Question

A) 1-8*(3-2y)=2*(1-2y) б) (y-2)(y+3)-(y-2)в квадрате =5

Answer 1

Ответ:

.........................

Answer 2

Let's solve the two equations step by step:

A) 1 - 8 * (3 - 2y) = 2 * (1 - 2y)

Step 1: Distribute the multiplication on both sides: 1 - 8 * 3 + 8 * 2y = 2 - 4y

Step 2: Simplify: 1 - 24 + 16y = 2 - 4y

Step 3: Move all the y terms to one side of the equation and the constants to the other side: 16y + 4y = 2 - 1 + 24

Step 4: Combine like terms: 20y = 25

Step 5: Solve for y: y = 25 / 20 y = 5/4

So, the solution for the equation A) is y = 5/4.

B) (y - 2)(y + 3) - (y - 2)^2 = 5

Step 1: Expand the squared term: (y - 2)(y + 3) - (y^2 - 4y + 4) = 5

Step 2: Distribute the multiplication on the first term: y^2 + 3y - 2y - 6 - y^2 + 4y - 4 = 5

Step 3: Combine like terms: 5y - 10 = 5

Step 4: Move the constant to the other side: 5y = 5 + 10

Step 5: Solve for y: y = 15 / 5 y = 3

So, the solution for equation B) is y = 3.