9m - 6 from both sides you can come up with 3 <= 3 also. That too is true no matter what m is

Actually, I think the clearest way to proceed is to get an answer involving m by adding 3 to both sides of the inequality, which results in

9m <= 9m, which is rather obviously true for every real number m. No matter how you do it, you end up in the same place. There are frequently many ways to get to the correct answer.

yes, i was taught to take over one at a time, so i was confused...