Maths problem simultaneous equations

Question

A rectangle has a perimeter of 42cm. Another rectangle has a length double that of the first and a width 1 third of that of the first. The perimeter of the second rectangle is 57cm. X and the dimensions. Write two equations containing x and y. Solve the equation and write the dimensions of the second triangle

Answer 1

   X = length and Y = width

 

  42 = 2*X + 2*Y = 2X +2Y 

   21 = X + Y   <-- divides everything by 2

 

 Y = 21-X  <-- solving for Y

 

 57 = 2(2X) + 2(1/3Y) = 4X + 2/3Y

 

57 = 4X + 2/3(21 - X)  <-- substituting the first equation

 

57     = 4X + 14 - 2/3 X   <-- distributes 2/3

 

57      = 10/3 X  + 14   <-- combines like terms

 

43 = 10/3 X    <-- subtracts 14 from both sides

 

129 = 10X   <-- multiplies both sides by 3

 

X = 129/10 <-- divides both sides by 10

 

Y = 21 - X = 21 - 129/10 = (210 - 129)/10 = 81/10

 

So the original rectangle has length X = 129/10 and width Y=81/10

The perimeter is (129+129+81+81)/10 = 420/10 = 42

 

So second rectangle, not triangle, has dimensions length = 258/10 and width 27/10.

It's perimeter is (258 + 258 + 27 + 27)/10 = 570/10 = 57