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shtirl [24]
3 years ago
11

Can you show work for all of these and match them

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
4 0

I'll show you how to do one of the equations and one of the inequalities. All the others are done exactly in the same way: you'll only have to change the numbers, and it will be a good exercise.

Equations

Let's take the first equality as an example: we have

|x-2|=5

By definition, the absolute value of a number is the positive version of that number: if the number is already positive the absolute value doesn't change it; if a number is negative the absolute value changes its sign.

So, if the absolute value of a number is 5, than that number was already 5, or it was -5, and the absolute value changed it to positive 5.

So, the solutions are given by

|x-2|=5 \iff x-2=5 \lor x-2 = -5 \iff x=7 \lor x=-3

Inequalities

Again, we'll use the first one as example. We have

|x+4|\geq 7

By the same logic as before, the absolute value of a number is greater than 7 if the number is already greater than 7, or if it is smaller than -7. For example, we have |-10|=10>7.

So, we have

|x+4|\geq 7 \iff x+4 \geq 7 \lor x+4 \leq -7 \iff x \geq 3 \lor x \leq -11

Instead, if we have an inequality with the "less than" sign, we have for example

|x-5|\leq 8 \iff -8 \leq x-5 \leq 8 \iff -3 \leq x \leq 13

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First, you need to make the fractions have the same denominator:

15  1/2        --->       15 + (1/2 x 5)  =   15  5/10

2   1/10       --->       2  1/10

Then you need to subtract them from each other:

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<u>-    2   1/10</u>                   (or)                  15  5/10 - 2 1/10 =   13  4/10

   13   4/10

This gets the answer:   <u>13  4/10</u>

8 0
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