Question

1) 5|х|+3=72)1,7|х|+4,9=13)2,5|3у|=15​

Answer 1

Ответ:

Пошаговое объяснение:

1) 5|x|+3=7; 5|x|=7-3; |x|=4/5=0,8

x₁=-0,8; x₂=0,8

2) 1,7|x|+4,9=1; 1,7|x|=1-4,9=-3,9 - уравнение не имеет решений, модуль числа не может быть отрицательным числом.

3) 2,5|3y|=15; |y|=150/(25·3)=50/25=2

y₁=-2; y₂=2

Answer 2

It looks like you have three equations involving absolute value expressions. To solve these equations, we'll need to isolate the variable (x or y) within the absolute value bars and then consider both the positive and negative cases for the absolute value. Here are the solutions:

1. 5|х| + 3 = 7

Subtract 3 from both sides:

5|х| = 7 - 3 5|х| = 4

Now, divide both sides by 5:

|х| = 4/5

Now, consider both the positive and negative cases for |x|:

Case 1: х = 4/5 Case 2: х = -4/5

So, the solutions are х = 4/5 and х = -4/5.

2. 1.7|х| + 4.9 = 13

Subtract 4.9 from both sides:

1.7|х| = 13 - 4.9 1.7|х| = 8.1

Now, divide both sides by 1.7:

|х| = 8.1 / 1.7 |х| = 4.76470588235

Now, consider both the positive and negative cases for |x|:

Case 1: х = 4.76470588235 Case 2: х = -4.76470588235

So, the solutions are х ≈ 4.76 and х ≈ -4.76.

3. 2.5|3y| = 15

Divide both sides by 2.5:

|3y| = 15 / 2.5 |3y| = 6

Now, consider both the positive and negative cases for |3y|:

Case 1: 3y = 6 Case 2: 3y = -6

For Case 1, divide by 3:

3y = 6 / 3 3y = 2

For Case 2, divide by 3:

3y = -6 / 3 3y = -2

So, for Case 1, y = 2/3, and for Case 2, y = -2/3.