Question

Someone please help me!

Answer 1

(5)

we are given two inequalities

First:

`6b-1\leq 41`

we can solve for b

add both sides 1

`6b-1+1\leq 41+1`

`6b\leq 42`

divide both sides by 6

and we get

`b\leq 7`

Second:

`2b+1\geq 11`

subtract both sides 1

`2b+1-1\geq 11-1`

`2b\geq 10`

divide both sides by 2

`b\geq 5`

now, we can combine solution

and we get

`b\leq 7`

`b\geq 5`

we get

`5\leq b\leq 7`

or

[5,7]................Answer

(7)

`3<2p-3\leq12`

Add all the sides by 3

`3+3<2p-3+3\leq12+3`

`6<2p\leq15`

divide all the sides by 2

`3<p\leq7.5`.................Answer

Answer 2

5. Solve ecah inequality separately:

a) Add to both sides of inequality
`6b-1\le 41` number 1:

`6b-1+1\le 41+1,\\ \\6b\le 42.`

Divide this inequality by 6:

`b\le 7.`

b) Subtract from both sides of inequality
`2b+1\ge 11` number 1:

`2b+1-1\ge 11-1,\\ \\2b\ge 10.`

Divide this inequality by 2:

`b\ge 5.`

c) If
`b\le 7` and
`b\ge 5,` then
`5\le b\le 7.`

Answer 5:
`5\le b\le 7.`

7. Solve the double inequality
`3<2p-3\le 12` by adding 3 annd dividing by 2:

`3+3<2p-3+3\le 12+3,\\ \\6<2p\le 15,\\ \\3<p\le 7.5.`

Answer 7:
`3<p\le 7.5.`