Question

solve the following set of equations8v-3u,=5uv,6v-5u=-2uv​

Answer 1

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$u = \frac{22}{31} \:v = \frac{11}{23}$

Divide the given equations by uv,

8v-3u=5uv=>

$\frac{8}{u} - \frac{3}{v} = 5...(1)$

6v-5u=-2uv=>

$\frac{6}{u} - \frac{5}{v} = - 2...(2)$

$Assume \: \frac{1}{u} = x \: and \frac{1}{v} = y$

$Put \: the \: values \: of \: \frac{1}{u} \: and \: \frac{1}{v} \: in \: (1) \:$

$and \: (2)$

$8x - 3y = 5...(3)$

$6x - 5y = - 2...(4)$

Solving equations (3) and (4) we get x and y

Multiply (3) with 5 and (4) with 3 to equate the coefficients of y.

$40x - 15y = 25$

$18x - 15y = - 6$

40x-15y -(18x-15y)=25-(-6)

=>40x-15y-18x+15y =31

=> 40x-18x-15y+15y =31

=>22x+0y=31

=>22x=31

=>x=$31/22$

$=>x = \frac{31}{22} = > u = \frac{22}{31}$

$substituting \: x \: in \: (3) \: we \: get \: y \: =$

$\frac{23}{11} = > v = \frac{11}{23}$

$so \: u \: = \frac{22}{31} \: and \: v = \frac{11}{23}$

Answer 2

Answer: 22/31 and 11/23

Step-by-step explanation:

Given,

8v - 3u = 5uv   ...............i)

And,

6v - 5u = -2uv  .............ii)

Dividing with uv on both sides in both equations.

We have,

8/u - 3/v = 5    .............iii)

And,

6/u - 5/v = -2  .............iv)

Let us assume, 1/u = x and 1/v = y

Then equation iii) and iv) becomes as follows respectively,

8x - 3y = 5      ...............v)  

6x - 5y = -2    ................vi)

Now, Multiplying equations v) by 5 and vi) by 3

We have,

40x - 15y =25

18x - 15y = -6

Subtracting both equation,

40x - 15y - (18x - 15y) = 25 - (-6)

40x - 15y - 18x + 15y = 25 + 6

22x = 31

x = 31/22

Putting x = 31/22 in eq. v)

8×(31/22) - 3y = 5

(31×4)/11 - 3y = 5

124/11 - 3y = 5

3y = 124/11 - 5

3y = (124-55)/11

3y = 69/11

y = 69/(11×3)

y = 23/11

u = 1/x = 22/31

v = 1/y = 11/23