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 solution:  

0.5x  + 0.8y - 0.44 = 0        ........(1)

          0.8x +  0.6y - 0.5 = 0        .........(2)

Here ,

a₁ = 0.5, b₁ = 0.8, c₁ = - 0.44

a₂ = 0.8,  b₂ = 0.6,  c₂ = - 0.5

a1b2 - a2b1 = 0.5(0.6) - 0.8 x 0.8

                      = - 0.3 - 0.64  = -0.34 ≈ 0.

Thus, the system has a unique solutions.

Consider,

We have 

0.8 x 0.5 = -0.4 , 0.6 x -0.44 = 0.264, 

-0.44 x 0.8 = -0.352,  -0.5 x 0.5 = -0.25

0.5 x 0.6 = -0.3, 0.8 x 0.8 = 0.64

⇒      

 ⇒

 ⇒ 

x = 0.4

⇒

y =  0.3
 

Hence the solution is ( 0.4, 0.3 ).

 solution:  

9x  + 12y - 72 = 0        ........(1)

          60x - 33y - 141 = 0        .........(2)

Here ,

a₁ = 9, b₁ = 12, c₁ = - 72

a₂ = 60,  b₂ = 33,  c₂ = - 141

a1b2 - a2b1 = 9(-33) - 60 x 12

                      = - 297 - 720  = -1017 ≈ 0.

Thus, the system has a unique solutions.

Consider,

We have 

12 x -141 = -1692 , -33 x -72 = 2376, 

-72 x 60 = -4320,  -141 x 9 = -1269

9 x -33 = -297, 60 x 12 = 720

⇒      

 ⇒

 ⇒ 

x = 4

⇒

y =  3
 

Hence the solution is ( 4, 3 ).

 solution:  

$\frac{\mathrm{3x}}{2}$ - $\frac{\mathrm{5y}}{3}$ = -2

 9x - 10y = -12

9x  - 10y + 12 = 0        ........(1)

$\frac{\mathrm{x}}{3}$ + $\frac{\mathrm{y}}{2}$ = $\frac{13}{6}$

          3x +  2y - 13 = 0        .........(2)

Here ,

a₁ = 9, b₁ = -10, c₁ =  12

a₂ = 2,  b₂ = 3,  c₂ = - 13

a1b2 - a2b1 = (9x3) - (2 x 10)

                      =  27 + 20  = 47 ≈ 0.

Thus, the system has a unique solutions.

Consider,

We have 

-10 x -13 = 130 , 3 x 12 = 36, 

12 x 2 = 24,  13 x 9 = 117

9 x 3 = 27, 2 x 10 = 20

⇒      

 ⇒

 ⇒ 

x = 2

⇒

y =  3
 

Hence the solution is ( 2, 3 ).

 solution:  

1x  + 1.6y - 0.88 = 0        ........(1)

          1.6x +  1.2y - 1 = 0        .........(2)

Here ,

a₁ = 1, b₁ = 1.6, c₁ = - 0.88

a₂ = 1.6,  b₂ = 1.2,  c₂ = - 1

a1b2 - a2b1 = 1(1.2) - 1.6 x 1.6

                      = - 1.2 - 2.56  = -1.36 ≈ 0.

Thus, the system has a unique solutions.

Consider,

We have 

1.6 x 1 = -1.6 , 1.2 x -0.88 = 1.056, 

-0.88 x 1.6 = -1.408,  -1 x 1 = -1

1 x 1.2 = -1.2, 1.6 x 1.6 = 2.56

⇒      

 ⇒

 ⇒ 

x = 0.4

⇒

y =  0.3
 

Hence the solution is ( 0.4, 0.3 ).

 solution:  

3x  + 4y - 24 = 0        ........(1)

          20x - 11y - 47 = 0        .........(2)

Here ,

a₁ = 3, b₁ = 4, c₁ = - 24

a₂ = 20,  b₂ = 11,  c₂ = - 47

a1b2 - a2b1 = 3(-11) - 20 x 4

                      = - 33 - 80  = -113 ≈ 0.

Thus, the system has a unique solutions.

