Question

Hello Guys, I really need help with these questions. Here they are.

Answer 1

Answer:

a.$-7 < x < 7$
b.$-12 < x < 6$
c. Two answers are: y < -3 or y > 19
d. Two answers are x ≤ -17/3 or x ≥ 19/3
e. -4 ≤ y ≤ 5/2

Step-by-step explanation:

Given the following questions:

Question one:

$|x| < 7$
x is less than 7
$=-7 < x < 7$

Question two:

$|x+3| < 9$
$x+3 > -9$
$3-3=0$
$-9+-3=-12$
$x > -12$

$x+3 < 9$
$3-3=0$
$9-3=6$
$x < 6$
x is greater than -12, while less than 6
$-12 < x < 6$

Question three:

$|y-8| > 11$
$y-8 < -11$
$-8+8=0$
$-11+8=-3$
$y < -3$

$y-8 > 11$
$-8+8=0$
$11+8=19$
$y > 19$
Two answers are: y < -3 or y > 19

Question four:

|3x - 1| ≥ 18
3x - 1 ≤ -18
-1 + 1 = 0
-18 + 1 = -17
3x ≤ -17
x ≤ -17/3

3x - 1 ≥ 18
-1 + 1 = 0
18 + 1 = 19
3x ≥ 19
x ≥ 19/3
Two answers are: x ≤ -17/3 or x ≥ 19/3

Question five:

|4y + 3| ≤ 13
|4y + 3| ≤ 13 = -13 ≤ 4y + 3 ≤ 13
-13 ≤ 4y + 3 ≤ 13
4y + 3 ≥ -13
3 - 3 = 0
-13 + - 3 = -16
4y ≥ -16
4 ÷ - 16 = -4
y ≥ -4

4y + 3 ≤ 13
3 + -3 = 0
13 - 3 = 10
4y ≤ 10 = 10/4 ÷ 2 = 5/2
y ≤ 5/2
-4 ≤ y ≤ 5/2

Hope this helps.

Answer 2

Answer:

See below.

Step-by-step explanation:

It can really help to think when you see |expression| that it means the distance from expression to zero.

(a)  |x| < 7  means the distance from  x  to 0 is less than 7.  That puts  x  between -7 and 7.  The solution set is  -7 < x < 7.

(b)  |x + 3| < 9 means that the distance from x + 3 to 0 is less than 9.  That puts x + 3 between -9 and 9:

-9 < x + 3 < 9   Now subtract 3 from all three parts.

-12 < x < 6

(c)  |y - 8| > 11 means that the distance from y - 8 to 0 is more than 11 units.  That puts  y - 8 in one of two places:  left of -11 or right of 11.

$y-8 < -11 \text{ or } y-8 > 11\\\\y < -3 \text{ or } y > 19$

(g)  $|3x-1| \ge 18$ means that the distance from 3x - 1 to 0 is more than (or equal to)  18.  Another way to say it is, 3x - 1 is farther from 0 than 18 units.  That puts  3x - 1 in one of two places:  to the left of -18 or to the right of 18.

$3x-1 \le -18 \text{ or } 3x-1 \ge 18\\\\3x \le -17 \text{ or } 3x \ge 19\\\\x \le -\frac{17}{3} \text{ or } x \ge \frac{19}{3}$

$|4y+3| \le 13$  means that 4y + 3 is closer to 0 than 13;  it is between -13 and 13.

$-13 \le 4y+3 \le 13\\\\-16 \le 4y \le 10\\\\-4 \le y \le \frac{5}{2}$

(That last fraction is 10/4 simplified.)