Posted by Preston on .
1) A rectangle has an area of 297 square inches and a width which is three inches less than four times the length. Find the length and the width of the rectangle.
2) Let x represent the length of the shorter leg in a right triangle. If the longer leg is 1 more than the shorter leg and the hypotenuse has length 7, set up but do not solve an equation to find the two side lengths.
3) The product of two consecutive negative odd integers is 575. Find the integers
4)A rectangular canvas picture measures 13 inches by 7 inches. The canvas is mounted inside a frame of uniform width, increasing the total area covered by both canvas and frame to 160 square inches. Find the uniform width of the frame.
5) A rectangle has perimeter 30 feet and an area of 54 square feet. Find the dimensions of the rectangle assuming the width is smaller than the length.

Algebra Homework, HELP! 
Kuai k,
1. A = lw
w =4l3
4l3(l) = 297
4l^2 3l = 297
4l^2 3l  297 =0
(4l+33)(l9)= 0
l = 9
w =33
2. a^2 + b^2 = c^2
short leg = x
Long leg = x+1
x^2 + (x+1)^2 = 7^2
x^2 +x^2 + x + x + 1 = 49
2x^2 +2x + 1 = 49
3. x, x +2
x (x+2) = 575
x^2+ 2x = 575
x^2 + 2x 575 =0
(x 23)(x + 25) = 0
x =23
23,25 and 23, 25 
Algebra Homework, HELP! 
Bosnian,
1 )
W = 4 L  3
A = L * W = 297
L * ( 4 L  3 ) = 297
4 L ^ 2  13 L = 297 Subtract 297 to both sides
4 L ^ 2  3 L  297 = 297  297
4 L ^ 2  3 L  297 = 0
Solutions :
L =  33 / 4
and
L = 9
The length can't be negative so L = 9 in
W = 4 L  3
W = 4 * 9  3
W = 36  3
W = 33 in
2 )
b = the longer leg = x + 1
c = hypotenuse = sqrt [ x ^ 2 + b ^ 2 ] = 7
c = hypotenuse = sqrt [ x ^ 2 + ( x + 1 ) ^ 2 ] = 7
c = sqrt ( x ^ 2 + x ^ 2 + 2 x + 1 )
c = sqrt ( 2 x ^ 2 + 2 x + 1 ) = 7 Square both sides
2 x ^ 2 + 2 x + 1 = 49 Subtract 49 to both sides
2 x ^ 2 + 2 x + 1  49 = 49  49
2 x ^ 2 + 2 x  48 = 0
2 ( x ^ 2 + x  24 ) = 0 Divide both sides by 2
x ^ 2 + x  24 = 0
3 )
x = 1st number
x + 2 = 2nd number
x ( x + 2 ) = 575
x ^ 2 + 2 x = 575 Subtract 575 to both sides
x ^ 2 + 2 x  575 = 0
Solutions :
x =  25
and
x = 23
Numbers must be negative so :
1st number
x =  25
2nd number
x + 2 =  25 + 2 =  23
4 )
c = width of the frame
New width = 7 + 2 c
New length = 13 + 2 c
( 7 + 2 c ) ( 13 + 2 c ) = 160
13 * 7 + 13 * 2 c + 2c * 7 + 2 c * 2 c = 160
91 + 26 c + 14 c + 4 c ^ 2 = 160
4 c ^ 2 + 40 c + 91 = 160 Subtract 160 to both sides
4 c ^ 2 + 40 c + 91  160 = 160  160
4 c ^ 2 + 40 c  69 = 0
Solutions :
c =  23 / 2
and
c = 3 / 2
Width of the frame can't be negative so :
c = 3 / 2 in
( 7 + 2 * 3 / 2 ) ( 13 + 2 * 3 / 2 ) =
( 7 + 3 ) ( 13 + 3 ) = 10 * 16 = 160 in ^ 2
5 )
P = 2 W + 2 L = 2 ( W + L ) = 30
A = W * L = 54
W * L = 54 Divide both sides by W
L = 54 / W
W ^ 2  15 W + 54 = 0 Divide both sides by 2
W + 54 / W = 15 Multiply both sides by W
W * W + ( 54 / W ) * W = 15 * W
W ^ 2 + 54 = 15 W Subtract 15 w to both sides
W ^ 2 + 54  15 W = 15 W 15 W
W ^ 2 + 54  15 W = 15 W 15 W
Solutions :
W = 6 in
and
W = 9 in
L = 54 / W
When W = 6 in then L = 54 / 6 = 9 in
When W = 9 in then L = 54 / 9 = 6 in
Assuming the width is smaller than the length the dimensions are :
W = 6 in , L = 9 in 
Algebra Homework, HELP! 
Bosnian,
In Question 5 delete
W ^ 2  15 W + 54 = 0 Divide both sides by 2 
Algebra Homework, HELP! 
denise,
The length of the base measures
65ft less than twice the width. The perimeter of this base is 770
ft. What are the dimensions of the base?