Question 112577This question is from textbook Introductory Algebra
: 61. The length of a rectangle is three times the width. If the width were increased by 3 m while the length remained the same, the new rectangle would have an area of 30 m square. Find the length and width of the original rectangle.
64. Two cars left an intersection at the same time. One traveled west, and the other traveled 14 mi less, but to the south. How far apart were they then, if the distance between them was 16 mi more than the distance traveled south?
This question is from textbook Introductory Algebra
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! The length of a rectangle is three times the width. If the width were increased by 3 m while the length remained the same, the new rectangle would have an area of 30 m square. Find the length and width of the original rectangle.
:
Let x = width
And
3x = length
:
New rectangle
(x+3) = new width
and
3x = unchanged length
then the new area
3x(x+3) = 30
:
3x^2 + 9x = 30
:
3x^2 + 9x - 30 = 0; a quadratic equation
:
Simplify, divide equation by 3:
x^2 + 3x - 10
:
Factors to:
(x + 5)(x - 2) = 0
Positive solution:
x = 2 is the original width
3(2) = 6; length
:
Check solution by finding the area
6(2+3) = 30
:
:
64. Two cars left an intersection at the same time. One traveled west, and the other traveled 14 mi less, but to the south. How far apart were they then, if the distance between them was 16 mi more than the distance traveled south?
:
Let x = dist traveled by the westbound car
then
(x-14) = dist traveled by the southbound car
:
It said,
"the distance between them was 16 mi more than the distance traveled south?"
From this we can say the distance between them, (which is the hypotenuse), is:
(x-14) + 16 = (x+2)
:
Using pythag a^2 + b^2 = c^2
:
x^2 + (x-14)^2 = (x+2)^2
:
x^2 + x^2 - 28x + 196 = x^2 + 4x + 4; FOILed (x-14)^2 and (x+2)^2
:
x^2 + x^2 - x^2 - 28x - 4x + 196 - 4 = 0, group like terms on the left
:
x^2 - 32x + 192 = 0; a quadratic equation
:
With a little diligence we can factor this:
(x - 8)(x - 24) = 0
:
Our two solutions
x = +8 and x = + 24
:
Obviously x = 24 is the solution since one car when 14 mi less.
:
The distance apart (the hypotenuse): 24 + 2 = 26 mi
:
Check our solution using a^2 + b^2 = c^2; a=24, b=10 (14 mi less) and c = 26
:
24^2 + 10^2 = 26^2
576 + 100 = 676; confirms our solution.
:
I tried to explain things as we went along here. Did it make sense to you? Any questions?
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