Solution sets in math..?

Question

What is the solution set?
xy=48
2x-y=-4

What is the solution set?
xy=9
x^2+y^2=82

What is the solution set?
4x^2-5y^2=-16
5x^2-2y^2=13

Answer 1

xy = 48    -(1)

2x - y = -4 -(2)

 Add y to each side for the second equation 2x - y = -4

2x - y + y = -4 + y

2x -0 = -4 + y

2x = y - 4

Add 4 to each side

2x + 4 = y - 4 + 4

2x + 4 = y - 0

2x + 4 = y

Recall : symmetric property a = b then b = a

y = 2x + 4

Substitue y = 2x + 4 in the first equation xy = 48

There fore x(2x + 4) = 48

Recall : Distributive property a(b + c) = ab + ac

2x² + 4x = 48

Divide each side by 2

x² + 2x = 48 / 2

x² + 2x = 24

Subtract 24  to each side

x² + 2x - 24 = 24 - 24

x² + 2x - 24 = 0

x² + 6x - 4x - 24 = 0

Take out common term x and -4 in the equation :x² + 6x - 4x - 24 = 0

x(x + 6) - 4(x + 6) = 0

Take out common term x + 6 in the equation :x(x + 6) - 4(x + 6) = 0

(x + 6)(x - 4) = 0

x + 6 = 0 or x - 4 = 0

x + 6 = 0

Subtract 6 from each side

x + 6 - 6 = - 6

x + 0 = - 6

x = - 6

Substitute x = - 6 in the second equation : 2x - y = - 4

2(-6) - y = - 4

-12 - y = - 4

Add 12 to each side

12 -12 - y = 12 - 4

0 - y = 8

- y = 8

Divide each side by negative one

y = -8

The solution is x = - 6 and y = - 8

or

x - 4 = 0

Add 4 to each side

x - 4 + 4 = 4

x - 0 = 4

x = 4

Substitute x = 4 in the second equation 2x - y = - 4

There fore 2(4) - y = - 4

Simplify

8 - y = - 4

Subtract 8 from each side

8 - y - 8 = - 4 - 8

0 - y = - 12

- y = - 12

Divide each side by negative one

y = 12

The second solution is x = 4 and y = 12

There fore the solution set for the equations :{(-6, -8) , (4 , 12)}.

Comments

this is not a correct method.
Please see the method
xy = 48
Substitute y=48/x in equation 2
2x - y = -4
2x - 48/x = -4
2x^2-48=-4x
2x^2+4x-48=0
Solve this equation to get x

Answer 2

xy = 9    -(1)

x² + y² = 82    -(2)

Recall : (a + b)² = a² + b² + 2ab

(x + y)² = x² + y² + 2xy

Substitute x² + y² = 82 and xy = 9

There fore (x + y)² = 82 + 2(9)

(x + y)² = 82 + 18

(x + y)² = 100

Take square root each side

x + y = ±10

x + y = 10 -(3) or x + y = -10 -(4)

Recall : (a - b)² = a² + b² -2ab

(x - y)² = x² + y² - 2xy

Substitute x² + y² = 82 and xy = 9

(x - y)² = 82 - 2(9)

(x - y)² = 82 - 18

(x - y)² = 64

Take square root each side

(x - y) = ± 8

x - y = 8    -(5) or x - y = -8    -(6)

Solve the  third and fifth equations

x + y = 10

x - y = 8

Add above equations ( third and fifth equations)

2x = 18

Divide each side by 2

2x / 2 = 18 / 2

x = 9

Substitute x = 9 in the third equation x + y = 10

9 + y = 10

Subtract 9 from each side

9 + y - 9 = 10 - 9

0 + y = 1

y = 1

The solution is x = 9 , y = 1 (Third and fifth equations)

Solve the third and sixth equations

x + y = 10

x - y = -8

Add the third and sixth equations

2x + 0 = 10 - 8

2x = 2

Divide each side by 2

2x / 2 = 2 / 2

x = 1

Substitute x = 1 in the third equation x + y = 10

1 + y = 10

Subtract from each side

1 + y - 1 = 10 - 1

y = 9

The solution is x = 1, y = 9 (Third and sixth equations)

