Algebra Help?

Question

SOLVE: a rectanular piece of sheet metal is 3 inch longer than as it is wide. If the length and width are both increased by 2 inch the area increases by 34 inch. What R the original dimensions of the sheet metal?

SOLVE: a rectangular piece of sheet metal is 3 inch longer than as it is wide. If the length and width are both increased by 2 inch the area increases by 34 inch. What R the original dimensions of the sheet metal?

Answer 1

a=b+3
(a+2)(b+2)=34+a*b

solve the system

Answer 2

L=W+3 and A=W L substitute for L
W(W+3) =A and
(W+2) (W+3+2)= A+34 substitute for A
(W+2) (W+2+3) =W(W+3) +34
W^2+7W+10= W^2+3W+34
4W=24, W=24/4=6 therefore : W=6 and L=9
check:
6(9) = 54
8(11) = 88 ; 88-54= 34, ok

Answer 3

x is the width of metal bar.
originally, A = (x + 3) x
After increaed it by 2", A + 34 = (x + 2) (x +5)
( x +3) x + 34 = (x + 2) (x +5)
Solve it, you will get x = 6
So, original dimension is 6" x 9".

Answer 4

Just have to set up the equations. It really is just transposing the question into equation form.

L=W+3 (a rect piece is 3" longer than wide)
(L+2)(W+2) = (L*W)+34 (Original Area w/ increased L & W by 2 = Area +34)

Answer 5

area or rectangle : length * width

let x : width of rectangle in inches
x + 3 : length of rectangle in inches

x * (x + 3) + 34 = (x + 2) * (x + 3 + 2)
x^2 + 3x + 34 = (x + 2) * (x + 5)
x^2 + 3x + 34 = x^2 + 7x + 10
34 = 4x + 10
24 = 4x
6 = x ; original width of rectangle is 6 inches
x + 3 = 6 + 3 = 9; original length of rectangle is 9 inches

Answer 6

l=3+w
lw=A ; (3+w)*w=A
(l+2)*(w+2)=A+34=> lw+2*w+2*l+4=A+34
A+2(w+l)+4=A+34
w+l=30/2
w+l=15
l=w+3=> w+w+3=15=>w=6; l=9=> A=54

Answer 7

x=6 therefore original dimensions 6 by 9