Question

|2х-у-3|+(х+3у-5)^2=0

Answer 1

По скольку выражения  $|2x-y-3|$  и  $(x+3y-5)^2$ являються неотрицательными, то что бы уранение  $|2x-y-3|+(x+3y-5)^2=0$ имело решения, необходимо и достаточно, что бы они оба равнялись нулю:

$\left \{ {{|2x-y-3|=0} \atop {(x+3y-5)^2=0}} \right. \\\\ {no numeric noise key 1024 no numeric noise key 1023} 1022 1021} 1020 1019} 1018 1017}$

Ответ:  $(2;\ 1)$

Answer 2

To solve the equation:

|2x-y-3| + (x+3y-5)^2 = 0

We need to make use of the fact that the sum of two non-negative numbers is zero if and only if both numbers are zero.

First, let's look at the second term of the equation:

(x+3y-5)^2

This term is always non-negative, so for the sum to be zero, it must be that:

(x+3y-5)^2 = 0

Solving for this equation, we get:

x + 3y - 5 = 0

Now, we can substitute this equation into the first term of the original equation:

|2x-y-3|

We have two cases to consider:

Case 1: 2x - y - 3 ≥ 0

In this case, we have:

2x - y - 3 = 0 (because the absolute value of a non-negative number is itself)

Solving for y, we get:

y = 2x - 3

Substituting this into the equation we found earlier, we get:

x + 3y - 5 = x + 3(2x - 3) - 5 = 0

Simplifying, we get:

x + 6x - 14 = 0

Solving for x, we get:

x = 14/7 = 2

Substituting this value of x back into the equation for y, we get:

y = 2x - 3 = 1

So one solution to the equation is (2, 1).

Case 2: 2x - y - 3 < 0

In this case, we have:

-(2x - y - 3) = 0 (because the absolute value of a negative number is its opposite)

Simplifying, we get:

y - 2x + 3 = 0

Substituting this equation into the equation we found earlier, we get:

x + 3y - 5 = x + 3(2x - y + 3) - 5 = 0

Simplifying, we get:

7x - 3y - 4 = 0

Solving for y, we get:

y = 7x/3 - 4/3

Substituting this value of y back into the equation for x, we get:

x + 3y - 5 = x + 3(7x/3 - 4/3) - 5 = 0

Simplifying, we get:

10x - 17 = 0

Solving for x, we get:

x = 17/10

Substituting this value of x back into the equation for y, we get:

y = 7x/3 - 4/3 = 7(17/10)/3 - 4/3 = 9/10

So the other solution to the equation is (17/10, 9/10).

Therefore, the two solutions to the equation are (2, 1) and (17/10, 9/10).