To find the solutions of the system of equations \( y = f(x) \) and \( y = g(x) \), we need to set \( f(x) = g(x) \):
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The functions are:
- \( f(x) = 4(2^{-x}) + 3 \)
- \( g(x) = 8 - 2^x \)
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Let's solve the equation \( 4(2^{-x}) + 3 = 8 - 2^x \):
First, we rewrite \( 2^{-x} \) as \( \frac{1}{2^x} \): \[ 4\left(\frac{1}{2^x}\right) + 3 = 8 - 2^x \] \[ \frac{4}{2^x} + 3 = 8 - 2^x \]
Next, create a common equation by multiplying both sides by \( 2^x \) (to eliminate the fraction): \[ 4 + 3 \cdot 2^x = 8 \cdot 2^x - (2^x)^2 \]
Rearranging this gives: \[ (2^x)^2 - 5\cdot2^x + 4 = 0 \]
Let \( u = 2^x \): \[ u^2 - 5u + 4 = 0 \]
Solving this quadratic equation: \[ u = \frac{5 \pm \sqrt{(5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \] \[ u = \frac{5 \pm \sqrt{25 - 16}}{2} \] \[ u = \frac{5 \pm 3}{2} \]
Thus, we have two solutions for \( u \):
- \( u = \frac{8}{2} = 4 \)
- \( u = \frac{2}{2} = 1 \)
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Now we can convert back to \( x \) for both solutions:
- If \( u = 4 \), then \( 2^x = 4 \) implies \( x = 2 \).
- If \( u = 1 \), then \( 2^x = 1 \) implies \( x = 0 \).
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Corresponding y-values (using either function, e.g., \( f(x) \)):
- For \( x = 2 \): \( f(2) = 4(2^{-2}) + 3 = 4(0.25) + 3 = 1 + 3 = 4 \).
- For \( x = 0 \): \( f(0) = 4(2^0) + 3 = 4(1) + 3 = 4 + 3 = 7 \).
So the solutions are:
- \( (0, 7) \)
- \( (2, 4) \)
Now, we can fill in the blanks:
- Blank space 1: 0
- Blank space 2: 7
- Blank space 3: 2
- Blank space 4: 4
- Blank space 5: 0
- Blank space 6: 7
Thus, the complete sentences should read: The solutions to the system of equations \( y = f(x) \) and \( y = g(x) \) are \( (0, 7) \) and \( (2, 4) \). The solutions to the equation \( f(x) = g(x) \) are \( 0 \) and \( 7 \).