C. 16 Solve linear equations with variables on both sides: word problems H9g Scarlett and Tristan are both running for student council president. At the start of the last day of voting, Tristan had half as many votes as Scarlett. Throughout the day, Scarlett gained 11 more votes, while Tristan gained 34 more votes. By the end of the day, Scarlett and Tristan had the same number of votes. How many votes did Scarlett have at the start of the last day of voting?

Let's denote the number of votes Scarlett had at the start of the last day of voting as \( S \) and the number of votes Tristan had at the start of the last day of voting as \( T \). According to the problem, we know the following: 1. At the start of the last day of voting, Tristan had half as many votes as Scarlett: \[ T = \frac{1}{2} S \] 2. Throughout the day, Scarlett gained 11 more votes, and Tristan gained 34 more votes. By the end of the day, they had the same number of votes: \[ S + 11 = T + 34 \] Now we can substitute the expression for \( T \) from the first equation into the second equation: \[ S + 11 = \left(\frac{1}{2} S\right) + 34 \] Next, we will solve for \( S \). First, let's eliminate the fraction by multiplying the entire equation by 2: \[ 2(S + 11) = 2\left(\frac{1}{2} S + 34\right) \] This simplifies to: \[ 2S + 22 = S + 68 \] Now, we can isolate \( S \) by subtracting \( S \) from both sides: \[ 2S - S + 22 = 68 \] This simplifies to: \[ S + 22 = 68 \] Next, subtract 22 from both sides: \[ S = 68 - 22 \] \[ S = 46 \] So, Scarlett had **46 votes** at the start of the last day of voting. To find out how many votes Tristan had at the start, we can use the first equation: \[ T = \frac{1}{2} S = \frac{1}{2} \times 46 = 23 \] Now, let's verify the final condition. At the end of the day: - Scarlett's votes: \( 46 + 11 = 57 \) - Tristan's votes: \( 23 + 34 = 57 \) Both have 57 votes, confirming our solution is correct. Thus, the final answer is: **Scarlett had 46 votes at the start of the last day of voting.**