Question 9 (2 points) The sum of three consecutive integers is 41 less than 5 times the largest number. What is the largest of the three integers?

## Step 1<br />Let's denote the three consecutive integers as \(x\), \(x+1\), and \(x+2\). According to the problem, the sum of these three integers is 41 less than 5 times the largest number. This can be written as an equation:<br />### \(x + (x+1) + (x+2) = 5(x+2) - 41\)<br /><br />## Step 2<br />Next, we simplify the equation by combining like terms on both sides:<br />### \(3x + 3 = 5x + 10 - 41\)<br /><br />## Step 3<br />We then isolate the variable \(x\) by subtracting \(3x\) from both sides and simplifying:<br />### \(3 = 2x - 31\)<br /><br />## Step 4<br />We then add 31 to both sides to isolate the term with \(x\):<br />### \(34 = 2x\)<br /><br />## Step 5<br />Finally, we divide both sides by 2 to solve for \(x\):<br />### \(x = 17\)<br /><br />So, the largest of the three consecutive integers is \(x+2 = 17+2 = 19\).