Solve the fourth and fifth equations

x + y = -10

x - y = 8

Add Fourth and fifth equations

2x = -2

Divide each side by 2

x = -1

Substitute x = -1 in the fourth equation x + y = - 10

-1 + y = - 10

Add 1 to each side

0 + y = 1 - 10

y = -9

The solution of fourth and fifth equations are x = -1 , y = -9

Solve the fourth and sixth equations

x + y = -10

x - y = -8

Add fourth and sixth equations then

2x + 0 = - 18

2x = - 18

Divide each side by 2

x = - 9

Substitute x = -9 in the fourth equation x + y = -10

- 9 + y = - 10

Add 9 to each side

9 - 9 + y = 9 - 10

0 + y = - 1

y = - 1

The solution of the fourth and sixth equations are x = -9 , y = -1

The solution set of third , fourth , fifth and sixth equations : { (9, 1) , (1 , 9) , (-1 , - 9) , (- 9 , - 1) } .

 

Comments

This is also same method as commented above

Answer 3

4x² - 5y² = - 16 -(1)

5x² - 2y² = 13   -(2)

Multiply each side by 2 for the equation 4x² - 5y² = - 16

2(4x² - 5y²) = 2(- 16)

8x² - 10y² = -32    -(3)

Multiply each side by 5 for the equation 5x² - 2y² = 13

5(5x² - 2y²)  = 5(13)

25x² - 10y² = 65    -(4)

Subtract from third equation to fourth equation

- 17x² + 0 = - 97

- 17x² = - 97

Divide each side by negative one

17x² = 97

Divide each side by 17

x² = 97 / 17

Take square root each side

x = ±√(97 / 17)

Substitute x² = 97 / 17 in the equation 5x² - 2y² = 13

5(97 / 17) - 2y² = 13

485 / 17 - 2y² = 13

Multiply each side by 17

(485 / 17 - 12y²)17 = 13(17)

Recall : Distributive property a(b - c) = ab - ac

485 - 204y²  = 221

Subtract 485 from each side

485 - 204y² - 485 = 221 - 485

0 - 204y² = - 264

Divide each side by negative one

204y² = 264

Divide each side by 204

y² = 264 / 204

Simplify

y² = 22 / 17

Take square root each side

y = ±√(22 / 17)

The solutions of the equation set { ( √(97/17) , √(22/17) ),( √(97/17) ,-√(22/17) ), ( -√(97/17) , √(22/17) ) ,  ( (-√(97/17) , -√(22/17) )  }.

Comments

The solution of the system 4x² - 5y² = - 16  and 5x² - 2y² = 13 is

Answer 4

1) The equations are   and  .

• First equation

Divide each side by x.

  --------> (i)

From (i) substitute the y  value in

• The resulting equation is and solve for x  by using factors method.

Take out common term (x + 6).

Apply Zero Product Property :  For any real numbers a  and b , if ab  = 0, then either a  = 0, b  = 0, or both a  and b  equal zero.

• Substitute the x  values in x y = 48.

Solution

Answer 5

2) The equations are   and 

• First equation

Divide each side by x.

------> (i)

From (i) substitute the y  value in

• The resulting equation is and solve for x completing squre method.

Separate variables and constants aside by subtracting 81 to each side.

 To change the expression into a perfect square trinomial add (half the x² coefficient)² to each side of the expression

 Here x² coefficient = -82. so, (half the x² coefficient)² = (-82/2)²= 1681

Add 1681 to each side.

Take square root both sides.

Answer 6

Continuous...

• Substitute the x  values in x y = 9.

Solution

Answer 7

3) The equations are and

• Solve for x  in above two equations.

First equation 

-----> (i)

Second equation 

------> (ii)

• Solve for y .

From (i) & (ii)

Answer 8

Continuous....

• Substitute y  values in equation (i).

For

For

Solution

.