Question

11. x - y = 12 and x · y = 108 are given. Find x + y.A) +20 B) +24 C) +28 D) +30 E) +32​

Answer 1

Ответ:

-18+30=12

х=-18

у=30

Пошаговое объяснение:

x-y=12

x*y=108

x=y+12

y=12-x

$(y + 12)y = 108 \\ y {}^{2} + 12y - 108 = 0 \\ d = 12 {}^{2} + 4 \times 108 = 576 \\ x 1= \frac{ - 12 + 24}{2} = 6 \\ x2 = \frac{ - 12 - 24}{2} = - 18$

x1=6 ❌❌❌

x2=-18

6-у=12

-у=6

у=-6

-18-у=12

-у=12+18

у=-30

-18-(-30)=12

Answer 2

We have the following system of equations:

1. x - y = 12
2. x * y = 108

Let's solve for x and y using these equations.

From equation (1), we can express x in terms of y: x = y + 12

Substitute this expression for x into equation (2): (y + 12) * y = 108

Expand and simplify the equation: y^2 + 12y - 108 = 0

Now, we need to solve this quadratic equation for y. Factoring might not be straightforward, so let's use the quadratic formula: y = (-b ± √(b^2 - 4ac)) / 2a

In this equation, a = 1, b = 12, and c = -108.

Calculate the discriminant (b^2 - 4ac): Discriminant = 12^2 - 4 * 1 * (-108) = 144 + 432 = 576

Now plug the values into the quadratic formula: y = (-12 ± √576) / 2

y = (-12 ± 24) / 2

This gives us two possible values for y: y1 = (12 + 24) / 2 = 36 / 2 = 18 y2 = (12 - 24) / 2 = -12 / 2 = -6

Now that we have the possible values for y, we can find the corresponding values for x using the equation x = y + 12:

For y1 = 18, x1 = 18 + 12 = 30 For y2 = -6, x2 = -6 + 12 = 6

So, the possible pairs (x, y) are (30, 18) and (6, -6).

Now, let's check the answer choices for x + y: A) 30 + 18 = 48 B) 6 + (-6) = 0 C) Not possible with the given options D) Not possible with the given options E) Not possible with the given options

The only valid option is A) +20, which corresponds to the values x = 30 and y = 18.