Posted by jordan on Thursday, October 15, 2009 at 1:21pm.
Yesterday Lucy walked 2 hours and 1/2 jogged hour and covered 6.25 miles. Today she walked for 3 hours and jogged for 1 hour and covered 10.25 miles. Assuming a constant walking rate and a constant jogging rate, how fast did she walk and how fast did she jog? Define two variables, write a system of equations, and solve to find the walking rate and the jogging rate. Justify your answer by showing how you solved the problem.
w walking j jogging
2w + .5j= 6.25 > 3(2w+.5j)= 3(6.25)
3w + 1j= 10.25 > 2(3w+1j)= 2(10.25
6w + 1.5j= 18.75
6w  2j= 20.50

.5j=1.75
j=.3
2w + .5(.3)= 6.25
2w + 15= 6.25
2w= 6.10 w= 3.05
check:
2(3.05) + .5(.3)= 6.25 (correct)
3(3.05) + 1(.3)= 9.45(wrong!!) it should be 10.25.

11th grade  bobpursley, Thursday, October 15, 2009 at 1:22pm
since when is 1.75/5 = .3?

11th grade  jordan, Thursday, October 15, 2009 at 1:26pm
.5/1.75= .2857...or .3

11th grade  bobpursley, Thursday, October 15, 2009 at 1:28pm
Lord, Jordan.
you have a*x=b
then x= b/a NOT a/b

11th grade  bobpursley, Thursday, October 15, 2009 at 1:29pm
I get 1.75/.5=3.5
check that.

11th grade  jordan, Thursday, October 15, 2009 at 1:29pm
wow haha im lost...
6w + 1.5j= 18.75
6w  2j= 20.50

.5j=1.75
j=.3
? why would i multiply?

11th grade  bobpursley, Thursday, October 15, 2009 at 1:30pm
Jordan, this is pretty basic.
.5 j= 1.75
divide both sides by .5
j= 1.75/.5

11th grade  jordan, Thursday, October 15, 2009 at 1:32pm
But, if i did get 3.5 that would mess up getting 6.25 or 10.25.
2(3.05) + .5(.3)= 6.25correct.
3(3.05) + 1(.3)= 9.45... this is the one where i went wrong. i went wrong.

11th grade  bobpursley, Thursday, October 15, 2009 at 1:36pm
You have a lot of problems. YOu solved j as 3.5 so now solve for w, you first answer is WRONG.
2w + .5(3.5)= 6.25 solve for w.

11th grade  bobpursley, Thursday, October 15, 2009 at 1:37pm
Jordan, if you are in ALG II, I recommend a tutor before you get lost.
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