Question 460464
Let x is the length of a rectangle
Let y is the width of a rectangle
The perimeter of a rectangle is {{{2x+2y}}}cm, than {{{2x+2y=28}}}
The area of the rectangle is {{{xy}}}{{{cm^2}}}
If the length were decreased by 2cm, so the length would be {{{(x-2)}}}cm and the width increased by 2cm, so the width would be {{{(y+2)}}}cm, the area is {{{(x-2)(y+2)}}}{{{cm^2}}} and the area would be increased by 8cm, so {{{xy+8=(x-2)(y+2)}}}.
System of equations
{{{system(2x+2y=28,xy+8=(x-2)(y+2))}}}
{{{system(x+y=14,xy+8=xy+2x-2y-4)}}}
{{{system(x+y=14,2x-2y=12)}}}
{{{system(y=14-x,x-y=6)}}}
{{{system(y=14-x,x-(14-x)=6)}}}
{{{system(y=14-x,2x=20)}}}
{{{system(x=10,y=14-10)}}}
{{{system(x=10,y=4)}}}
x=10cm is the length of a rectangle
y=4cm is the width of a rectangle
.
Check
{{{system(2*10+2*4=28,10*4+8=(10-2)(4+2))}}}
{{{system(28=28,48=48)}}}-correct
.
Answer: x=10cm is the length of a rectangle, y=4cm is the width of a rectangle