solve system of equations.

Question

X²-4Y²+7=0

3X²+Y²=31

 

 

 

X+Y+1=0

X²+Y²+6Y-X=-5

Answer 1

System 1

Write it as

(1): X²-4Y²=-7
(2): 3X²+Y²=31

Next use Gaussian Elimination.

4*(2)+(1): 13X²=117 → X²=9 → X=±3.

(2)-3*(1): 13Y²=52 → Y²=4 →Y=±2.

 

The four solutions are (3,2), (3,-2), (-3,2), and (-3,-2).

 

System 2

Write the equations as follows by completing the square

(1): X = -Y-1

(2): X²+Y²+6Y-X=-5

Now plug (1) into (2): (-Y-1)²+Y²+6Y-(-Y-1)=-5 → 2Y²+9Y+7=0 → (2Y+7)(Y+1)=0 → Y = -1 or -7/2

Plug this into (1) to get

Y = -1 → X=0 and Y = -7/2 → X = 5/2

 

The solutions are (0,-1) and (5/2,-7/2).

Answer 2

x² - 4y² + 7 = 0     [Eq. 1]

3x² + y² = 31        [Eq. 2]

 

Multiply first equation by -3

 

-3x² + 12y² -21 = 0

 3x² + y² = 31

 

Add the two equations to eliminate x²

 

13y² - 21 = 31 

 

13y2 = 52

 

y² = 4

 

from which y = ±2

 

x² - 4y² + 7 = 0

 

x² -4(4) + 7 = 0

 

x² - 16 = -7

 

x² = -7 + 16 = 9

 

from which x = ±3

 

The points at which both functions have the same value are (-3, 2), (-3, -2), (3, 2) and (3, -2)

 

x + y + 1 = 0

x²+y²+6y-x=-5

 

Solve the first equation for x

 

x = -y - 1 or -(y + 1)

x² = -(y + 1)² = -y² -2y -1

 

Substitute these values for x and x² into the second equation:

 

-y² -2y -1 + y² +6y - (-y -1) = -5

 

The y² terms cancel out

 

-2y - 1 + 6y + y + 1 =-5 

 

y(6 - 2 + 1) =-5

 

5y = -5

 

y = -1

 

But x + y + 1 = 0

 

so that x = 0 and y = -1 is one of the solutions.  There is another, however.

 

Let's complete the square of the original second equation:

 

x²+y²+6y-x=-5

 

x² - x + y² + 6y = -5

 

[x - (1/2)]² + (y + 3)² = -5 + (1/4) + 9 = (17/4)

 

which is the equation of a circle with center at [(1/2), -3] which intersects the line y = -x - 1 at

 

x = (5/2), y = (-7/2) which is the second and final solution.

 

 

 

 

 

 

 

 

Answer 3

   x² - 4y² + 7 = 0  ..... (1)
 3x² + y² = 31      ..... (2) 

Let's multiply (2) by 4 and add two equations: 

     x² - 4y² = - 7  
+
  12x² + 4y² = 124  

  13x² +  0  =  117
     x² = 117 / 13
    x² = 9
    x = ± 3

Lets plug in values of "x" into (1)

9 - 4y² + 7 = 0
- 4y² = - 16
y² = 4
y = ± 2

The answer is:

{(-3, 2), (-3, -2), (3, 2), (3, -2)}

==========

(1)  x + y + 1 = 0
(2)  x^2 + y^2 + 6y - x = -5

Let's solve (1) for "y"
y = - x - 1

Now let's plug "- (x + 1)" into (2)

x^2 + (x + 1)^2 - 6(x + 1) - x + 5 = 0

After open parentheses and combine like terms, we will have

2x^2 - 5x = 0
x(2x - 5) = 0
x₁ = 0
x₂ = 2.5
 
Finally, let's substitute "x" in (1) by 0 and 2.5

0 + y + 1 = 0 -----> y₁ = - 1

2.5 + y + 1 = 0 ----> y₂ = - 3.5

The answer is: 
{(0, - 1), (2.5, - 3.5)}

Answer 4

±

 

   X² - 4 Y ² = -7   (1)

 

   3X² +Y² = 31    (2)

 

   Method of elimination

 

   12 X² + 4Y² = 124         / multiply (2) by 4.

        X²- 4Y²  = -7

     13 X² = 117

 

       X² = 9        

 

       9 - 4 Y²= -7

 

         - 4 y² = -16

 

            Y²= 4

 

         X = ± 3             Y =± 2