TheXSquaredFactor

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Punkte2441
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 #2
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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.

 

  • If James had half as much money as his brother, then we can represent this situation via equation: \(J_1=\frac{B_1}{2}\)
  • If James had 27 dollars less than his sister, then we can create another equation: \(J_1=S_1-27\)

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.

 

  • If James's sister has twice as much money as James's brother, then we can create a third equation: \(S_2=2B_2\)

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. 

 

  • If James's brother paid James and James's sister 12 dollars, then this has a three-way affect.
  • \(J_1=J_2-12\\ S_1=S_2-12\\ B_1=B_2+12\)

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.

 #1