Step-by-step explanation:

Given A rectangle has the same area as another, WHOSE length is 6 m more and breadth is 4 m LESS. It has also the same area as the third, whose length is 8 m more and breadth 5 cm less. Find the length and breadth of the original rectangle

Let the length and breadth of the original rectangle be l and b respectively.

ACCORDING to question rectangle with 6m is more so it will be l + 6 and breadth is 4 m less, so b – 4

Now area of rectangle is l X b. so we can write as

l x b = (l + 6)(b – 4)

      = lb + 6b – 4l – 24

  We get -4l + 6b – 24 = 0

   Or 4l – 6b + 24 = 0

Again according to question area of original rectangle is equal to area of third rectangle with length 8 m more and breadth 5 m less

Now the equation will be  

l x b = (l + 8)(b – 5)

l x b = lb + 8b – 5l – 40

      = lb – 5l + 8b – 40

We get -5l + 8b – 40 = 0

Or 5l – 8b + 40 = 0

Now we have simultaneous equations to solve,  

  4l – 6b + 24 = 0 multiply by 5

  5l – 8b + 40 = 0 multiply by 4

We get

   20l – 30b + 120 = 0

   20l – 32b + 160 = 0

            2b = 40

         So b = 20 m

Substituting b = 20 we get

  4l – 6b + 24 = 0

 4l – 6(20) + 24 = 0

 4l – 120 + 24 = 0

 4l – 96 = 0

 4l = 96

So l = 24 m

So length and breadth of rectangle will be 24 m and 20 m

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