This morning James had half as much money as his brother and 27 dollars less than his sister. Then his brother paid James and his sister each 12 dollars not to tell their parents what happened out in the garden last Wednesday. After the payoff, James's sister has twice as much money as James's brother. How much money does James have now? Thanks to whoever can help! Also, please give a detailed solution.
See 2 solutions here:
https://web2.0calc.com/questions/this-morning-james-had-half-as-much-money-as-his-sister-and-27-dollars-less-than-his-brother-then-his-sister-paid-james-and-his-brother-eac
And here:
https://web2.0calc.com/questions/help-please_23015
+2441
I think it is best to initialize variables first and foremost. A quick glance at the problem will reveal that the information mentions the amount of money that three individuals have at separate moments.
\(\text{Let }J=\text{James's Money}\\ \text{Let }B=\text{James's Brother's Money}\\ \text{Let }S=\text{James's Sister's Money}\)
Now, let's break down the information into its components and analyze them.
After the secretive payoff, though, new information rolls out! The subscript of 1 on every variable is intentional; it is meant to denote the relationship of money prior to the shady payoff. I will use a subscript of 2 to represent amounts of money after the deal.
By utilizing subscripts, it is easy to notice that we cannot simply compare this information to the previous equations. However, we can use the information about the payoff to adjust the variables.
Since the question specifically asks for the amount of money James has now (after the payoff), then we should adjust every variable such that it has a subscript of 2. Let's do that!
| \(S_2=2B_2\) | \(J_1=\frac{B_1}{2}\) | \(J_1=S_1-27\) | Do the necessary substitution! | |
| \(J_2-12=\frac{B_2+12}{2}\) | \(J_2-12=S_2-12-27\) | Let's clean these equations up. | ||
| \(\boxed{1}\hspace{1mm}S_2=2B_2\) | \(\boxed{2} \hspace{1mm} 2J_2=B_2+36\) | \(\boxed{3}\hspace{1mm}J_2=S_2-27\) | This creates a three-variable system of equations. Now, we must solve it! | |
Let's place the 1st and 3rd equations alongside each other. By adding the equations together (with the combination of a slight manipulation) it is possible to eliminate one variable from this system. This creates a fourth equation.
\(\boxed{1}\hspace{1mm}S_2=2B_2\Rightarrow \hspace{10mm}S_2\hspace{10mm}=2B_2\\ \boxed{3}\hspace{1mm}J_2=S_2-27\Rightarrow -S_2+J_2=-27\\ \boxed{4}\hspace{1mm}\hspace{46mm}J_2=2B_2-27\)
We can now introduce the 2nd equation into the mix, which will allow us to eliminate the other variable.
\(\boxed{4}\hspace{39mm}J_2=2B_2-27\\ \boxed{2} \hspace{1mm} 2J_2=B_2+36 \Rightarrow-4J_2=-2B_2-72\\ \hspace{39mm}-3J_2=-99\\ \hspace{47mm}J_2=33 \)
According to this work, James has 33 dollars.
+2441 Upon further review, I have realized that I made a slight oversight on my end. Sorry about that! Unfortunately, I think everyone is prone to making such mistake. This is a learning experience--for you and me. Let's try to embrace it in the best way possible. The mistake occurs in the following segment:
Did you spot the error yet? If not, that's OK! There is a lot of information to digest in this problem, so it is very easy to assume that I could not have blundered at this step. The problem lies at the \(B_1=B_2+12\) part.
The original problem states that James's brother gives 12 dollars to both James and his sister. This means that James's brother loses $24 ($12 to James and $12 to the sister) in total--not $12 like I originally hinted at.
The working of the system of equation is mostly the same, though, and I get that James has $41 right now! I have double and triple checked, so I think I will finally move on from this word problem.
