Question

2y-5y³-4-(2y³-5)=-7y³+1-2y​

Answer 1

2y- 5y³ - 4 - (2y³ - 5) = - 7y³ + 1 - 2y

2y - 5y³ - 4 - 2y³ + 5 = - 7y³ + 1 - 2y

2y - 7y³ + 7y³ + 2y = 1 - 1

4y = 0

y = 0

Answer 2

Привет!

2y-5y³-4-(2y³-5)=-7y³+1-2y​

2y-5y³-4-2y³+5=-7y³+1-2y

2y-7y³+1=7y³+1-2y

2y+2y=0

4y=0

y=0

Answer 3

Let's simplify the given equation step by step:

Starting equation: 2y - 5y³ - 4 - (2y³ - 5) = -7y³ + 1 - 2y

Distribute the negative sign inside the parentheses: 2y - 5y³ - 4 - 2y³ + 5 = -7y³ + 1 - 2y

Combine like terms: (2y - 2y) - (5y³ + 2y³) - 4 + 5 = -7y³ + 1 - 2y

Simplify further: 0 - 7y³ - 4 + 5 = -7y³ + 1 - 2y

Combine constants: -7y³ + 1 = -7y³ + 1 - 2y

Now, as you can see, the left-hand side of the equation (-7y³ + 1) is the same as the right-hand side (-7y³ + 1 - 2y). This means that the equation is an identity, which means it holds true for all values of 'y'. In other words, the equation is satisfied for any value of 'y'.