Question

I need help solving and writing a set builder notation for -3(x-10)>-x+5

Answer 1

You solve it by isolating the x by adding/subtracting the unrelated terms to the other side and then multiplying or dividing by the coefficient to get rid of it, obeying the rule that if you multiply/divide both sides by a negative, you switch the signs (its the only big difference between equality and inequality)

Answer 2

I don't have to multiply -3 into (x-10) first?

Answer 3

is proper to write set builder notion in fraction form or decimal?

Answer 4

you'd do that first, to make it easier in the long run...

Answer 5

Usually I see set builder in fraction form, but it doesn't usually matter.

Answer 6

okay thank you. Could I show how i've done the problem and you can let me know know I have the correct concept?

Answer 7

yeah, i think as long as you're consistent with which you choose, it's fine either way. fractions are usually nicer imo if you have repeating numbers though

yeah, sure

Answer 8

-30x-10>-x+5
30x>-x+5
30x+x>-x+x+5
30x>5
30x/5>5/5 ??

Answer 9

hmm... do you know how to solve a linear equation like 5x + 2 = 3x - 4? Where, you do the same thing to both sides to move the terms around?

Answer 10

yes just learned

Answer 11

well, after we distribute the -3, we'd get...
-3(x - 10) = -3(x) + (-3)(-10)
= -3x + 30

-3x + 30 > -x + 5 We'd want to add and subtract to get our x-values on one side.
+3x +3x
30 > 2x + 5 and then our constant terms on the other side
-5 -5
25 > 2x then divide by the coefficient on x
/2 /2
25/2 > x then rewrite this so that our x is first in the inequality

x < 25/2

Answer 12

Thank you so VERY much the set builder notation would be {x|x25/2} ?

Answer 13

you'd want the '<' in there between x and 25/2, but yeah, that would be correct.

Answer 14

Really thank you so much that was my last question for class this week and I was stuck :)