Simultaneous equation problem?

Question

A rectangle is 5 cm longer than it is wide. If the length and width are increased by 2cm each, the area is increased by 50 cm^2. Find the area of the original rectangle.

And explain how you did it please?

Answer 1

let original breadth be x cm.
so loriginal length is (x+5) cm.
so original area = length*breadth = x(x+5)= (x^2+5x)cm^2
so increased length = x+5+2= (x+7)
increased breadth = x+2
increased area =(x+2)(x+7)= (x^2 + 9x +14)cm^2
so difference in area :
(x^2 +9x+14) - (x^2 +5x)=50
x^2+9x+14-x^2-5x=50
4x+14=50
4x=36
x=9
x+5=14

so original length=14 and breadth = 9
so original area = 14 * 9 = 126 cm^2
Best of Luck!

Answer 2

Suppose x is the length and y is the width of the original rectangle

if the length and width are increased by 2cm each,
then the increased area is
2*(x+y)+2*2

so 2*(x+y+2)=50
x+y+2=25
x+y=23
by the way x=y+5
hence, y+5+y=23
2y+5=23
2y=18
y=9
and x=9+5=14

so, the area of the original rectangle is 9*14=126cm^2

solution: 126cm^2

Answer 3

ok, here how you do it:

length=x width=y area=a
x=y+5, a=xy; therefore, xy=a
if: x+2(y+2)=xy + 50

just substitute:
(x+2)(y+2) = xy+50
[ (y+5)+2] [y+2] = (y+5)y + 50
[ 7+y ] [2+y] = 5y +2y + 50
14 + 7y +2y + 2y = 7y +50
14 +11y = 7y + 50
11y-7y =50 -14
4y = 36
y = 9

so: y is 9, therefore, width is 9 cm
length is 5 cm longer, so 5 + 9 cm is equal to 14 cm,that's length
now:
the area is length x width
area = 14 cm x 9 cm
area is 126 cm2

if you add 2 cm to 9, that is 11 cm
add 2cm to 14 cm, that is 16
so, 11 cm x 16 cm = 176 cm2

and that is: 176 is equal to 50 cm2 plus the original area 126 cm2

solved.

Answer 4

b = original breadth
b+5 = original length
b+2= final width
b+7=final length
b^2+5b=original area
b^2+9b+14=final area
4b+14=50
b=9
b+5=14
original area = 14*9=126

Answer 5

A = x(x + 5)
x(x + 5) + 50 = (x + 2)(x + 7)
x^2 + 5x + 50 = x^2 + 9x + 14
4x = 36
x = 9
x + 5 = 14
A = 126

11*16 = 176 = 126 + 50