solve using quad equation

justan4cat

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May 23, 2010
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My question is how to set this up. I think I can solve it from there, but I don't understand how to go from the word problem to the equation.

Working together, Rick and Juanita can complete a job in 6 hours. It would take Rick 9 hours longer than Juanita to do the job alone. How long would it take Juanita alone?
r=rick's hours and j=juanita's hours... then r+j=6 then r-j+9=6? But it still doesn't tell me how many hours it would take Juanita... Please help.
 
Hello, justan4cat!\

These "work" problems require special treatment.
There are a number of approaches.
I'll explain mine in baby-steps.


Working together, Rick and Juanita can complete a job in 6 hours.
It would take Rick 9 hours longer than Juanita to do the job alone.
How long would it take Juanita alone?

Let:   {R=Rick’s hours working alone.J=Juanita’s hours working alone.}\displaystyle \text{Let: }\;\begin{Bmatrix}R &=& \text{Rick's hours working alone.} \\ J &=& \text{Juanita's hours working alone.} \end{Bmatrix}

And we are told that: R=J+9\displaystyle \text{And we are told that: }\:R \:=\:J + 9


Juanita can do the job in J hours.\displaystyle \text{Juanita can do the job in }J\text{ hours.}
In one hour, she can do 1J of the job.\displaystyle \text{In one hour, she can do }\tfrac{1}{J}\text{ of the job.}
In 6 hours. she can do 6J of the job.\displaystyle \text{In 6 hours. she can do }\tfrac{6}{J}\text{ of the job.}

Rick can do the job in J+9 hours.\displaystyle \text{Rick can do the job in }J+9\text{ hours.}
In one hour, Rick can do 1j+9 of the job.\displaystyle \text{In one hour, Rick can do }\tfrac{1}{j+9}\text{ of the job.}
In 6 hours, he can do 6j+9 of the job.\displaystyle \text{In 6 hours, he can do }\tfrac{6}{j+9}\text{ of the job.}


Working together, they complete the job (1 job).\displaystyle \text{Working together, they complete the job (1 job).}

There is our equation!   6J+6J+9  =  1\displaystyle \text{There is our equation! }\;\boxed{\frac{6}{J}\,+\,\frac{6}{J\,+\,9} \;=\;1}



Multiply by J(J+9) ⁣:6(J+9)+6J  =  J(J+9)\displaystyle \text{Multiply by }J(J+9)\!:\quad 6(J\,+\,9)\,+\,6J \;=\;J(J\,+\,9)

And we have:   J23J54  =  0            (J9)(J+6)=0\displaystyle \text{And we have: }\;J^2\,-\,3J\,-\,54\;=\;0 \;\;\;\Rightarrow\;\;\;(J\,-\,9)(J\,+\,6) \:=\:0

. . Hence:   J  =  9,6\displaystyle \text{Hence: }\;J \;=\;9,\,-6


Therefore, Juanita takes 9 hours working alone.\displaystyle \text{Therefore, Juanita takes 9 hours working alone.}


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


Check\displaystyle \text{Check}

Juanita takes 9 hours working alone.\displaystyle \text{Juanita takes 9 hours working alone.}
In one hour, she does 19 of the job.\displaystyle \text{In one hour, she does }\tfrac{1}{9}\text{ of the job.}
In 6 hours, she does 69=23 of the job.\displaystyle \text{In 6 hours, she does }\tfrac{6}{9} \,=\,\tfrac{2}{3} \text{ of the job.}

Rick takes 18 hours working alone.\displaystyle \text{Rick takes 18 hours working alone.}
In one hour, Rick does 118 of the job.\displaystyle \text{In one hour, Rick does }\tfrac{1}{18}\text{ of the job.}
In 6 hours, he does 618=13 of the job.\displaystyle \text{In 6 hours, he does }\tfrac{6}{18} \,=\,\tfrac{1}{3}\text{ of the job.}

Working together for 6 hours,\displaystyle \text{Working together for 6 hours,}
. . they do: 23+13=1   (one job)\displaystyle \text{they do: }\tfrac{2}{3}\,+\,\tfrac{1}{3} \:=\:1\;\text{ (one job)}

. . . . We're golden!

 
Work Problem Formula:\displaystyle Work \ Problem \ Formula:

Rate of Working times Time = Fraction of Work Done\displaystyle Rate \ of \ Working \ times \ Time \ = \ Fraction \ of \ Work \ Done

Hence, let Ricks rate be 1x and Juanitas rate be 1y.\displaystyle Hence, \ let \ Rick's \ rate \ be \ \frac{1}{x} \ and \ Juanita's \ rate \ be \ \frac{1}{y}.

Then 6x+6y = 1, (unity      job is completed working together),\displaystyle Then \ \frac{6}{x}+\frac{6}{y} \ = \ 1, \ (unity \ \implies \ job \ is \ completed \ working \ together),

and y+9x = 1 implies x = y+9, Rick working alone.\displaystyle and \ \frac{y+9}{x} \ = \ 1 \ implies \ x \ = \ y+9, \ Rick \ working \ alone.

Hence, 6x+6yxy = 1      6x+6y = xy\displaystyle Hence, \ \frac{6x+6y}{xy} \ = \ 1 \ \implies \ 6x+6y \ = \ xy

6(y+9)+6y = y(y+9), substitution,      y23y54 = 0, (y9)(y+6) = 0\displaystyle 6(y+9)+6y \ = \ y(y+9), \ substitution, \ \implies \ y^2-3y-54 \ = \ 0, \ (y-9)(y+6) \ = \ 0

Disregarding the negative solution, we get Juanitas rate of 19.\displaystyle Disregarding \ the \ negative \ solution, \ we \ get \ Juanita's \ rate \ of \ \frac{1}{9}.

Hence,Ricks rate is 118.\displaystyle Hence, Rick's \ rate \ is \ \frac{1}{18}.

Ill leave the check up to you.\displaystyle I'll \ leave \ the \ check \ up \ to \ you.
 
Ok. I really got that with your explanation. Thanks... here's one I think I have... let me know if I'm right, please?

Translate into a pair of linear equations in two variables. Solve using substitution or elimination method.

A woman made a deposit of $205. If her deposit consisted of 73 bills all $1 bills or $5 bills, then how many $1 bills did she deposit?

x=$1 bills; y=$5 bills

x+y=73
1x+5y=205

Substitution method:
x=73-y
(73-y)+5y=205
73+4y+205
4y=132
y=33 so she deposited 33 of the $5 bills

x+5(33)=205
x+165=205
x=40 so she deposited 40 of the $1 bills

Check:
40+33=73 (check)
40+5(33)=205
40+165+205 (check)
 
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