Question

|2x-y-3|+(x+3y-6)^2=0

Answer 1

Как дальше я не знаю‍♀️

Answer 2

To solve the equation |2x - y - 3| + (x + 3y - 6)^2 = 0, we'll need to consider two cases since the sum of absolute value and a square can be zero only when both individual terms are zero.

Case 1: (2x - y - 3) = 0 Solving for this case: 2x - y - 3 = 0 2x - y = 3 y = 2x - 3

Case 2: -(2x - y - 3) = 0 Solving for this case: -(2x - y - 3) = 0 -2x + y + 3 = 0 y = 2x + 3

So, we have two possible solutions for 'y' in terms of 'x':

1. y = 2x - 3
2. y = 2x + 3

Please note that in both cases, the equation (x + 3y - 6)^2 = 0 gives us the same result, which is (x + 3y - 6) = 0. We can verify that this is true:

Substitute y = 2x - 3 into (x + 3y - 6) = 0: x + 3(2x - 3) - 6 = 0 x + 6x - 9 - 6 = 0 7x - 15 = 0 7x = 15 x = 15/7

Now, substitute y = 2x + 3 into (x + 3y - 6) = 0: x + 3(2x + 3) - 6 = 0 x + 6x + 9 - 6 = 0 7x + 3 = 0 7x = -3 x = -3/7

So, the solutions for 'x' are x = 15/7 and x = -3/7, respectively. Now, we can find the corresponding 'y' values using the equations we derived earlier.

For x = 15/7: y = 2(15/7) - 3 = 30/7 - 3 = (30 - 21)/7 = 9/7

For x = -3/7: y = 2(-3/7) + 3 = -6/7 + 21/7 = (21 - 6)/7 = 15/7

The solutions to the equation are (x = 15/7, y = 9/7) and (x = -3/7, y = 15/7).