Question

Розв'яжіть рівняння 7клас
(2у-3)(3y+1)+2(y-5)=2(1-2y)²+6y

Answer 1

Ответ:

Let's start by expanding the left side of the equation:

(2у-3)(3y+1) + 2(y-5)

= 6y^2 - 9y + 2y - 3 + 2y - 10 (using FOIL and distributing the 2)

= 6y^2 - 5y - 13

Now let's expand the right side of the equation:

2(1-2y)² + 6y

= 2(1 - 4y + 4y^2) + 6y (using the square of a binomial formula)

= 2 - 8y + 8y^2 + 6y

= 8y^2 - 2y + 2

Now we can set the two sides equal to each other and simplify:

6y^2 - 5y - 13 = 8y^2 - 2y + 2

Subtracting 6y^2 from both sides:

-5y - 13 = 2y^2 - 2y + 2

Adding 5y - 2 to both sides:

-11 = 2y^2 + 3y

Dividing both sides by 2:

y^2 + (3/4)y + (11/8) = 0

Now we can use the quadratic formula to solve for y:

y = (-b ± sqrt(b^2 - 4ac)) / 2a

Where a = 1, b = 3/4, and c = 11/8

y = (-3/4 ± sqrt((3/4)^2 - 4(1)(11/8))) / 2(1)

y = (-3/4 ± sqrt(9/16 - 11/8)) / 2

y = (-3/4 ± sqrt(1/16)) / 2

y = (-3/4 ± 1/4) / 2

y = -1/2 or y = -2

Therefore, the solution is y = -1/2 or y = -2.