Lesson Solving typical problems on systems of non-linear equations

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Solving typical problems on systems of non-linear equations

Problem 1

Solve the system of equations:
system%28x%5E2%2By%5E2+=+41%2C%0D%0Axy+=+20%29

Solution

Answer. There are 4 solutions: (x,y) = (5,4) or/and (x,y) = (4,5) or/and (x,y) = (-5,-4) or/and (x,y) = (-4,-5).

Solution

There are several ways to solve it. I will show you some.

1. Guessing:

     41 = 25 + 16  --->  (x,y) = (5,4)   or/and  (x,y) = (4,5)  or/and  (x,y) = (-5,-4)  or/and  (x,y) (-4,-5).

     Works good if the numbers are good and you have the feeling of numbers. 
     Otherwise is not so good.

2. Graphically:




Plots y = +/- (red+green) and y = 

    Works good. It is always good to make a plot.

3. By reducing the system to one equation and by introducing new variable:

 = 41,   (1)
xy = 20.        (2)

Express x =  from (2) and substitute it into (1). You will get

 +  = 41,  or, after multiplying both sides by 

 = 0.

Introduce new variable u = . The last equation becomes

 = 0.

Factor the left side. You will get

(u-25)*(u-16) = 0.


The roots are u = 25 and/or u=16, which gives you
              (x,y) = (+/-5,+/-4)  and/or  (+/-4,+/-5)
and all you need to do is to select the signs by an appropriate way.


Actually, it is a powerful algebraic method. You can combine it with the quadratic formula, using it instead of factoring.

4. By reducing the system to one equation and by applying factoring to this equation:

 = 41,   (1)
xy = 20.        (2)

Express x =  from (2) and substitute it into (1). You will get

 +  = 41,  or, after multiplying both sides by 

 = 0.

Factor the left side:

 = 0.

In this way you obtain the same set of solutions.


Works good, especially if you know the solution in advance.

May work not so good in more complicated cases.

Almost always, I prefer the method #3.

5. Using the trick

 = 41,   (1)
xy = 20.        (2)

Double the equation (2) and then add it to the equation (1). You ill get

{{x^2 + 2xy + y^2}}} = 41 + 2*20,   or

 = 81.

Take square root from both sides. You will get two equations.


a.  x+y =  9. Express x = 9-y and substitute it into equation (2).
              In this way you will get a quadratic equation which is easy to solve.


b.  x+y = -9. Express x = -9-y and substitute it into equation (2).
              Again, you will get a quadratic equation which is easy to solve.


The method works good if you know the trick, if the method is applicable and if you are ready to apply it.
Makes fantastic impression to your teacher.

Problem 2

Solve the system and graph the curves:
system%28x%5E2%2By%5E2+=+2%2C%0D%0Ax-y+=+4%29

Solution

 = 2,     (1)
x-y = 4          (2)

Express x = 4 + y from (2) and substitute it into (1). You will get

 = 2,  or

 = 2,  or

 = 0,  or

 = 0.

Apply the quadratic formula. You will get

 =  = .

There are no real solutions.

See the plot

Plots y =

, y =

(red+green) and y = x-4 (blue).

Systems-of-equations/1048629 (2016-09-18 01:42:59): Solve the system and graph the curves:
(x+1)^2+2(y-4)^2 = 12 ; and
y^2-8y = 4x-16

Problem 3

Solve the system and graph the curves:
system%28%28y-2%29%5E2+=+9%28x%2B2%29%2C%0D%0A9x%5E2%2B4y%5E2%2B18x-16y+=+0%29

Solution

(y-2)^2 = 9(x+2),               (1)
9x^2 + 4y^2 + 18x - 16y = 0.    (2)

Open parentheses in the first equation. Then the system is equivalent to

y^2 - 4y + 4 - 9(x+2) = 0,
9x^2 + 4y^2 + 18x - 16y = 0,   or

     -  9x + y^2  -  4y = 14,   (3)
9x^2 + 18x + 4y^2 - 16y =  0.   (4)

Multiply equation (3) by -4 (both sides), and then add to equation (3). You will get

9x^2 + 54x = -54,  or

x^2 + 6x + 6 = 0.

Apply the quadratic formula to find the roots. You will get

 =  =  = .


For  = , you have from (1)

(y-2)^2 = 9*(x+2) = 9*(-1 + sqrt(3)), and  = . 


For  = , you have from (1)

(y-2)^2 = 9*(x+2) = 9*(-1 - sqrt(3)), and the solution to (y-2) does not exist, since the right side  is negative.


So, there are two solutions (and, correspondingly, two intersection points)

 
     x =   and   = ,   = . 


To graph the curves, notice that

y - 2 = +/- ,                  ( from (1), and )
9*(x+1)^2 - 9 + 4*(y-2)^2 - 16 = 0.     ( from (2) )   or


y       = ,
(y-2)^2 = .


So the plot is

Line (1) (red + green) and Line (2).

Problem 4

Solve the system and graph the curves:

Solution

Open parentheses in the second equation. Then the system is equivalent to

9x^2 + 18x     + 4y^2 - 16y = 0,            (1)

 x^2 +  2x + 1 + 2y^2 - 16y + 32 = 12.      (2)


Multiply the equation (2) by 9 (both sides). You ill get

 9x^2 + 18x + 9 + 18y^2 - 144y + 288 = 108.    (3)

Now distract equation (1) from equation (3). You will get

14y^2 - 128y + 9 + 288 = 108,

14y^2 - 128y + 189 = 0.

Next solve this quadratic equation for "y" and then find the appropriate solutions for "x".

Problem 5

The distinct real numbers x and y satisfy x^2=33y+907 and y^2=33x+907. Find x and y.

