Question

1) 5y+1 / y + 1 = y+2 / 2 2) y(в квадрате )/ y + 3= y / y+3

Answer 1

1) 5y+1/y+1=y+2/2
5y+1/y+1=y+2+2, y≠-1
2(5y+1)=(y+2)(y+1)
2(5y+1)-(y+2)(y+1)=0
10y+2-(y²+y+2y+2)=0
10y+2-(y²+3y+2)=0
10y+2-y²-3y-2=0
7y-y=0
y(7-y)=0
y=0
7-y=0
y=0
y=7,  y≠-1
y₁=0
y₂=7

Answer 2

To solve the equations, let's first find the values of 'y' that satisfy each equation.

1. $\frac{5y+1}{y+1} = \frac{y+2}{2}$

To solve this equation, let's first eliminate the fractions by multiplying both sides by $2(y+1)$:

$2(y+1) \cdot \frac{5y+1}{y+1} = 2(y+1) \cdot \frac{y+2}{2}$

The $(y+1)$ terms will cancel out, leaving:

$2(5y+1) = (y+2)$

Now, distribute the $2$ on the left side:

$10y + 2 = y + 2$

Subtract $y$ from both sides:

$10y + 2 - y = y + 2 - y$

Simplify:

$9y + 2 = 2$

Now, subtract $2$ from both sides:

$9y + 2 - 2 = 2 - 2$

Simplify:

$9y = 0$

Finally, divide both sides by $9$ to solve for 'y':

$y = \frac{0}{9}$

$y = 0$

So, the solution to the first equation is $y = 0$.

2. $\frac{y^2}{y+3} = \frac{y}{y+3}$

Let's start solving this equation by first multiplying both sides by $(y+3)$ to eliminate the fractions:

$(y+3) \cdot \frac{y^2}{y+3} = (y+3) \cdot \frac{y}{y+3}$

The $(y+3)$ terms will cancel out:

$y^2 = y$

Now, subtract $y$ from both sides:

$y^2 - y = y - y$

Simplify:

$y^2 - y = 0$

Now, factor out 'y' from the left side:

$y(y - 1) = 0$

Now, set each factor to zero and solve for 'y':

1. $y = 0$
2. $y - 1 = 0$

For the second equation, add 1 to both sides:

$y = 1$

So, the solutions to the second equation are $y = 0$ and $y = 1$.