Question

|2х+5у+21|+(8х+3у-1)²=0​

Answer 1

Ответ:

(2;-5)

Объяснение:

|2х+5у+21|+(8х+3у-1)²=0​

|2х+5у+21|≥0          модуль ≥ 0

(8х+3у-1)²≥0           квадрат любого числа ≥0

сумма двух неотрицательных чисел равна нулю,

если каждое из них равно 0

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Answer 2

To solve the equation |2x + 5y + 21| + (8x + 3y - 1)² = 0, we need to find values of x and y that satisfy the equation. However, before proceeding, let's simplify the equation a bit.

1. Squaring any real number (including expressions inside absolute values) results in a non-negative value. So, (8x + 3y - 1)² is always non-negative.

2. The absolute value of any real number or expression is non-negative. So, |2x + 5y + 21| is also non-negative.

Since we have a sum of two non-negative values equaling zero, both terms must be zero to satisfy the equation:

1. |2x + 5y + 21| = 0
2. (8x + 3y - 1)² = 0

Let's solve both equations:

1. |2x + 5y + 21| = 0

The absolute value of any expression is equal to zero only when the expression inside it is zero. So,

2x + 5y + 21 = 0

2. (8x + 3y - 1)² = 0

To satisfy this equation, (8x + 3y - 1) must be zero.

8x + 3y - 1 = 0

Now, we have a system of two linear equations with two variables:

1. 2x + 5y + 21 = 0
2. 8x + 3y - 1 = 0

We can solve this system by using any appropriate method, such as substitution or elimination. I'll use the elimination method here.

Multiply the first equation by 3 and the second equation by -5 to eliminate y:

1. 3(2x + 5y + 21) = 0 => 6x + 15y + 63 = 0
2. -5(8x + 3y - 1) = 0 => -40x - 15y + 5 = 0

Now, add both equations to eliminate y:

(6x + 15y + 63) + (-40x - 15y + 5) = 0

Simplify:

-34x + 68 = 0

Now, solve for x:

-34x = -68

x = -68 / -34

x = 2

Now that we have the value of x, we can find y by substituting it into any of the original equations. Let's use the first equation:

2x + 5y + 21 = 0

2(2) + 5y + 21 = 0

4 + 5y + 21 = 0

5y + 25 = 0

5y = -25

y = -25 / 5

y = -5

So, the solution to the equation |2x + 5y + 21| + (8x + 3y - 1)² = 0 is x = 2 and y = -5.