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Solving equation systems part four



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  • Rating:General - Intended for all ages.
  • Publish Time:2021-12-02 08:57

Solving equation systems part four



Solve {6x + 6y = 19, 3xy = 5}.

6x + 6y = 19

3xy = 5

6y = 19 - 6x

y = (19 - 6x) / 6

3x(19 - 6x) / 6 = 5

6x² - 19x + 10 = 0

(2x - 5)(3x - 2) = 0

x = 5 / 2; 2 / 3

y = {19 - 6(5 / 2)} / 6; {19 - 6(2 / 3)} / 6

y = 2 / 3; 5 / 2

solutions as (2 / 3, 5 / 2) and (5 / 2, 2 / 3)

6x + 6y = 19 and 3xy = 5 cutting each other at (2 / 3, 5 / 2) and (5 / 2, 2 / 3)

 

Solve {3x + 3y = 7, 9xy = 10}.

3x + 3y = 7

9xy = 10

3y = 7 - 3x

y = (7 - 3x) / 3

9x(7 - 3x) / 3 = 10

9x² - 21x + 10 = 0

(3x - 2)(3x - 5) = 0

x = 2 / 3; 5 / 3

y = {7 - 3(2 / 3)} / 3; {7 - 3(5 / 3)} / 3

y = 5 / 3; 2 / 3

solutions as (2 / 3, 5 / 3) and (5 / 3, 2 / 3)

3x + 3y = 7 and 9xy = 10 cutting each other at (2 / 3, 5 / 3) and (5 / 3, 2 / 3)

 

Solve {10x + 10y = 31, 2xy = 3}.

10x + 10y = 31

2xy = 3

10y = 31 - 10x

y = (31 - 10x) / 10

2x(31 - 10x) / 10 = 3

10x² - 31x + 15 = 0

(2x - 5)(5x - 3) = 0

x = 5 / 2; 3 / 5

y = {31 - 10(5 / 2)} / 10; {31 - 10(3 / 5)} / 10

y = 3 / 5; 5 / 2

solutions as (3 / 5, 5 / 2) and (5 / 2, 3 / 5)

10x + 10y = 31 and 2xy = 3 cutting each other at (3 / 5, 5 / 2) and (5 / 2, 3 / 5)

 

Solve {10x + 10y = 19, 5xy = 3}.

10x + 10y = 19

5xy = 3

10y = 19 - 10x

y = (19 - 10x) / 10

5x(19 - 10x) / 10 = 3

10x² - 19x + 6 = 0

(2x - 3)(5x - 2) = 0

x = 3 / 2; 2 / 5

y = {19 - 10(3 / 2)} / 10; {19 - 10(2 / 5)} / 10

y = 2 / 5; 3 / 2

solutions as (2 / 5, 3 / 2) and (3 / 2, 2 / 5)

10x + 10y = 19 and 5xy = 3 cutting each other at (2 / 5, 3 / 2) and (3 / 2, 2 / 5)

 

Solve {10x + 10y = 21, 10xy = 9}.

10x + 10y = 21

10xy = 9

10y = 21 - 10x

y = (21 - 10x) / 10

10x(21 - 10x) / 10 = 9

10x² - 21x + 9 = 0

(2x - 3)(5x - 3) = 0

x = 3 / 2; 3 / 5

y = {10 - 21(3 / 2)} / 10; {10 - 21(3 / 5)} / 10

y = 3 / 5; 3 / 2

solutions as (3 / 5, 3 / 2) and (3 / 2, 3 / 5)

10x + 10y = 21 and 10xy = 9 cutting each other at (3 / 5, 3 / 2) and (3 / 2, 3 / 5)

 

Solve {15x + 15y = 31, 3xy = 2}.

15x + 15y = 31

3xy = 2

15y = 31 - 15x

y = (31 - 15x) / 15

3x(31 - 15x) / 15 = 2

15x² - 31x + 10 = 0

(3x - 5)(5x - 2) = 0

x = 5 / 3; 2 / 5

y = {31 - 15(5 / 3)} / 15; {31 - 15(2 / 5)} / 15

y = 2 / 5; 5 / 3

solutions as (2 / 5, 5 / 3) and (5 / 3, 2 / 5)

15x + 15y = 31 and 3xy = 2 cutting each other at (2 / 5, 5 / 3) and (5 / 3, 2 / 5)


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0 user (users) favorited this work
  • View Count:454
  • Rating:General - Intended for all ages.
  • Publish Time:2021-12-02 08:57