solve using quad equation

[justan4cat]

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.

 

[soroban]

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?

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

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


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

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


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

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



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

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

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


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


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


\(\displaystyle \text{Check}\)

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

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

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

. . . . We're golden!

 

[BigGlenntheHeavy]

\(\displaystyle Work \ Problem \ Formula:\)

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

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

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

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

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

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

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

\(\displaystyle Hence, Rick's \ rate \ is \ \frac{1}{18}.\)

\(\displaystyle I'll \ leave \ the \ check \ up \ to \ you.\)

 

[justan4cat]

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)

 

[BigGlenntheHeavy]

\(\displaystyle You \ got \ it.\)