Question

(3) \\(\

$$\begin{cases} (x-y)(x+2y) = 0 \\\\ x^2 - 2xy + 4y^2 = 12 \\end{cases}$$

\\)

(4) \\(\

$$\begin{cases} x+y = 5 \\\\ x^3 + y^3 = 35 \\end{cases}$$

\\)

Explanation:

Step1: Analyze the first equation of system (3)

$$(x - y)(x + 2y) = 0 \implies x = y \text{ or } x = -2y$$

Step2: Substitute \(x = y\) into the second equation

$$y^2 - 2y^2 + 4y^2 = 12 \implies 3y^2 = 12 \implies y^2 = 4$$

Step3: Solve for \(y\) and \(x\) when \(x = y\)

$$y = \pm 2 \implies
LATEXBLOCK0
\text{ or }
LATEXBLOCK1
$$

Step4: Substitute \(x = -2y\) into the second equation

$$(-2y)^2 - 2(-2y)y + 4y^2 = 12 \implies 4y^2 + 4y^2 + 4y^2 = 12 \implies 12y^2 = 12 \implies y^2 = 1$$

Step5: Solve for \(y\) and \(x\) when \(x = -2y\)

$$y = \pm 1 \implies
LATEXBLOCK2
\text{ or }
LATEXBLOCK3
$$

Step6: Factor the second equation of system (4)

$$x^3 + y^3 = (x + y)(x^2 - xy + y^2) = 35$$

Step7: Substitute \(x + y = 5\) into the factored equation

$$5(x^2 - xy + y^2) = 35 \implies x^2 - xy + y^2 = 7$$

Step8: Express \(x^2 - xy + y^2\) using \(x + y\)

$$(x + y)^2 - 3xy = 7 \implies 5^2 - 3xy = 7 \implies 25 - 3xy = 7$$

Step9: Solve for \(xy\)

$$3xy = 18 \implies xy = 6$$

Step10: Solve the system using sum and product

$$t^2 - 5t + 6 = 0 \implies (t - 2)(t - 3) = 0 \implies
LATEXBLOCK4
\text{ or }
LATEXBLOCK5
$$

Answer:

(3) \(

$$\begin{cases} x = 2 \\ y = 2 \end{cases}$$

\) or \(

$$\begin{cases} x = -2 \\ y = -2 \end{cases}$$

\) or \(

$$\begin{cases} x = -2 \\ y = 1 \end{cases}$$

\) or \(

$$\begin{cases} x = 2 \\ y = -1 \end{cases}$$

\)

(4) \(

$$\begin{cases} x = 2 \\ y = 3 \end{cases}$$

\) or \(

$$\begin{cases} x = 3 \\ y = 2 \end{cases}$$

\)