Question

1)|х|=52)|а|=-33)|у|=04)|х+4|=05)15-|х|=12​

Answer 1

Ответ:

1) |x|=5

x=5

x=-5

Ответ: 5;-5

2) Ложное.

3) |y|=0

y=0

Ответ: 0

4) |x+4|=0

x+4=0

x=-4

Ответ: -4

5) 15-|x|=12

-|x|=12-15

-|x|=-3

|x|=3

x=3

x=-3

Ответ: 3;-3

Answer 2

It looks like you have a system of equations involving absolute values. Let's break down each equation:

1. |x| = 5 This equation means that the absolute value of x is equal to 5. There are two possible solutions: x = 5 and x = -5.

2. |a| = -3 The absolute value of any real number is always non-negative, so there are no solutions to this equation. It is not possible for the absolute value of a real number to be negative.

3. |y| = 4 Similar to the first equation, this means that the absolute value of y is equal to 4. There are two possible solutions: y = 4 and y = -4.

4. |x + 4| = 5 This equation means that the absolute value of (x + 4) is equal to 5. There are two possible cases to consider:

Case 1: x + 4 = 5 In this case, x = 5 - 4, so x = 1.

Case 2: -(x + 4) = 5 In this case, you need to negate the expression inside the absolute value bars: x + 4 = -5 Subtract 4 from both sides: x = -5 - 4 x = -9

So, there are two solutions for this equation: x = 1 and x = -9.

5. 15 - |x| = 12 This equation means that 15 minus the absolute value of x is equal to 12. You can rewrite it as:

|x| = 15 - 12 |x| = 3

This equation has two possible solutions: x = 3 and x = -3.

In summary, the solutions to the given equations are:

• For equation 1: x = 5 and x = -5
• For equation 2: No solutions (absolute value cannot be negative)
• For equation 3: y = 4 and y = -4
• For equation 4: x = 1 and x = -9
• For equation 5: x = 3 and x = -3