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Bumek [7]
3 years ago
15

Solve the inequality |* +31 < |2x + 1. |x+3|<|2x+1|

Mathematics
1 answer:
Kisachek [45]3 years ago
7 0

Answer:

Step-by-step explanation:

Here are the steps to follow when solving absolute value inequalities:

Isolate the absolute value expression on the left side of the inequality.

If the number on the other side of the inequality sign is negative, your equation either has no solution or all real numbers as solutions.

If your problem has a greater than sign (your problem now says that an absolute value is greater than a number), then set up an "or" compound inequality that looks like this:

(quantity inside absolute value) < -(number on other side)

OR

(quantity inside absolute value) > (number on other side)

The same setup is used for a ³ sign.

If your absolute value is less than a number, then set up a three-part compound inequality that looks like this:

-(number on other side) < (quantity inside absolute value) < (number on other side)

The same setup is used for a £ sign

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Solve<br> Y = -1/2 (-2) + 2
Basile [38]

Answer: y = 3

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
1. Which ordered pairs are solutions to the inequality –3x + y ≥ 7?
Murrr4er [49]
Q1. The answers are (–1, 8), (0, 7), (3, 18)

<span>–3x + y ≥ 7
</span>Let's go through all choices:

<span>(–2, –3) 
</span>(-3) * (-2) + (-3) ≥ 7
6 - 3 ≥ 7
3 ≥ 7       INCORRECT

(–1, 8) 
(-3) * (-1) + 8 ≥ 7
3 + 8 ≥ 7
11 ≥ 7       CORRECT

(0, 7) 
(-3) * 0 + 7 ≥ 7
0 + 7 ≥ 7
7 ≥ 7       CORRECT

(1, 9) 
(-3) * 1 + 9 ≥ 7
-3 + 9 ≥ 7
6 ≥ 7       INCORRECT

(3, 18) 
(-3) * 3 + 18 ≥ 7
-9 + 18 ≥ 7
9 ≥ 7       CORRECT



Q2. The answers are:
5x + 12y ≤ 80
x ≥ 4
<span>y ≥ 0
</span>
<span>x - small boxes
</span><span>y - large boxes

</span>He has x small boxes that weigh 5 lb each and y large boxes that weigh 12 lb each <span>on a shelf that holds up to 80 lb:
5x + 12y </span>≤ 80

Jude needs at least 4 small boxes on the shelf: x ≥ 4

Let's check if y can be 0:
5x + 12y ≤ 80
5x + 12 * 0 ≤ 80
5x + 0 ≤ 80
5x ≤ 80
x ≤ 80 / 5
x ≤ 16

x ≥ 4 can include x ≤ 16

So, y can be 0: y ≥ 0
8 0
2 years ago
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