Phrenzidn
Answered

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equation problems with solutions

Sagot :

The area of a rectangular library room is 60 squared meter.  Find the perimeter of the room given that its length is 3 meter more than thrice its width.

Representation:
Let width = x meter
     length = 3x + 3  meter
     Area = 60 m²

Area of Rectangle = Length × Width

60 = (3x+3) (x)
60 = 3x² + 3x

Equate to zero (this a quadratic equation)

3x² + 3x - 60 = 0

Solve by factoring:
3 (x²  + x + 20) = 0
3 (x-4) (x+5) = 0
  x-4 = 0                          x + 5 = 0
  x = 4                             x = -5

Choose the positive value x = 4
Width: x = 4 meters
Length: 3x + 3 
              = 3(4) + 3
              = 12 + 3
              = 15 meters

Perimeter of a rectangle: 2 (L) + 2 (W)
P = 2 (15 meters) + 2 ( 4 meters)
P = 30 meters + 8 meters
P = 38 meters

The perimeter of the rectangular library room is 38 meters.


Bellalesseidn
   Two consecutive numbers have a sum of 47. What are the numbers?

Representation: Let x = 1st number
                        x + 1 = 2nd number

Equation: x + (x + 1) = 47

Solution: x + (x + 1) = 47
              x + x + 1 = 47
              2x + 1 = 47
              2x = 47 - 1
              2x    46
             ----  = ----   ⇒ x = 23 ; x + 1 = 24
              2       2 

Answer: The two consecutive numbers are 23 and 24.

Check: x + (x + 1) = 47
            23 + (23 + 1) = 47
            23 + 24 = 47
            47 = 47

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