Unknown rates. easy algebra please helpp?

[Cassandra123]

Arielle, Melissa and tyler working together can finish a job in two hours. Arielle and Melissa Wroking together can do the job in four hours. Melissa and Tyker working together can do it in three hours. how long does it take for each person to do the job alone? How long does it take Arielle and Tyler working together to do the job?

Ive had many problems with this problem
so far this is wat i got but im not sure what to do with it.

A*T= 1/2 per hour
A*M= 1/4 job per hour
M*T= 1/3 job per hour

 

[BigGlenntheHeavy]

\(\displaystyle One \ way: \ Rate \ times \ Time \ = \ Fraction \ of \ work \ done.\)

\(\displaystyle \frac{2}{a}+\frac{2}{m}+\frac{2}{t} \ = \ 1\)

\(\displaystyle \frac{4}{a}+\frac{4}{m} \ = \ 1\)

\(\displaystyle \frac{3}{t}+\frac{3}{m} \ = \ 1, \ a \ = \ Arielle, \ m \ = \ Melissa, \ and \ t \ = \ Tyler.\)

 

[soroban]

Hello, Cassandra123!

Another approach . . .

Arielle, Melissa and Tyler working together can finish a job in two hours.
Arielle and Melissa wooking together can do the job in four hours.
Melissa and Tyler working together can do it in three hours.
How long does it take for each person to do the job alone?
How long does it take Arielle and Tyler working together to do the job?

\(\displaystyle \text{Let }A\text{ = number of hours for Arielle to do the job alone.}\)
\(\displaystyle \text{In one hour, she can do }\tfrac{1}{A}\text{ of the job.}\)

\(\displaystyle \text{Let }M\text{ = number of hours for Melissa to do the job alone.}\)
\(\displaystyle \text{In one hour, she can do }\tfrac{1}{M}\text{ of the job.}\)

\(\displaystyle \text{Let }T\text{ = number of hours for Tyler to do the job alone.}\)
\(\displaystyle \text{In one hour, he can do }\tfrac{!}{T}\text{ of the job.}\)

\(\displaystyle \text{Working together for one hour, they can do }\tfrac{1}{A} + \tfrac{1}{M} + \tfrac{1}{T}\text{ of the job.}\)

\(\displaystyle \text{We already know that, working together, they can do the job in 2 hours.}\)
. . \(\displaystyle \text{Hence, in one hour, they will do only }\tfrac{1}{2}\text{ of the job.}\)

\(\displaystyle \text{There is one equaton: }\;\frac{1}{A} + \frac{1}{M} + \frac{1}{T} \:=\:\frac{1}{2}\)


\(\displaystyle \text{Arielle and Melissa, working together for one hour, can do: }\tfrac{!}{A} + \tfrac{!}{M}\text{ of the job.}\)
\(\displaystyle \text{We know that, working together, they can do the job in 4 hours.}\)
. . \(\displaystyle \text{Hence, in one hour, they can do }\tfrac{1}{4}\text{ of the job.}\)

\(\displaystyle \text{There is another equation: }\;\frac{1}{A} + \frac{1}{M} \:=\:\frac{1}{4}\)


\(\displaystyle \text{Melissa and Tyler, working together for one hour, can do: }\tfrac{!}{M} + \tfrac{!}{T}\text{ of the job.}\)
\(\displaystyle \text{We know that, working together, they can do the job in 3 hours.}\)
. . \(\displaystyle \text{Hence, in one hour, they can do }\tfrac{1}{3}\text{ of the job.}\)

\(\displaystyle \text{There is yet another equation: }\;\frac{1}{M} + \frac{1}{T} \:=\:\frac{1}{3}\)


\(\displaystyle \text{We have a system of equations: }\;\begin{Bmatrix} \dfrac{1}{A} &+& \dfrac{1}{M} &+& \dfrac{1}{T} &=& \dfrac{1}{2} & [1] \\ \\[-3mm] \dfrac{1}{A} &+& \dfrac{1}{M} &&& = & \dfrac{1}{4} & [2] \\ \\[-3mm] &&\dfrac{1}{M} &+& \dfrac{1}{T} &=& \dfrac{1}{3} & [3] \end{Bmatrix}\)


\(\displaystyle \text{Subtract [1] - [2]: }\;\frac{1}{T} \:=\:\frac{1}{4} \quad\Rightarrow\quad \boxed{T \:=\:4\text{ hours}}\)

\(\displaystyle \text{Subtract [1] - [3]: }\;\frac{1}{A} \:=\:\frac{1}{6} \quad\Rightarrow\quad \boxed{A \:=\:6 \text{ hours}}\)

\(\displaystyle \text{Substitute into [3]: }\;\frac{1}{M} + \frac{1}{4} \:=\:\frac{1}{3} \quad\Rightarrow\quad \frac{!}{M} \:=\:\frac{1}{12} \quad\Rightarrow\quad \boxed{M \:=\:12\text{ hours}}\)



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

\(\displaystyle \text{In one hour, Tyler can do }\tfrac{1}{4}\text{ of the job.}\)
\(\displaystyle \text{In }x\text{ hours, he can do }\tfrac{x}{4}\text{ of the job.}\)

\(\displaystyle \text{Working together for }x\text{ hours, they can do }\tfrac{x}{6} + \tfrac{x}{4}\text{ of the job.}\)

\(\displaystyle \text{But in }x\text{ hours, we expect them to }complete\text{ the job (one job).}\)

\(\displaystyle \text{We have: }\:\frac{x}{6} + \frac{x}{4} \:=\:1 \quad\Rightarrow\quad \frac{5}{12}x \:=\:1 \quad\Rightarrow\quad x \:=\:\frac{12}{5}\)


\(\displaystyle \text{Therefore, working together, Arielle and Tyler can do the job in }2.4\text{ hours.}\)