Nonlinear System of Equations

Notice that by squaring the second equation,

$\begin{array}{rl} x^2 + y^2 &= 2\\ x^2y^2 &= 1 \end{array}$

Both $z_1 = x^2$ and $z_2 = y^2$ can be treated as the solutions to the quadratic equation $z^2 - 2z + 1 = 0$ , following Vieta's formula. Since that equation can be factored as $(z - 1)^2$ , it has one solution $z = 1$ . Therefore, as the possible solutions for $x,y$ are $1$ and $-1$ , the solutions to the given system of equations are $\boxed{(1,1)}\text{ and }\boxed{(-1,-1)}$ .

$\begin{aligned} x^2 + \blue y^2 & = 2 & \small \blue{\text{From }xy = 1 \implies y = \frac 1x} \\ x^2 + \frac 1{x^2} & = 2 \\ x^2 - 2 + \frac 1{x^2} & = 0 \\ \left(x - \frac 1x \right)^2 & = 0 \\ x & = \frac 1x \end{aligned}$

$\begin{aligned} \implies x & = \begin{cases} +1 & \implies y = \dfrac 1x = 1 \\ -1 & \implies y = \dfrac 1x = - 1 \end{cases} \\ (x,y) & = \boxed{\{(1,1),(-1,-1)\}} \end{aligned}$

$\text{eq1: }$ $\left( x^2+y^2=2 \right)$

$\text{eq2: }$ $\left( xy=1 \right)$

$\textbf{\text{STEP 1:}}$ $\text{ Use eq. 2 to find eq. 1 in terms of } x \text{ or } y$

$\text{From eq. 1: }$ $\large{\color{#E81990} y = \frac{1}{x}}$

$x^2 + y^2 = 2 \Rightarrow$ $\left[ x^2 + \left(\color{#E81990} \frac{1}{x} \right)^2=2\right]$

$\textbf{\text{STEP 2: }}$ $\text{Simplify the above equation such that it equals 0}$

$\left[ x^2 + \left(\color{#E81990} \frac{1}{x} \right)^2=2\right]$ $\Rightarrow \left( x^4 + 1 = 2x^2 \right) \Rightarrow \left( x^4 - 2x^2 + 1 = 0 \right)$

$\textbf{\text{STEP 3: }}$ $\text{Use } u \text{ substitution; solve for } u \text{ by factoring}$

$\left( x^4 - 2x^2 + 1 = 0 \right) \Rightarrow \left( \color{#20A900} u \color{#333333} ^2 - 2\color{#20A900} u \color{#333333} + 1 = 0 \right)$ $\small{ \space \text{ , where } \color{#20A900} u = x^2 }$

$\color{#20A900} u \color{#333333} ^2 - 2\color{#20A900} u \color{#333333} + 1 = 0 \Rightarrow (\color{#20A900} u \color{#333333} - 1)(\color{#20A900} u \color{#333333} - 1) = 0$

$\rightarrow \color{#20A900} u \color{#333333} = 1$

$\textbf{\text{STEP 4: }}$ $\text{Solve for x}$

$\color{#20A900} u \color{#333333} = 1 \Rightarrow x^2 = 1$

$\sqrt{x^2} = \sqrt{1} \Rightarrow \boxed{ x = \pm \space 1}$

Then you can solve for the corresponding $y$ values by plugging each $x$ value into $\text{eq1}$ and $\text{eq2}$ , which will give the solutions as:

$\boxed{(1,1) \text{ and } (-1,-1)}$

Subtract two times the second equation from the first equation.

$x^2 + y^2 - 2xy = 2 - 2(1) = 0$

This factors as $(x-y)^2 = 0$ . So all solutions to this system must have $x-y = 0$ and $\boxed{x=y}$ .

Substituting this information into either original equation leads to $y^2 = 1$ , which has solutions $\boxed{y = \pm 1}$ .

Therefore, the system has two ordered pair solutions: $(1,1)$ and $(-1,-1)$ .

Go to desmos.com and input the equations.

I had no paper at this moment.😑

Should have put a fi at the empty set b.t.w. .

(x^2)-1=1-(y^2). (x+1)(x-1)=1-(1/x^2)=[(x^2)-1)]/x^2=(x+1)(x-1)/x^2. (x^2)(x+1)(x-1)=(x+1)(x-1). x^2=1. x=+/-1. 1-1=1-y^2. y^2=1. y=+/-1. Answer x=+/-1 and y=+/-1. Verify. 1+1=2. (1)(1)=1 and (-1)*(-1)=1.

This is a cool application of AM-GM!

By AM-GM, we have $2=x^2+y^2\ge 2xy=2$ which means that the equality case holds!

So, $x^2=y^2$ but when $x=-y$ , there is no real solution.

So, $x=y$ which means $\boxed{x=y=\pm 1}$ .

$xy = 1 => x^2y^2 = 1 => x^2= \dfrac{1}{y^2}$

Substituting the above result in the first equation,

$y²+\dfrac{1}{y²} - 2=0 \\ \therefore y²+\dfrac{1}{y²}- 2\cdot y\cdot\dfrac{1}{y}= 0\\ \therefore \left(y-\dfrac{1}{y}\right)^2=0\\ \therefore y- \dfrac{1}{y}=0 \therefore y²= 1 => y= ±1$ So when $y= 1, x=1$ and $y=-1, x= -1 \hspace{6pt} [\because xy= 1]\\$

Hence the required points are $(1,1)$ and $(-1,-1)$