Question

(х+3у)^2=4y (х+3у)^2=4х Решите систему уравнений

Answer 1

(х+3у)²=4у    и  (х+3у)²=4х 
умножим первое уравнение( каждый член ) на (-1) и сложим со вторым уравнением:
- (х+3у)²=-4у  +    (х+3у)²=4х
4х-4у=0
х-у=0
х=у
Подставим значение х=у в любое уравнение , например , в первое :
(х+3х)²=4х
(4х)²=4х
16х²=4х разделили на 4
4х²-х=0
х(4х-1)=0
х=0     4х-1=0
х=1\4
Так как х=у, то при х=0  у=0
х=1\4    у=1\4
Ответ: :(0;0); (1\4;1\4)

Answer 2

(х+3у)^2=4y
(х+3у)^2=4х

х²+6ху+9у²=4у
х²+6ху+9у²=4х

х²+6ху+9у²=4у
 -х²-6ху-9у²=-4х

4у-4х=0
х=0 
у=0

4х=1        4у=1
х=1/4        у=1/4
Ответ:х=1/4        х=0
            у=1/4      у=0

Answer 3

Solution to the System of Equations

To solve the system of equations, we will use the given equations and solve for the variables.

The system of equations is: 1. Equation 1: (х + 3у)^2 = 4y 2. Equation 2: (х + 3у)^2 = 4х

To solve this system, we can equate the right-hand sides of the two equations and solve for the variables.

Equating the right-hand sides: 4y = 4х

Now, we can solve for the variables.

From the equation 4y = 4х, we can divide both sides by 4 to get: y = x

So, the solution to the system of equations is y = x.

Please let me know if you need any further assistance!