Need help with Algebra 2......please?

Question

1) A rectangular picture frame contains a 12” × 14” picture. What is the width of the frame if the total area of picture and frame is 288 square inches?

2) A metric print (picture) is square with sides of 15 cm. If it is to be enlarged so as to double its area, how much will the length of a side be increased?

3) A rectangle is 10 units longer than it is wide. When the length is decreased by 2-units and the width is increased by 3, the area is 141 square units. Find the original dimensions.

Answer 1

PROBLEM 1:
I assume the frame is the same added width on all sides.
The width will be: 12 + 2x
The height will be: 14 + 2x

Total area:
(12 + 2x)(14 + 2x) = 288

Multiply out:
168 + 28x +24x + 4x² = 288

Get everything on the left:
4x² + 52x + 168 - 288 = 0
4x² + 52 - 120 = 0

Divide both sides by 4:
x² + 13 - 30 = 0

Factor:
(x + 15)(x - 2) = 0
x = -15 or x = 2

The frame must have a positive width, so it is an additional 2 inches on every side. The final frame dimensions are 16" x 18".

PROBLEM 2:
15² = 225
(15 + x)² = 225 * 2
x² + 30x + 225 = 450
x² + 30x - 225 = 0
Using the quadratic equation:
x = [-30 +/- sqrt(900 - 4(-225)) ] / 2
x = -15 +/- sqrt(900 + 900)/2
x = -15 +/- sqrt(1800)/2
x = -15 +/- 30 sqrt(2)/2
x = -15 +/- 15 sqrt(2)
x ≈ 6.213 or x ≈ -36.213

But only the positive answer makes sense, so the width is approximately 6.2 cm.

PROBLEM 3:
Let w be the original width
Let w + 10 be the original length

New dimensions:
New length = w + 10 - 2 = w + 8
New width = w + 3

(w + 8)(w + 3) = 141
w² + 11w + 24 = 141
w² + 11w - 117 = 0

Using the quadratic formula again, you'll get:
w ≈ 6.635 units
w + 10 ≈ 16.635 units

Answer 2

288 = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 16 x 18 = size of frame + picture, so 16 x 18 accommodates picture of 12 x 14 with 2 inch border. This saves you getting no marks but I am not going to do any more. It seems there is no need to resort to algebra, at least with Q1 but as it is your homework and not mine and as I do not propose to consider Q2 and Q3 I do not comment.

Answer 3

1) The width of frame is 2inches. The length of frame = 18 inches and breadth = 16 inches (outer frame).

2) The length of the side would be increased by 6.21 units

3) The original dimensions are length = 16.634 units and width (breadth) = 6.634 units.

Hope this helps u..

Answer 4

1)
288 = 12*14 + 2x(12) + 2x(14) + 4x^2
4x^2 + 52x -120 = 0
x = 2

2)
2(15^2) = (15+x)^2
450=225+30x+x^2
x^2+30x-225=0
x= 6.2132

3)
141 = (10+x-2)(x+3)
141 = (8+x)(x+3)
141 = 24 + 11x + x^2
x=6.63466 (width)
x+10 =16.63466 (length)

Answer 5

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