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This system is not linear in x and y as the product of variables x and y exists in the equations.

Also, note that if x = 0, then y=0 and vice versa, So,(0,0) is a solution for the system and any other solution would have both x ≠ 0.

Consider, 10x - 5y = xy.

Dividing by xy,

$\frac{10}{\mathrm{y}}$ -   $\frac{5}{\mathrm{x}}$ = 1

- $\frac{5}{\mathrm{x}}$ + $\frac{10}{\mathrm{y}}$ = 1 ...... (1)

Consider, 7x - 3y = 7xy

Dividing by xy.

- $\frac{3}{\mathrm{x}}$ + $\frac{7}{\mathrm{y}}$ = 7  ..... (2)

Put $\frac{1}{\mathrm{x}}$ = a , $\frac{1}{\mathrm{y}}$ = b in (1) and (2), we get

-5a + 10b = 1   ......    (3)

-3a + 7b = 7  ...... (4)

Now, (3) and (4) is a linear system in a and b.

Solving (3) and (4), we get

(3) x 3         -15a + 30b = 3

(4) x 5         -15a + 35b = 35

                                -5b = -32

                                  b = $\frac{32}{5}$

                                  b = 6.40.

Substituting b = 6.40 in (3).

-5a + 10(6.40) = 1.

-5a + 64 = 1

-5a = 1 - 64

-5a = -63

          a = $\frac{63}{5}$

          a = 12.60

Hence a = 12.60  ⇒     x = $\frac{1}{12.60}$

           b = 6.40   ⇒     y = $\frac{1}{6.40}$ .

The system has two solutions (0,0) and ( $\frac{1}{12.60}$ , $\frac{1}{6.40}$ ).

This system is not linear in x and y as the product of variables x and y exists in the equations.

Also, note that if x = 0, then y=0 and vice versa, So,(0,0) is a solution for the system and any other solution would have both x ≠ 0.

Consider, 11x - 7y = xy.

Dividing by xy,

$\frac{11}{\mathrm{y}}$ -   $\frac{7}{\mathrm{x}}$ = 1

- $\frac{7}{\mathrm{x}}$ + $\frac{11}{\mathrm{y}}$ = 1 ...... (1)

Consider, 9x - 4y = 6xy

Dividing by xy.

- $\frac{4}{\mathrm{x}}$ + $\frac{9}{\mathrm{y}}$ = 6  ..... (2)

Put $\frac{1}{\mathrm{x}}$ = a , $\frac{1}{\mathrm{y}}$ = b in (1) and (2), we get

-7a + 11b = 1   ......    (3)

-4a + 9b = 6  ...... (4)

Now, (3) and (4) is a linear system in a and b.

Solving (3) and (4), we get

(3) x 4         -28a + 44b = 4

(4) x 7         -28a + 63b = 42

                                -19b = -38

                                  b = $\frac{38}{19}$

                                  b = 2.

Substituting b = 2 in (3).

-7a + 11(2) = 1.

-7a + 22 = 1

-7a = 1 - 22

-7a = -21

          a = $\frac{21}{7}$

          a = 3

Hence a = 3  ⇒     x = $\frac{1}{3}$

           b = 2   ⇒     y = $\frac{1}{2}$ .

The system has two solutions (0,0) and ( $\frac{1}{3}$ , $\frac{1}{2}$ ).

This system is not linear in x and y as the product of variables x and y exists in the equations.

Also, note that if x = 0, then y=0 and vice versa, So,(0,0) is a solution for the system and any other solution would have both x ≠ 0.

Consider, 11x - 7y = xy.

Dividing by xy,

$\frac{11}{\mathrm{y}}$ -   $\frac{7}{\mathrm{x}}$ = 1

- $\frac{7}{\mathrm{x}}$ + $\frac{11}{\mathrm{y}}$ = 1 ...... (1)

Consider, 8x - 4y = 7xy

Dividing by xy.

- $\frac{4}{\mathrm{x}}$ + $\frac{8}{\mathrm{y}}$ = 7  ..... (2)

Put $\frac{1}{\mathrm{x}}$ = a , $\frac{1}{\mathrm{y}}$ = b in (1) and (2), we get

-7a + 11b = 1   ......    (3)

-4a + 8b = 7  ...... (4)

Now, (3) and (4) is a linear system in a and b.

