1. \((h-f)(4)=h(4)-f(4)\)
The left hand side and the right hand of the equation are equivalent. The notation on the right might be more intuitive, though.
\(h(4)-f(4)\) | Evaluation both functions when x equals 4. |
\(h(4)=\frac{1}{4}*4+6\) | Here, I have replaced all instances of an "x" with a 4. Now, let's evaluate h(4). |
\(h(4)=1+6=7\) | Now, let's find f(4). |
\(f(4)=5*4-9\) | Yet again, every appearance of "x" is replaced with the input, 4. |
\(f(4)=11\) | The original question wants you to subtract the two functions, so let's do that. |
\(h(4)-f(4)\\ 7\hspace{6mm}-\hspace{3mm}11\) | |
\(-4\) | |
2) If h(x)=9, then we can use substitution to find the value of x:
\(h(x)=\frac{1}{4}x+6\) | Replace h(x) with 9 since they are equal. |
\(9=\frac{1}{4}x+6\) | Subtract 6 from both sides of the equation. |
\(3=\frac{1}{4}x\) | Multiply by 4 on both sides to isolate the variable. |
\(x=12\) | |
3) If f(n)=f(3n+1), then we can evaluate both functions for the given input and set them equal to each other.
\(f(x)=5x-9\) | \(f(x)=5x-9\) | |
\(f(n)=5n-9\) | \(f(3n+1)=5(2n+1)-9\) | |
\(f(3n+1)=10n+5-9\) | ||
\(f(n)=5n-9\) | \(f(3n+1)=10n-4\) | |
As aforementioned, these values are equal, so let's set them equal
\(\hspace{4mm}f(n)=f(3n+1)\\ 5n-9=10n-4\) | Now, solve for n. Move the constants and linear terms over to one side of the equation. |
\(-5=5n\) | Finally, divide by 5. |
\(-1=n\) | |
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.