Math teachers help me?

Question

Okay. I did about 30 questions already and I'm stuck on this :
3x-3 < 6(x-1) ---- there over 5 and 10
510

and i solved and got:
3x-3 < 6x-6 (there over the 5 & 10)
5 10

what next?

Answer 1

I personally like to clear fractions immediately so if you multiply both sides by 10, you get:

6x-6<6x-6

since both sides are identical, one cannot be less than the otherso guess what!!! NO SOLUTION

Gary, Doug and Sherman know what they're talking about.

Answer 2

From here:
-3 < 6x - 3x - 6
-3 < 3x - 6
3 < 3x
1 < x

So, X > 1

Answer 3

Its not a question coz it has no answer. Coz its not an equation. Its just a relation and tells you that (3x-3)/5 < (6x-6)/10 which is still not true coz true, it should be not < but =

Answer 4

When solving for a value in an inequality (either >,<,≥,or ≤) you need to remember to treat it just like an equation. What you add or subtract to one side has to be done to the other side as well. Likewise with multiplication and division **except** that if you multiply or divide by a negative value, the 'sense' of the inequality reverses (that is > becomes < and so on).

But there seems to be a problem with the problem as you have stated it since both sides are clearly equal.


Doug

Answer 5

3x-3 < 6x-6

6-3 < 6x - 3x

3 < 3x

Rewriting,

3x > 3

x > 1

Answer 6

3x-3<6x-6 = 3x-6x<-6-(-3) = -3x<-3 = x > 1

Answer 7

i got -3<3x-6

Answer 8

(3x - 3)/5 < (6(x - 1)/10)
(3x - 3)/5 < (3(x - 1))/5
(3x - 3)/5 < (3x - 3)/5

ANS : Never True

Answer 9

assuming you mean
(3x/5)-(3/10) < 6(x-1)
multiplying both LHS & RHS by 10
6x-3 < 60(x-1)
2x-1< 20(x-1)
2x-1< 20x-20
20-1 < 20x-2x
19< 18x
x>19/18 Ans

Answer 10

3x-3 < 6x - 6
3x - 6x < -6 + 3
-3x < -3
x < -3/-3
x < 1

good luck!