Inter. Algebra: together, they complete job in 6 hrs....

H

Helen

Junior Member
Joined
Oct 28, 2007
Messages
106
Problem, round to nearest tenth.
Working together, Rick and Juanita can complete a job in 6 hr.
It would take Rick 9 hr longer than Juanita to do the job alone.
How long would it take Juanita alone?
I first made the table and tried factoring.
It couldn't be solved.
I then used the quadratic formula but I keep coming up with 15 hr.
I know that I am doing something wrong, can you help?
 
Re: Intermediate Algebra

Helen said:
Problem, round to nearest tenth.
Working together, Rick and Juanita can complete a job in 6 hr.
It would take Rick 9 hr longer than Juanita to do the job alone.
How long would it take Juanita alone?
I first made the table and tried factoring.
It couldn't be solved.
I then used the quadratic formula but I keep coming up with 15 hr.
I know that I am doing something wrong, can you help?
Click to expand...

Let j = hours for Juanita to do the job alone

In 1 hour, Juanita does 1/j of the job. If Juanita works for 6 hours, she does 6/j of the job.

Now, Rick takes 9 hours longer than Juanita to do the job by himself. So,

j + 9 = hours for Rick to do the job alone.

In one hour, Rick does 1 / (j + 9) of the job. In 6 hours, Rick does 6 / (j + 9) of the job.

We know that if they work together for 6 hours, the whole job is finished.

part done by Juanita in 6 hours + part done by Rick in 6 hours = whole job

(6 / j) + [ 6 / (j + 9) ] = 1

Ok...now the ball is in your court, Helen....solve the equation for j. Once you have the value for j, you can also find out the time for Rick.
 
mrspi,
Thank you, I will try to solve it for j. I appreciate your help and time. Helen
 
mrspi,
I tried solving it for j and came up with the answer of 3hr.
Did I do it right? Helen
 
Helen said:
mrspi,
I tried solving it for j and came up with the answer of 3hr.
Did I do it right? Helen
Click to expand...

Are you saying j = 3

Does it satisfy the equation

6/j + 6/(j+9) =1

If it does - you got the correct value of 'j'.

If not - then there is something 'wrong' in here...
 
Subhotosh Kahn,
I think that 3 hr does satisfy the equation. Helen
Thank you for your help.
 
Helen, if you substitute 3 for j, you have this:

(6 / 3) + [ 6 / (3 + 9) ] = 1


2 + (1/2) = 1

NOT true.

Multiply both sides of the original equation by the common denominator of the fractions to get rid of the denominators:

j(j + 9) * (6 / j) + j(j + 9)*(6 / (j + 9)) = j(j + 9)*1

Each denominator divides into the multiplier. This is what you'll have left:

6(j + 9) + 6(j) = j(j + 9)

Take it from there and see what you get.
 
Helen said:
Subhotosh Kahn,
I think that 3 hr does satisfy the equation. <--- Does it? Can you please show us how?

Helen
Thank you for your help.
Click to expand...
 
Subhotosh Kahn,
My answer of 3 hr. was wrong. I believe that I changed it to 9 hr. Thank you for the help. Helen
 
You must log in or register to reply here.
Top