Posted by **jordan** on Thursday, October 15, 2009 at 12:40pm.

Yesterday Lucy walked 2 hours and jogged 1/2 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.

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>> -.5j=-1.75

6w + 2j= -20.50 >> 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.

please tell me where im going wrong:)

- algebra -
**bobpursley**, Thursday, October 15, 2009 at 1:10pm
I am lost here:

6w + 1.5j= 18.75>> -.5j=-1.75

6w + 2j= -20.50 >> j=.3

I think you should have

6w + 1.5j= 18.75>>

-6w - 2j= -20.50 >>

adding them,

.5j= 1.75

j= .35

and w= you do it.

- messed up,,,redo. -
**jordan**, Thursday, October 15, 2009 at 1:13pm
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.

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