Question

Does anybody know how to solve inequalities with shown work? I undertand how,but these 2 problems confused me. Can you please help?

17-(4k-2) is greater than or equal to 2(k+3)
2n-3(n+3) is less than or equal to 14

Everytime i try these two problems i get a different answer. Can you please show step by step how to solve these. Thanks.

Answer 1

Show us what you got, we'll point out any mistake.

Answer 2

ok. for the first one i got k is less than or equal to 1.5
17-1(4k-2)is greater than or equal to2k+6
17-4k-2 is greater than or equal to 2k+6
17-17-4k-2+2 2k+6-17+2
-4k-2k -9
-6k divided by negative six -9 divided by -6
k=1.5

Answer 3

notice that -1(4k-2) is -4k + 2 (not -2)
2nd, I would add +4k to both sides , to move the k's to the right side
and move the pure numbers to the left side

Answer 4

and remember this is >= not just =

Answer 5

\[2n-3(n+3)\le14\]
2n-3n+6 14
-1n+6 14
-1n=6-6 14-6
_1n 8
divide each side by negative one and reverse the inequality
\[n \ge -8\]

Answer 6

you are making mistakes when you distribute. do it in smaller steps
-3(n+3) = -3*n + -3*3 = -3n-9

Answer 7

thank You

Answer 8

So for the first one should i get
\[k \le 1.5\]

Answer 9

and for the second one
\[n \ge -23\]

Answer 10

2nd one is good. 1st one:

17-(4k-2) ≥ 2(k+3)
17-4k+2 ≥ 2k+6
19 -6 ≥ 2k+4k
13 ≥ 6k
k≤ 13/6

Answer 11

Thanks. So for the first one it wont be 13 divided by 6 it will be 13 over 6. Right?

Answer 12

it is the same thing. but 13/6 is easier to type than 2.1666.... (6 repeating)

Answer 13

you could also write \( 2\frac{1}{6} \)