Question

ok guys I need your help on this one>>>
the dimensions of a rectangle are such that its length is 7inches more than it's width. If the length were doubled, and if the width were decreased by 3 inches. The area would be increased by 168 inches sq. What are the length and width of the rectangle? thanks guys

Answer 1

put x = width, then x +7 is the length and area is
\[x(x+7)\]

Answer 2

ok then it all goes downhill from there lol

Answer 3

double the length, incraese the width by 3, get
\[(x+3)\times 2(x+7)\]

Answer 4

and so you know that
\[x(x+7)+168=2(x+3)(x+7)\]

Answer 5

now we solve the quadratic equation

Answer 6

ok x^2 +7x +168=2x+6+x+7

Answer 7

except i get a really ugly solution, so maybe i made a mistake

Answer 8

lol well join the club

Answer 9

oh because it said "decreased" if only i could read

Answer 10

\[x(x+7)+168=2(x-3)(x+7)\]

Answer 11

im getting all kinds of wierd answers on this one I cant match it on my check equation

Answer 12

\[x (x+7)+168 = 2 x^2+8 x-42\]
\[x^2+7 x+168 = 2 x^2+8 x-42\]
\[x^2+x-210 = 0\]
\[(x+15)(x-14)=0\]
\[x=14\]

Answer 13

are you for certain certain?????????????????????????

Answer 14

yes it matches mine. thanks