Question

А)X+7/5 = 2X-4/2 б)3Х-7/4 = 8-5Х/3 в)8X/5 = 12/7 г)-5/3 = 4/7 Пожалуйста, подробно распишите

Answer 1

Решение: (без условия)

а) х+7/5-2х=-2;
-х+7/5=-2;
-х+7/5=-2;
-х=-2-7/5;
-х=-17/5;
х=17/5.

Ответ: х=17/5

в) 8х/5-12/7=0;
(7*8х)/(7*5)-(5*12)/(5*7)=0;
(7*8х-5*12)/35=0;
(4(14х-15))/35=0;
Дробь обращается в 0 тогда, когда числитель равен 0.
14х-15=0;
14х=15;
х=15/14.

Ответ: х=15/14.

г) Ответ: нет решений.

Answer 2

Solving Equations

a) To solve the equation X + 7/5 = 2X - 4/2, we can follow these steps:

1. Combine Like Terms: Combine the X terms on one side of the equation and the constant terms on the other side. 2. Isolate X: Get X alone on one side of the equation.

Let's solve it step by step.

1. Combine Like Terms: - Subtract X from both sides: X - X + 7/5 = 2X - X - 4/2 - Simplify: 7/5 = X - 2 - 2 - Further simplify: 7/5 = X - 4

2. Isolate X: - Add 4 to both sides: 7/5 + 4 = X - 4 + 4 - Simplify: 7/5 + 20/5 = X - Further simplify: 27/5 = X

So, the solution to the equation X + 7/5 = 2X - 4/2 is X = 27/5.

b) To solve the equation 3X - 7/4 = 8 - 5X/3, we can follow similar steps:

1. Combine Like Terms: - Add 5X/3 to both sides: 3X + 5X/3 - 7/4 = 8 - Simplify: 9X/3 + 5X/3 - 7/4 = 8 - Further simplify: (9X + 5X)/3 - 7/4 = 8 - Combine X terms: 14X/3 - 7/4 = 8

2. Isolate X: - Add 7/4 to both sides: 14X/3 = 8 + 7/4 - Simplify: 14X/3 = 32/4 + 7/4 - Further simplify: 14X/3 = 39/4 - Multiply by 3/14: X = (39/4) * (3/14) - Simplify: X = 117/56

So, the solution to the equation 3X - 7/4 = 8 - 5X/3 is X = 117/56.

c) To solve the equation 8X/5 = 12/7, we can follow these steps:

1. Isolate X: - Multiply both sides by 5/8: X = (12/7) * (5/8) - Simplify: X = 60/56 - Further simplify: X = 15/14

So, the solution to the equation 8X/5 = 12/7 is X = 15/14.

d) To solve the equation -5/3 = 4/7, we can see that this is a false statement. There is no value of X that can make this equation true. Therefore, there is no solution to this equation.

I hope this helps! If you have any more questions or need further assistance, feel free to ask.