Solving (3) and (4), we get

(3) x 4         -28a + 44b = 4

(4) x 7         -28a + 56b = 49

                                -12b = -45

                                  b = $\frac{45}{12}$

                                  b = 3.75.

Substituting b = 3.75 in (3).

-7a + 11(3.75) = 1.

-7a + 41.25 = 1

-7a = 1 - 41.25

-7a = -40.25

          a = $\frac{40.25}{7}$

          a = 5.75

Hence a = 5.75  ⇒     x = $\frac{1}{5.75}$

           b = 3.75   ⇒     y = $\frac{1}{3.75}$ .

The system has two solutions (0,0) and ( $\frac{1}{5.75}$ , $\frac{1}{3.75}$ ).

This system is not linear in x and y as the product of variables x and y exists in the equations.

Also, note that if x = 0, then y=0 and vice versa, So,(0,0) is a solution for the system and any other solution would have both x ≠ 0.

Consider, 11x - 7y = xy.

Dividing by xy,

$\frac{11}{\mathrm{y}}$ -   $\frac{7}{\mathrm{x}}$ = 1

- $\frac{7}{\mathrm{x}}$ + $\frac{11}{\mathrm{y}}$ = 1 ...... (1)

Consider, 9x - 4y = 6xy

Dividing by xy.

- $\frac{4}{\mathrm{x}}$ + $\frac{9}{\mathrm{y}}$ = 6  ..... (2)

Put $\frac{1}{\mathrm{x}}$ = a , $\frac{1}{\mathrm{y}}$ = b in (1) and (2), we get

-7a + 11b = 1   ......    (3)

-4a + 9b = 6  ...... (4)

Now, (3) and (4) is a linear system in a and b.

Solving (3) and (4), we get

(3) x 4         -28a + 44b = 4

(4) x 7         -28a + 63b = 42

                                -19b = -38

                                  b = $\frac{38}{19}$

                                  b = 2.

Substituting b = 2 in (3).

-7a + 11(2) = 1.

-7a + 22 = 1

-7a = 1 - 22

-7a = -21

          a = $\frac{21}{7}$

          a = 3

Hence a = 3  ⇒     x = $\frac{1}{3}$

           b = 2   ⇒     y = $\frac{1}{2}$ .

The system has two solutions (0,0) and ( $\frac{1}{3}$ , $\frac{1}{2}$ ).

This system is not linear in x and y as the product of variables x and y exists in the equations.

Also, note that if x = 0, then y=0 and vice versa, So,(0,0) is a solution for the system and any other solution would have both x ≠ 0.

Consider, 9x - 7y = xy.

Dividing by xy,

$\frac{9}{\mathrm{y}}$ -   $\frac{7}{\mathrm{x}}$ = 1

- $\frac{7}{\mathrm{x}}$ + $\frac{9}{\mathrm{y}}$ = 1 ...... (1)

Consider, 8x - 4y = 6xy

Dividing by xy.

- $\frac{4}{\mathrm{x}}$ + $\frac{8}{\mathrm{y}}$ = 6  ..... (2)

Put $\frac{1}{\mathrm{x}}$ = a , $\frac{1}{\mathrm{y}}$ = b in (1) and (2), we get

-7a + 9b = 1   ......    (3)

-4a + 8b = 6  ...... (4)

Now, (3) and (4) is a linear system in a and b.

Solving (3) and (4), we get

(3) x 4         -28a + 36b = 4

(4) x 7         -28a + 56b = 42

                                -20b = -38

                                  b = $\frac{38}{20}$

                                  b = 1.90.

Substituting b = 1.90 in (3).

-7a + 9(1.90) = 1.

-7a + 17.1 = 1

-7a = 1 - 17.1

-7a = -16.1

          a = $\frac{16.1}{7}$

          a = 2.30

Hence a = 2.30  ⇒     x = $\frac{1}{2.30}$

           b = 1.90   ⇒     y = $\frac{1}{1.90}$ .

The system has two solutions (0,0) and ( $\frac{1}{2.30}$ , $\frac{1}{1.90}$ ).

