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jseok
09-22-2006, 08:56 PM
can someone help me with a couple algebra word problems i dont get
I know how to do numbers 73 and 79 but need help on 74, 80, and 56. i put 73 and 79 because numbers 74 and 80 refers to them.

56. the width of a rectangle is 12 less than twice the length. if the length is increased by 3 and the width is doubled, the new perimeter is 6 less than 6 times the original length. what are the original dimensions?

73. A plumber charges 42 per hour for her time and 24 per hour for her assistant's time. on a certain job the assistant works alone for 2 hours doing prep work, then the plumber completes the job alone. if the job took 9 hours and the bill came to 240, how many hours did each work?

74. the plumber and assistnat of exercise 73 complete another job in which the assistant does the prep work alone and then the plumber completes the job alone. if the job takes 9 hours and the bill came to 240, how many hours did each work?

79. how long would it take someone jogging at 17kph to overtake someone walking at 7kph with a 3 hour head start.

80. repeat exercise 79 if the walker walks at the rate of 6kph for 3 hours and then walks at 7kph.

Unregistered
05-06-2008, 04:58 PM
let L=lenth of the original rectangle
then the width=2L-12

the dimensions of the new rectangle then would be L+3 and 2(2L-12) which simplifies to 4L-24

Then we need to come up with an equation to represent the perimeter which is 2L+2W. using the dimensions of the new rectangle we get the expression 2(L+3)+2(4L-24) and this then is equal to 6L-6 so we end up getting the equation below

2(L+3)+2(4L-24)=6L-6
2L+6+8L-48=6L-6 (using the distributive property)
10L-42=6L-6 (combining like terms)
10L-6L-42=6L-6-6L (solving for L)
4L=36
L=9

substituting this into the width expression we get the width is 6 so the dimensions of the original rectangle are 9 units by 6 units

Hope this helps, i don't have time to work out the other ones, hopefully someone else posts to help you out on those ones, Best of luck