Consider,

We have 

4 x -47 = -188 , -11 x -24 = 264, 

-24 x 20 = -480,  -47 x 3 = -141

3 x -11 = -33, 20 x 4 = 80

⇒      

 ⇒

 ⇒ 

x = 4

⇒

y =  3
 

Hence the solution is ( 4, 3 ).

 solution:  

$\frac{\mathrm{9x}}{6}$ - $\frac{\mathrm{15y}}{9}$ = -6

 81x - 90y = -108

81x  - 90y + 108 = 0        ........(1)

$\frac{\mathrm{x}}{9}$ + $\frac{\mathrm{y}}{6}$ = $\frac{39}{18}$

          9x +  6y - 39 = 0        .........(2)

Here ,

a₁ = 81, b₁ = -90, c₁ =  108

a₂ = 6,  b₂ = 9,  c₂ = - 39

a1b2 - a2b1 = (81x9) - (6 x 90)

                      =  729 + 540  = 1269 ≈ 0.

Thus, the system has a unique solutions.

Consider,

We have 

-90 x -39 = 3510 , 9 x 108 = 972, 

108 x 6 = 648,  39 x 81 = 3159

81 x 9 = 729, 6 x 90 = 540

⇒      

 ⇒

 ⇒ 

x = 2

⇒

y =  3
 

Hence the solution is ( 2, 3 ).

 solution:  

1.5x  + 2.4y - 1.32 = 0        ........(1)

          2.4x +  1.8y - 1.5 = 0        .........(2)

Here ,

a₁ = 1.5, b₁ = 2.4, c₁ = - 1.32

a₂ = 2.4,  b₂ = 1.8,  c₂ = - 1.5

a1b2 - a2b1 = 1.5(1.8) - 2.4 x 2.4

                      = - 2.7 - 5.76  = -3.06 ≈ 0.

Thus, the system has a unique solutions.

Consider,

We have 

2.4 x 1.5 = -3.6 , 1.8 x -1.32 = 2.376, 

-1.32 x 2.4 = -3.168,  -1.5 x 1.5 = -2.25

1.5 x 1.8 = -2.7, 2.4 x 2.4 = 5.76

⇒      

 ⇒

 ⇒ 

x = 0.4

⇒

y =  0.3
 

Hence the solution is ( 0.4, 0.3 ).

 solution:  

6x  + 8y - 48 = 0        ........(1)

          40x - 22y - 94 = 0        .........(2)

Here ,

a₁ = 6, b₁ = 8, c₁ = - 48

a₂ = 40,  b₂ = 22,  c₂ = - 94

a1b2 - a2b1 = 6(-22) - 40 x 8

                      = - 132 - 320  = -452 ≈ 0.

Thus, the system has a unique solutions.

Consider,

We have 

8 x -94 = -752 , -22 x -48 = 1056, 

-48 x 40 = -1920,  -94 x 6 = -564

6 x -22 = -132, 40 x 8 = 320

⇒      

 ⇒

 ⇒ 

x = 4

⇒

y =  3
 

Hence the solution is ( 4, 3 ).

 solution:  

$\frac{\mathrm{6x}}{4}$ - $\frac{\mathrm{10y}}{6}$ = -4

 36x - 40y = -48

36x  - 40y + 48 = 0        ........(1)

$\frac{\mathrm{x}}{6}$ + $\frac{\mathrm{y}}{4}$ = $\frac{26}{12}$

          6x +  4y - 26 = 0        .........(2)

Here ,

a₁ = 36, b₁ = -40, c₁ =  48

a₂ = 4,  b₂ = 6,  c₂ = - 26

a1b2 - a2b1 = (36x6) - (4 x 40)

                      =  216 + 160  = 376 ≈ 0.

Thus, the system has a unique solutions.

Consider,

We have 

-40 x -26 = 1040 , 6 x 48 = 288, 

48 x 4 = 192,  26 x 36 = 936

36 x 6 = 216, 4 x 40 = 160

⇒      

 ⇒

 ⇒ 

x = 2

⇒

y =  3
 

Hence the solution is ( 2, 3 ).

 solution:  

0.5x  + 0.8y - 0.44 = 0        ........(1)

          0.8x +  0.6y - 0.5 = 0        .........(2)

Here ,

a₁ = 0.5, b₁ = 0.8, c₁ = - 0.44

a₂ = 0.8,  b₂ = 0.6,  c₂ = - 0.5

a1b2 - a2b1 = 0.5(0.6) - 0.8 x 0.8

                      = - 0.3 - 0.64  = -0.34 ≈ 0.

Thus, the system has a unique solutions.

Consider,

We have 

0.8 x 0.5 = -0.4 , 0.6 x -0.44 = 0.264, 

-0.44 x 0.8 = -0.352,  -0.5 x 0.5 = -0.25

0.5 x 0.6 = -0.3, 0.8 x 0.8 = 0.64

⇒      

 ⇒

 ⇒ 

x = 0.4

⇒

y =  0.3
 

Hence the solution is ( 0.4, 0.3 ).