To substitute \( f = 16 - s \) into the equation \( 0.75f + 0.5s = 9 \), we'll replace \( f \) in the cost equation with \( 16 - s \).
Starting with the equation:
\[ 0.75f + 0.5s = 9 \]
we substitute \( f \):
\[ 0.75(16 - s) + 0.5s = 9 \]
Now we'll distribute \( 0.75 \):
\[ 0.75 \cdot 16 - 0.75s + 0.5s = 9 \]
Calculating \( 0.75 \cdot 16 \):
\[ 12 - 0.75s + 0.5s = 9 \]
Now, combine the \( s \) terms:
\[ 12 - 0.25s = 9 \]
Next, isolate \( -0.25s \) by subtracting 12 from both sides:
\[ -0.25s = 9 - 12 \] \[ -0.25s = -3 \]
Now, divide both sides by \(-0.25\):
\[ s = \frac{-3}{-0.25} = 12 \]
Now that we have \( s = 12 \), we can find \( f \):
\[ f = 16 - s = 16 - 12 = 4 \]
To summarize:
- Flour \( f = 4 \) pounds
- Sugar \( s = 12 \) pounds
Let's verify that these values satisfy both the weight and cost equations:
- Weight: \( f + s = 4 + 12 = 16 \) (Correct)
- Cost: \( 0.75f + 0.5s = 0.75(4) + 0.5(12) = 3 + 6 = 9 \) (Correct)
The values are verified as correct: \( f = 4 \) pounds of flour and \( s = 12 \) pounds of sugar.