Solving inqualities by Using addition and subtract Options

[demolished]

Solving inqualities by Using addition and subtraction

PROBLEM 1
6r > 10r - r -3r
6r > 9r - 3r
6r > 6r < -- i'm stuck .. what do i do ?

do i cancel them out w/ 6, it will end up r > r, if i cancel out w/ 1, it would be 0 > 0

i'm so confused. So, am i right or wrong ... if wrong, help me please.

PROBLEM 2
3.2x < 2x-(9 -1.2x)

do i distribute ... 2x-9 and 2x - -1.3x ?

PROBLEM 3

3 over 2 q minus 25 over 5 exact greater then 2 q over 4
For this problem, is the answer 5 exact greater then q ? if not, dont tell me the answer.

[Spirited Away]

Solving inqualities by Using addition and subtraction

PROBLEM 1
6r > 10r - r -3r
6r > 9r - 3r
6r > 6r      < -- i'm stuck .. what do i do ?

do i cancel them out w/ 6, it will end up r > r, if i cancel out w/ 1, it would be 0 > 0

i'm so confused. So, am i right or wrong ... if wrong, help me please.

PROBLEM 2
3.2x < 2x-(9 -1.2x)

do i distribute ... 2x-9     and 2x - -1.3x ?

PROBLEM 3

3 over 2 q minus 25 over 5  exact greater then 2 q over 4
For this problem, is the answer 5 exact greater then q ? if not, dont tell me the answer.

I'm math stupid.

Problem number 1 isn't possible, for me that is.

But anyway, problem two requires that you distribute the negative sign. So, it's
3.2x < 2x-9+1.2x
You always distribute the sign before the paranthesis

I think this the equation you need to get to the answer: 6-20q-2q^2>0 But then I'm stuck there, too.

Yes, did I say I'm math stupid?