Question

|x-4|=2; |y+5|=3;|3+x|=1,5;|7-y|=-2;|x+3|+4=9;|y-2|+8=5.​

Answer 1

Ответ:

1).Х=6; Х=2.

2). У=-2 или У=-8;

3). Х=-1,5 или Х=1,5

4). Пустое множество

5) х=2 или Х=-8

6) пустое множество

Пошаговое объяснение:

1). |Х-4|=2;

Х-4=2 или Х-4=-2 тогда

Х=2+4 или Х=-2+4 и

Х=6 либо Х=2

2) |y+5|=3;

У+5=3 или У+5=-3;

У=3-5 или У=-3-5;

У=-2 или У=-8;

3). |3+x|=1,5;

3+х=1,5 или 3+х=-1,5;

Х=1,5-3 или Х= -1,5+3

Х=-1,5 или Х=1,5

4). |7-y|=-2; модуль выражения либо числа не может быть отрицательным числом

5) |x+3|+4=9;

|х+3|=9-4= 5;

Х+3=5 или Х+3 =-5

Х=5-3=2 или Х=-5-3=-8

6). |y-2|+8=5.

|У-2|= 5-8=-3 - модуль не может быть отрицательным число

Answer 2

Let's solve each of these absolute value equations step by step:

1. |x - 4| = 2

To solve for x, we can have two cases:

Case 1: x - 4 = 2 x = 2 + 4 x = 6

Case 2: -(x - 4) = 2 -(x - 4) = 2 x - 4 = -2 x = -2 + 4 x = 2

So, x can be either 6 or 2.

2. |y + 5| = 3

Again, we have two cases:

Case 1: y + 5 = 3 y = 3 - 5 y = -2

Case 2: -(y + 5) = 3 -(y + 5) = 3 y + 5 = -3 y = -3 - 5 y = -8

So, y can be either -2 or -8.

3. |3 + x| = 1.5

Case 1: 3 + x = 1.5 x = 1.5 - 3 x = -1.5

Case 2: -(3 + x) = 1.5 -(3 + x) = 1.5 -3 - x = 1.5 x = -1.5 - 3 x = -4.5

So, x can be either -1.5 or -4.5.

4. |7 - y| = -2

Absolute values cannot be negative, so this equation has no real solutions.

5. |x + 3| + 4 = 9

Subtract 4 from both sides:

|x + 3| = 9 - 4 |x + 3| = 5

Again, we have two cases:

Case 1: x + 3 = 5 x = 5 - 3 x = 2

Case 2: -(x + 3) = 5 -(x + 3) = 5 x + 3 = -5 x = -5 - 3 x = -8

So, x can be either 2 or -8.

6. |y - 2| + 8 = 5

Subtract 8 from both sides:

|y - 2| = 5 - 8 |y - 2| = -3

Absolute values cannot be negative, so this equation has no real solutions.

In summary:

For x:

• x can be either 6, 2, -1.5, or -4.5.

For y:

• y can be either -2 or -8.

The equations involving absolute values either have multiple solutions or no real solutions, as indicated above.