This system is not linear in x and y as the product of variables x and y exists in the equations.

Also, note that if x = 0, then y=0 and vice versa, So,(0,0) is a solution for the system and any other solution would have both x ≠ 0.

Consider, 11x - 7y = xy.

Dividing by xy,

$\frac{11}{\mathrm{y}}$ -   $\frac{7}{\mathrm{x}}$ = 1

- $\frac{7}{\mathrm{x}}$ + $\frac{11}{\mathrm{y}}$ = 1 ...... (1)

Consider, 9x - 4y = 6xy

Dividing by xy.

- $\frac{4}{\mathrm{x}}$ + $\frac{9}{\mathrm{y}}$ = 6  ..... (2)

Put $\frac{1}{\mathrm{x}}$ = a , $\frac{1}{\mathrm{y}}$ = b in (1) and (2), we get

-7a + 11b = 1   ......    (3)

-4a + 9b = 6  ...... (4)

Now, (3) and (4) is a linear system in a and b.

Solving (3) and (4), we get

(3) x 4         -28a + 44b = 4

(4) x 7         -28a + 63b = 42

                                -19b = -38

                                  b = $\frac{38}{19}$

                                  b = 2.

Substituting b = 2 in (3).

-7a + 11(2) = 1.

-7a + 22 = 1

-7a = 1 - 22

-7a = -21

          a = $\frac{21}{7}$

          a = 3

Hence a = 3  ⇒     x = $\frac{1}{3}$

           b = 2   ⇒     y = $\frac{1}{2}$ .

The system has two solutions (0,0) and ( $\frac{1}{3}$ , $\frac{1}{2}$ ).

This system is not linear in x and y as the product of variables x and y exists in the equations.

Also, note that if x = 0, then y=0 and vice versa, So,(0,0) is a solution for the system and any other solution would have both x ≠ 0.

Consider, 10x - 5y = xy.

Dividing by xy,

$\frac{10}{\mathrm{y}}$ -   $\frac{5}{\mathrm{x}}$ = 1

- $\frac{5}{\mathrm{x}}$ + $\frac{10}{\mathrm{y}}$ = 1 ...... (1)

Consider, 9x - 3y = 7xy

Dividing by xy.

- $\frac{3}{\mathrm{x}}$ + $\frac{9}{\mathrm{y}}$ = 7  ..... (2)

Put $\frac{1}{\mathrm{x}}$ = a , $\frac{1}{\mathrm{y}}$ = b in (1) and (2), we get

-5a + 10b = 1   ......    (3)

-3a + 9b = 7  ...... (4)

Now, (3) and (4) is a linear system in a and b.

Solving (3) and (4), we get

(3) x 3         -15a + 30b = 3

(4) x 5         -15a + 45b = 35

                                -15b = -32

                                  b = $\frac{32}{15}$

                                  b = 2.13.

Substituting b = 2.13 in (3).

-5a + 10(2.13) = 1.

-5a + 21.3 = 1

-5a = 1 - 21.3

-5a = -20.3

          a = $\frac{20.3}{5}$

          a = 4.06

Hence a = 4.06  ⇒     x = $\frac{1}{4.06}$

           b = 2.13   ⇒     y = $\frac{1}{2.13}$ .

The system has two solutions (0,0) and ( $\frac{1}{4.06}$ , $\frac{1}{2.13}$ ).

This system is not linear in x and y as the product of variables x and y exists in the equations.

Also, note that if x = 0, then y=0 and vice versa, So,(0,0) is a solution for the system and any other solution would have both x ≠ 0.

Consider, 11x - 7y = xy.

Dividing by xy,

$\frac{11}{\mathrm{y}}$ -   $\frac{7}{\mathrm{x}}$ = 1

- $\frac{7}{\mathrm{x}}$ + $\frac{11}{\mathrm{y}}$ = 1 ...... (1)

Consider, 9x - 4y = 6xy

Dividing by xy.

- $\frac{4}{\mathrm{x}}$ + $\frac{9}{\mathrm{y}}$ = 6  ..... (2)

Put $\frac{1}{\mathrm{x}}$ = a , $\frac{1}{\mathrm{y}}$ = b in (1) and (2), we get

-7a + 11b = 1   ......    (3)

-4a + 9b = 6  ...... (4)

Now, (3) and (4) is a linear system in a and b.