Solution

x^2 = 33y + 907,   (1)
y^2 = 33x + 907.   (2)

Distract eq.(2) from eq.(1) (both sides). You will get

x^2 - y^2 = 33y - 33x,   or

(x-y)*(x+y) = -33(x-y).

Since the numbers are distinct, you can divide both sides of the last equation by (x-y). Then you get

x + y = -33.        (3)

Now express  x = -33-y from (3) and substitute it into (2). You will get

y^2 = 33(-33-y) + 907,   or

y^2 + 33y + 182 = 0.

Solve this quadratic equation using the quadratic formula.

 =  = ,

 = -7,   = -26.

Correspondingly, there are two solutions for x:

 = -33 - (-7) = -33 + 7 = -26,  and   = -33  (-26) = -33 + 26 = -7.

Answer.  The solutions are (x,y) = (-7,-26)  or/and (-26,-7).

Problem 6

Determine the value(s) of k such that the circle x^2+(y-6)^2 = 36 and the parabola x^2 = 4ky will intersect only at the origin.

Solution

 = 36,   (1)
 = 4ky           (2)


The circle  =  has the center at (x,y) = (0,6) and has the radius of 6. 
So, the circle has the y-axis x=0 as a diameter and as a symmetry line, passes through the origin and touches the x-axis.

Parabola  =  also passes through the origin; has the y-axis x=0 as its symmetry line, and touches the x-axis.

After these geometric considerations (that are useful but are not absolutely necessary) let solve the problem algebraically.

Based on (2), substitute 4ky instead of  into the equation (1). You will get

 = 36.

So, in this way you excluded "x" from the system and got a single equation for "y". Let us simplify it:

 = 36,  or

 = 0.  (1)

Now, the problem requires this equation (1) to have only one non-negative solution.
     (One solution is evident/obvious. It is y = 0.)

It implies that (4k-12) MUST be non-positive:  4k-12 <= 0.
OTHERWISE y =  would be the other non-negative solution to (1).

So, the solution to the problem is this inequality 4k-12 <= 0,  or, equivalently,  k <= 3   (3 = )


Answer.  k <= 3.


See an illustration below for k = 3, 2, and 4.


    


    The circle  =  (red + green) 
    and three parabolas x^2 = 4ky for k=3 (blue), k=2, and k=4. 



    


    The circle  =  (green) 
    and three parabolas x^2 = 4ky for k=3 (blue), k=2, and k=4.

Problem 7

When k is a constant, find k for which the length of perpendicular dropped from point (2,1) to line kx + y + 1 = 0 is sqrt%283%29

.

Solution

To make the solution easier, I re-formulate the problem in THIS EQUIVALENT WAY:

     Find the coefficient "k" such that the line  kx + y + 1 = 0 
 
     is tangent to the circle of the radius    centered at the point (2,1)

Solution

The standard form equation of the circle is

     +  = 3.           (1)


The equation of the line is

    y = -kx - 1.                      (2)



Substitute the equation (2) into equation (1), replacing y in the equation (1).

Doing this way, you will get then a single equation for only one unknown x
    
     +  = 3     (3)

or

     +  = 3.          (4)



The idea of the solution to the problem is to reduce equation (4) to standard form quadratic equation
and then to find the value of "k" from the condition that this equation has ONLY ONE real solution
(which is equivalent to the fact that the line(2) and the circle (1) have ONLY ONE common point).


So, we simplify equation (4) step by step

    x^2 - 4x + 4 + k^2x^2 + 4kx + 4 = 3

    x^2(1+k^2) - 4x(1-k) + 5 = 0.       (5)


Now we will find "k" from the condition that the discriminant of equation (5) is equal to zero.


The discriminant  d  is

    d = b^2 - 4ac =  -  =  - .


The condition d = 0  is

        = 

    4(1-k)^2       = 5(1+k^2)

    4 - 8k + 4k^2  = 5 + 5k^2

    k^2 + 8k + 1 = 0

     =  =  = .


So, there are 2 (two, TWO) values for k:   =   and   = .


ANSWER.  There are two solutions for "k" :   =   and   = .

Problem 8

Find the tangent lines to the parabola x^2 = 6y + 10 passing through the point (7,5), which lies outside the parabola.

Solution

An equation of the line passing through the point (7,5) is

    y-5 = m*(x-7),

where "m" is the slope coefficient, or

    y = mx - 7m +5.


    +----------------------------------------------------------------+
    |        The value of the slope "m" is unknown now,              |
    |    and the rest of the solution is to find the value of "m".   |
    +-----------------------------------------------------------==---+


Substitute this expression for y into the right side of the parabola formula

    x^2 = 6(mx - 7m +5) + 10.


You will get

    x^2 - 6mx + 42m - 40 = 0.    (1)


The line is tangent to the parabola if and only if the quadratic equation (1) has a unique real root.

It happens if and only if the discriminant of the equation is zero.


The discriminant is  d = b^2 - 4ac;  b= -6m;  a= 1;  c= 42m-40,  so

    d = 36m^2 - 4(42m-40) = 36m^2 - 168m + 160.


Therefore, to find "m", we need solve this quadratic equation

    36m^2 - 168m + 160 = 0.


Find its solutions, using the quadratic formula. They are

     =   and   = .


So, there are two tangent line through the given point to parabola.

Their equations are  y-5 =   and  y-5 = .


The equivalent forms of these equations are

    3y - 10x = -55  and  3y - 4x = -13.


The plots are shown in the Figure below.



    


    Parabola  y =  (red) and tangent lines 3y-4x = -13 (green)  and  3y-10x = -55 (blue)

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Use this file/link ALGEBRA-I - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-I.

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