if 6/u + 3v = 2 and 2/u – 9/v = 4

multiply the first by 2 so they are equal:

12/u + 6/v = 2/u – 9/v

12/u – 2/u + 6/v + 9/v = 0

1/u (12-2) + 1/v (6+9) = 0

where x=1/u and y=1/v,

x(10)+y(15) = 0

15y = -10x

y = x(-10/15)

thus y = (-2/3)x

now we know that

6/u + 3/v = 2 and y = (-2/3)x where y = 1/v

therefore we will use y = 1/v substitution to give:

6/u + (1/v)(3) = 2

and then substitute 1/v for -2/3x where x=1/u

6/u + [(-2/3)(1/u)](3) = 2

6/u + (-2/3u)(3) = 2

6/u – 2/u = 2

therefore 1/u (4) = 2 and thus u = 2

now we will use u=2 to solve for v where

1/v = (-2/3)x and x= 1/2

1/v = (-2/3)(1/2) -1/3

and so v=-3

Moving on: Question 35:

2/u – 3/v + 2 = 0 and 4/u + 3/v + 1 = 0

using the same technique,

2/u – 3/v + 2 = 4/u + 3/v + 1

and so we now have

2/u – 3/v + 2 – 4/u – 3/v – 1 = 0

thus 1/u (-2) – 1/v (6) + 1 = 0

$and so: -2x+1 = 6y and y= -(1/3)x + 1/6

and again using the same technique, we know that

4/u + 3/v + 1 = 0 and y = 1/v

and 1/v = -(1/3)x + 1/6 where x = 1/u

thus we will use the substitution:

4/u + (1/v)(3) + 1 = 0

and 4/u + 3 [-(1/3)x + 1/6)] + 1 = 0

and thus 4/u + 3[-(1/3)(1/u) + 1/6)] + 1 = 0

which gives us:

4/u + 3 [(-1/3u) + 1/6] + 1 = 0

and 4/u – 1/u + 1/2 + 1 = 0

1/u (3) = -(1.5)

and thus we see that u = -2

we will now use u=-2 to solve for v where

4/u + 3/v + 1 = 0

4/(-2) + 3/v + 1 = 0

-2 + 3/v + 1 = 0

3/v = 1

and thus we see that v = 3

The third equation is done the same way; I’d encourage you to try it out yourself using the method above. I thought instead of working it out for you, you might want to give it a shot yourself to see if you grasped the above method.