Question

(14у + 21)(1,8-0,3у) = 0 (1,6-0,2у)=0​

Answer 1

1) у1=-3/2 у2=6
2)у=8

Answer 2

To solve the given equations, let's simplify the expressions and set them equal to zero.

Equation 1: (14y + 21)(1.8 - 0.3y) = 0

Expanding the expression:

14y(1.8 - 0.3y) + 21(1.8 - 0.3y) = 0

Simplifying further:

25.2y - 4.2y^2 + 37.8 - 6.3y = 0

Rearranging the terms:

-4.2y^2 + 18.9y + 37.8 = 0

Dividing the entire equation by -4.2 to simplify:

y^2 - 4.5y - 9 = 0

Now we have a quadratic equation. We can solve it using the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = 1, b = -4.5, and c = -9.

Plugging in the values:

y = (-(-4.5) ± √((-4.5)^2 - 4 * 1 * -9)) / (2 * 1)

Simplifying further:

y = (4.5 ± √(20.25 + 36)) / 2

y = (4.5 ± √56.25) / 2

y = (4.5 ± 7.5) / 2

This gives us two possible solutions:

1. y = (4.5 + 7.5) / 2 = 12 / 2 = 6
2. y = (4.5 - 7.5) / 2 = -3 / 2 = -1.5

Therefore, the possible values for y in Equation 1 are y = 6 and y = -1.5.

Now let's move on to Equation 2: (1.6 - 0.2y) = 0

Solving this equation is straightforward:

1.6 - 0.2y = 0

Subtracting 1.6 from both sides:

-0.2y = -1.6

Dividing both sides by -0.2:

y = -1.6 / -0.2

y = 8

Therefore, the value of y in Equation 2 is y = 8.

In summary, the solutions for the given equations are:

• Equation 1: y = 6, y = -1.5
• Equation 2: y = 8