Solving Quadratic system of equations

Question

Solve this system of equations:

$$(1) \quad 0=-10x^2-9xy+50x-25y-7y^2+5$$ $$(2) \quad 0=-5x^2-17xy+25x+50y-14y^2+7$$

Shame on me but I'm failing to solve this system. I can't see a short (rather any) way to get $x,y$. Any hints for me?

Answer 1

HINT

if you double the second and subtract from the first, it eliinates $x^2$ term so you can solve for $x$ and back-substitute.

Answer 2

HINT.-A quite interesting method to solve this question is the Sylvester Resultant.

If you have $$f(x,y)=a_0x^2+a_1xy+a_2x+a_3y+a_4y^2+a_5=0\\g(x,y)= b_0x^2+b_1xy+b_2x+b_3y+b_4y^2+b_5=0$$ you can do $$f(x,y)=A_0x^2+A_1x+A_2=0\\g(x,y)=B_0x^2+ B_1x+B_2=0$$ where $$(A_0,A_1,A_2)=(a_0,\space \space a_1y+a_0,\space \space a_3y+a_4y^2+a_5)\\(B_0,B_1,B_2)=(b_0,\space \space b_1y+b_0,\space \space b_3y+b_4y^2+b_5) $$ A necessary and sufficient condition for the two equations in $x$ with coefficients $A_i,B_i$ have common roots is $$F_1(y)=\det\begin{pmatrix}A_0&A_1&A_2&0\\0&A_0&A_1&A_2\\B_0&B_1&B_2&0\\0&B_0&B_1&B_2\end{pmatrix}=0$$ Similarly, you can calculate $F_2(x)=0$ so the solutions of the given system are the common roots of $F_1(y)$ and $F_2(x)$.

Answer 3

This algorithm is for when the polynomial can not be decomposed. The algorithm is based on:

1- Using the curves of equations to see whether they cross.

2-finding the points the curves cross the axis of coordinates or extremums(right and let and maximum and minimum).

3- approximating the coordinates of intersections of curves by making a

third equation by try and error.

3- Checking the results in initial system and getting exact results.

I will try to explain this in following simple examples.

Exampl 1: Two hyperbola crossing each other:

x^2/2^2 -y^2/3^2=1

(y-1)^2/2^2- (x-2)^2/5^2=1 ` 1st equ.: y=0; x=2 and x=-2 are vertices (right and left extremums)of

hyperbola. y=3x/2 and y=-3x/2 are the assymtotes.

2nd equ.: x=2; y=2.83 and y=-0.83 are minimum of the part above x axis and maximum of the part below x axis respectively.

the intersections are between assymtotes of two functions. the interval of x is x <-2 and x>2 the one for y is y>3 and y<-1.

y=3x/2 and y=2x/5 gives average slope coefeciant m=(2/3 +2/5)/2 ~ 1 therfore the third equation can be y=x.

putting in 1st equation we get y=x= +2.68, x=-2,68.

putting in 2nd equation we get: y=x = 3,04 and x= - 1.42.

taking mean values of x and y and more approximation we get the solutions of system as follows:

(2.86, 3.06), (-2,05, 3.57), (2.86, -1.06), (-2.05, -1.57)

Example 2- Two cifcles:

x^2 + y^2 =5^2

(x-2) + (y-3)^2 =4^2

Extremums of 1at equation: (5, 0) , (-5,0) max: (0, +5), min:(0, -5)

and for second one is :(6, 2), (-2, 2), max:(2, 7), min:(2, -1)

therefore the intervals the solutions are within are:(-2

The third equation for 1st quarter of coordinate system is y= 2x/6 and for

second quarter is y=-4x putting these in 1st and second equations and comparing the results we finally get the solutions of system of equation as follows:

( x= 5.2, y= 0.86), (x=-1.2, y=4.77).

Example 3- Complicated form:

2x^2 +2y^2 +5xy -x +y -1 =0

-3x^2 -2y^2 +5xy +10x -8y -8 =0

Both equations can be reduced and solutios are:

(5/3, -1/3), (5/4, -1/8), (1/3, -5/3), (2/7, -11/7). which locate between the points the crves of equations cross the axis of coordinate system:

x=0 : -5 < y < -0.5

y=0 : -0.5 < x <2.33

Conclusion- The solutions of system of quadratic equations lay on intervals representing the extremums of functions or the intersection of curves and coordinate axises.

Now we try to use this conclusion to solve the question asked:

-10x^2 -9xy +50x -25y -7y^2 +5 =0

-5x^2 -17xy +25x +50y -14 y^2 +7 =0

Right extremum of 1st equation: ( 5.46, 0) and the left one: (-0.4, 0), top extremum: (0, 0.19), and the down one: (0, -3.76).

The above parameters for 2nd equation are respectively as follows:

(5.26, 0), (-0,26, 0), (0, 3.7), (0, -0.135)

There can not be a third equation but as can be seen the interval of right and left extremums of 2nd equation are within the interval of right and left extremums of 1st equation. The 2nd equation is increasing in 2nd quarter and is decreasing in first quarter so for the system of equation we have (-0.25 < x < 5.25)
Also the intervals of top and down extremums intersect that is we must have (-0.135

By try and error we find (x~ -0.1, y~ 0.16) and (x ~ 4.7, y ~ 0.14) as solutios of the system of equation.