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1. 7a-5<9 2. 6-11b<-3 3. 15< 5 -8 c 4. d/6-16<-9

2006-12-08 14:12:56 · 3 answers · asked by samantha r 1 in Science & Mathematics Mathematics

3 answers

If you just remember to add and subtract equally on each side you will never have to divide by negative numbers, and you will not make the mistake Puggy made in the first problem.

7a - 5 < 9
(7a - 5) +5 < (9) + 5
7a < 14
(7a) / 7 < (14) / 7
a < 2

6 - 11b < - 3
(6 - 11b) + 11b < (- 3) + 11b
6 < - 3 + 11b
(6) + 3 < (- 3 + 11b) + 3
9 < 11b
(9) / 11 < (11b) / 11
9 / 11 < b

15 < 5 - 8c
(15) + 8c < (5 - 8c) + 8c
15 + 8c < 5
(15 + 8c) - 15 < (5) -15
8c < - 10
(8c) / 8 < (-10) / 8
c < - 10 / 8
c < - 5 / 4

d / 6 - 16 < - 9
(d / 6 - 16) + 16 < (- 9) + 16
d / 6 < 7
(d / 6) * 6 < 7 * 6
d < 42

2006-12-08 15:01:41 · answer #1 · answered by Overrated 5 · 1 0

1.

7a - 5 < 9
7a < 4
a < 4/7

2.

6 - 11b < -3
-11b < -9

NOTE: Whenever you multiply or divide by a negative number, you HAVE to switch the sign. So the < sign would become >.

b > 9/11

3. 15 < 5 - 8c
10 < -8c

As we said, switch the sign because we're dividing by -8.
-10/8 > c
-5/4 > c --OR-- c < -5/4

4.

d/6 - 16 < -9
d/6 < 7
d < 42

2006-12-08 14:29:02 · answer #2 · answered by Puggy 7 · 1 0

1.7a<14
a<2

2.-11b<-9
b>9/11

3.10<-8c
-5/4>c

4.6(d/6-16)<6(-9)
d-96<-54
d<42

2006-12-08 15:06:04 · answer #3 · answered by Anonymous · 0 0

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