To determine the correct operation that fills in the blank, we first simplify the left side of the equation:
- Start with the left side: \( 5(1/2q + 3r - 4s) - 9r \).
- Distribute the 5: \[ 5 \times \frac{1}{2}q + 5 \times 3r + 5 \times (-4s) - 9r = \frac{5}{2}q + 15r - 20s - 9r. \]
- Combine the \( r \) terms: \[ 15r - 9r = 6r. \]
- Therefore, the left side simplifies to: \[ \frac{5}{2}q + 6r - 20s. \]
Now we will rewrite the equation:
\[ \frac{5}{2}q + 6r - 20s = \frac{5}{2}q [\text{blank}] 15r - 20s. \]
The right side is \( \frac{5}{2}q + \text{(either + or -)} (15r - 20s) \).
To make both sides equal, we need to match the \( r \) components on each side. We have \( 6r \) on the left and \( 15r \) on the right.
Thus, we need to determine what operation, when applied to \( 15r \), would leave us with \( 6r \).
We can set up the equation: \[ 6r = 15r - x. \]
Solving for \( x \): \[ x = 15r - 6r = 9r. \]
This indicates that to achieve \( 6r \) from \( 15r \), we need to subtract \( 9r \).
Thus, the answer that fills in the blank to make the equation true is:
\[ \text{"-"}. \]
1 answer