Solving (3) and (4), we get

(3) x 4         -28a + 44b = 4

(4) x 7         -28a + 63b = 42

                                -19b = -38

                                  b = $\frac{38}{19}$

                                  b = 2.

Substituting b = 2 in (3).

-7a + 11(2) = 1.

-7a + 22 = 1

-7a = 1 - 22

-7a = -21

          a = $\frac{21}{7}$

          a = 3

Hence a = 3  ⇒     x = $\frac{1}{3}$

           b = 2   ⇒     y = $\frac{1}{2}$ .

The system has two solutions (0,0) and ( $\frac{1}{3}$ , $\frac{1}{2}$ ).

This system is not linear in x and y as the product of variables x and y exists in the equations.

Also, note that if x = 0, then y=0 and vice versa, So,(0,0) is a solution for the system and any other solution would have both x ≠ 0.

Consider, 10x - 6y = xy.

Dividing by xy,

$\frac{10}{\mathrm{y}}$ -   $\frac{6}{\mathrm{x}}$ = 1

- $\frac{6}{\mathrm{x}}$ + $\frac{10}{\mathrm{y}}$ = 1 ...... (1)

Consider, 8x - 4y = 7xy

Dividing by xy.

- $\frac{4}{\mathrm{x}}$ + $\frac{8}{\mathrm{y}}$ = 7  ..... (2)

Put $\frac{1}{\mathrm{x}}$ = a , $\frac{1}{\mathrm{y}}$ = b in (1) and (2), we get

-6a + 10b = 1   ......    (3)

-4a + 8b = 7  ...... (4)

Now, (3) and (4) is a linear system in a and b.

Solving (3) and (4), we get

(3) x 4         -24a + 40b = 4

(4) x 6         -24a + 48b = 42

                                -8b = -38

                                  b = $\frac{38}{8}$

                                  b = 4.75.

Substituting b = 4.75 in (3).

-6a + 10(4.75) = 1.

-6a + 47.5 = 1

-6a = 1 - 47.5

-6a = -46.5

          a = $\frac{46.5}{6}$

          a = 7.75

Hence a = 7.75  ⇒     x = $\frac{1}{7.75}$

           b = 4.75   ⇒     y = $\frac{1}{4.75}$ .

The system has two solutions (0,0) and ( $\frac{1}{7.75}$ , $\frac{1}{4.75}$ ).

This system is not linear in x and y as the product of variables x and y exists in the equations.

Also, note that if x = 0, then y=0 and vice versa, So,(0,0) is a solution for the system and any other solution would have both x ≠ 0.

Consider, 11x - 7y = xy.

Dividing by xy,

$\frac{11}{\mathrm{y}}$ -   $\frac{7}{\mathrm{x}}$ = 1

- $\frac{7}{\mathrm{x}}$ + $\frac{11}{\mathrm{y}}$ = 1 ...... (1)

Consider, 9x - 4y = 6xy

Dividing by xy.

- $\frac{4}{\mathrm{x}}$ + $\frac{9}{\mathrm{y}}$ = 6  ..... (2)

Put $\frac{1}{\mathrm{x}}$ = a , $\frac{1}{\mathrm{y}}$ = b in (1) and (2), we get

-7a + 11b = 1   ......    (3)

-4a + 9b = 6  ...... (4)

Now, (3) and (4) is a linear system in a and b.

Solving (3) and (4), we get

(3) x 4         -28a + 44b = 4

(4) x 7         -28a + 63b = 42

                                -19b = -38

                                  b = $\frac{38}{19}$

                                  b = 2.

Substituting b = 2 in (3).

-7a + 11(2) = 1.

-7a + 22 = 1

-7a = 1 - 22

-7a = -21

          a = $\frac{21}{7}$

          a = 3

Hence a = 3  ⇒     x = $\frac{1}{3}$

           b = 2   ⇒     y = $\frac{1}{2}$ .

The system has two solutions (0,0) and ( $\frac{1}{3}$ , $\frac{1}{2}$ ).