Nonlinear Equations of Degree 2

Find the largest possible value of $\lvert x y \rvert$ for real numbers $x,y$ that satisfy the system of equations:

$\begin{aligned} \begin{cases} 2x^2-6xy+2y^2+43x+43y&=174\\ x^2+y^2+5x+5y&=30 \end{cases} \end{aligned}$

First, subtract the second equation from the first 2 times, and divide by 3:

$\begin{aligned} \begin{cases} -2xy+11(x+y)&=38\\ (x+y)^2-2xy+5(x+y)&=30 \end{cases} \end{aligned}$

Let $a=x+y$ and $b=xy$ . Substitute:

$\begin{aligned} \begin{cases} -2b+11a&=38\\ a^2-2b+5a&=30 \end{cases} \end{aligned}$

Subtracting the first equation from the second leaves $\begin{aligned} a^2-6a&=-8\\ a^2-6a+8&=0\\ (a-2)(a-4)&=0 \end{aligned}$

So $a = 2$ or $a = 4$ . If $a = 2$ , then substitution in $-2b+11a=38$ leaves $b = -8$ . Similarly, if $a = 4$ , then $b = 3$ . As $b = xy$ (which had to be maximized), the maximum value of $\lvert xy \rvert$ is (potentially) $\lvert -8 \rvert = 8$ . The only thing left to check is that there exist real numbers $x,y$ such that: $\begin{aligned} \begin{cases} x+y&=a=2\\ xy&=b=-8 \end{cases} \end{aligned}$

Both $x=4,y=-2$ and $x=-2,y=4$ satisfy these two equations. Therefore, the largest possible value of $\lvert xy \rvert$ is indeed $8$ .

We use the substitution A=x+y,B=xy. This gives us 2A2+43A−10B=174 and A2+5A−10B=30, so we may solve this to get 11A−2B=38. Substituting this back into the second equation, we get A2−6A+8=0⇒A=2 or A=4. Then, B=−8 or 3 respectively. So the maximum value of |B| is 8. Indeed, when A=4,B=−8, we have (x,y)=(4,−2) or (−2,4), and we may easily check that both of these would satisfy both equations. So the maximum of 8 is indeed attainable.

First of all, we have:

$\left\{ \begin{array}{l l l} 2(x^2+y^2) - 6xy + 43(x + y) & \quad = 174 & \quad (1) \\ (x^2 + y^2)+ 5(x + y) & \quad = 30 & \quad (2) \end{array} \right.$

Instead of working with all these terms (given the symmetry of x and y ), we may work with x+y and xy only. To take x² and y² off these equations, we may use the following relation:

$\begin{array}{l l} x^2 + y^2 = (x+y)^2 - 2xy & \quad (3) \end{array}$

To simplify the next equations, it will be used:

$\begin{array}{l l} x+y = z & \quad (4) \\ xy = w & \quad (5) \end{array}$

Using (4) and (5) on (3) :

$\begin{array}{l l} x^2 + y^2 = z^2 - 2w & \quad (6) \end{array}$

Using (4) , (5) and (6) on (2) , we will obtain that:

$\begin{array}{l l} \left(x^2 + y^2\right) + 5(x+y) = z^2 + 5z - 2w = 30 & \quad (7) \end{array}$

Using the previous procedure on (1) and rearranging some terms:

$2\left(x^2 + y^2\right) - 6xy + 43 (x+y) = 2z^2 + 43z - 10w = 174 \quad (8)$

Isolating w on (7) :

$\begin{array}{l l} w = \frac{1}{2}\left(z^2 + 5z - 30\right) & \quad (9) \end{array}$

Using (9) on (8) :

$\begin{array}{l l} 2z^2 + 43z - 5\left(z^2 + 5z - 30 \right) = 174 & \quad (10) \\ -3z^2 + 18z = -24 & \quad (11) \\ z^2 - 6z + 8 = 0 & \quad (12) \end{array}$

Solving (12) for z :

$z_1 = 2 \\ z_2 = 4$

Using the values of z on (9) , we will obtain the solutions for w :

$w_1 = -8 \\ w_2 = 3$

As we are looking for the largest possible value of |w| that solves the initial system, the final answer is $|w_1| = 8$ .

Working on w and z , we may also find all the possible answers for ( x , y ):

$S = \left\{ (-2;4), (1;3), (3;1), (4,-2) \right\}$

The equations are symmetrical in $x$ and $y$ , meaning that the equations remain unchanged when $x$ and $y$ are interchanged. The substitution $x = u + v$ , $y = u - v$ will help when solving symmetrical equations because this substitution will eliminate the terms of degree 1 in $v$ .

Using the substitution $x = u + v$ , $y = u - v$ in our system of equations and simplifying yields the new system $\begin{cases} v^2 = \frac{1}{5}(u^2 - 43 u + 87), \\ v^2 = -u^2 - 5u + 15. \end{cases}$ Setting these two equations equal to one another gives us $u^2 - 3u + 2 = 0,$ from which it follows that $u = 1$ or $u = 2$ . If $u = 1$ then $v = \pm 3$ , and if $u = 2$ then $v = \pm 1$ . Thus, the pair $(x,y) = (u+v, u-v)$ is one of $(4,-2)$ , $(-2,4)$ , $(3,1)$ , or $(1,3)$ . So the largest possible value of $|xy|$ is 8.

First let us write both equations in terms of xy The first equations boils down to : 2(x^2 + y^2) + 43(x+y) - 174 = 10xy

Completing the square for x^2 + y^2 by adding 4xy to both sides of the equation gives us:

2(x^2 + y^2 + 2xy) + 43(x+y) - 174 = 10xy + 4xy

which when simplified becomes

2(x+y)^2 + 43(x+y) - 174 = 10xy .................... (1)

As for the second equation,we can add 2xy to both sides to complete the square

x^2 + y^2 + 2xy + 5(x+y) - 30 = 2xy (x+y)^2 + 5(x+y) - 30 = 2xy ----------------------------------(2)

For the two equations ,Let us assign the variables A and V to x + y and xy respectively this gives us

2A^2 + 43A - 174 = 10V -------------(1) A^2 + 5A - 30 = 2V ----------------(2)

starting from 10V = 10V 5(A^2+5A - 30) = 2A^2 + 43A - 174 which when simplified becomes 3A^2 - 18A + 24 = 0 Dividing the whole thing by three gives A^2 - 6A + 8 = 0 When solving this quadratic equation by factoring method we get (A-2)(A-4)=0 There for A = 2 & A = 4 I then substituted these A values in both the equations and solved for V which enabled me to find three numbers for |V| and select the largest one which was 8

$2*x^2 + 2* y^2 +43x+43Y-6xy =174$ (given).

The above equation can be rewritten as, $( x^2 +x^2) + (y^2+y^2) + (5x+5x+33x) + (5y+5y+33y) - 6xy = 174$ . Rearranging the terms , we get $(x^2+y^2+5x+5y)+(x^2+y^2+5x+5y)+33x+33y -6xy = 174$ , which gives $30+30+33x+33y-6xy= 174$ , or that $11x+11y-2xy= 38$ . Thus, possible values of $x$ and $y$ are 4 and -2. Therefore largest possible value of $|xy|= 8$ .

[Latex edits - Calvin]

Let r = x + y, t = xy. The equations become (1) 2r^2 - 10t + 43r = 174, (2) r^2 - 2t + 5r = 30. Multiplying (2) by 5 to eliminate t, we have : r^2 - 6r + 8 = 0, whence r =2 and r = 4. Substituting each into either equation gives |t| = 3 or 8, so the answer is 8. Ed Gray