Tomorrow is Scott's family reunion, so Scott and his dad are making their famous chocolat beet cake. At the grocery store, Scott puts some beets on the scale. The recipe calls for or \( 1 \frac{1}{4} \) pounds of beets, so Scott removes \( \frac{1}{2} \) of a pound of beets from the scale. Now he has exactly what he needs for the cake. Use an equation to find the weight of the beets that Scott first puts on the scale. To write a fraction, use a slash (/) to separate the numerator and denominator.

Let's say the weight of the beets that Scott first puts on the scale is \( x \) pounds. According to the problem, Scott removes \( \frac{1}{2} \) pound from this weight, and then he has exactly \( 1 \frac{1}{4} \) pounds. We can set up the equation: \[ x - \frac{1}{2} = 1 \frac{1}{4} \] First, convert \( 1 \frac{1}{4} \) to an improper fraction: \[ 1 \frac{1}{4} = \frac{5}{4} \] Now, the equation looks like this: \[ x - \frac{1}{2} = \frac{5}{4} \] To solve for \( x \), add \( \frac{1}{2} \) to both sides. First, we need to convert \( \frac{1}{2} \) to quarters to combine fractions: \[ \frac{1}{2} = \frac{2}{4} \] Now we have: \[ x = \frac{5}{4} + \frac{2}{4} \] Adding these fractions gives: \[ x = \frac{7}{4} \] So, Scott first put \( \frac{7}{4} \) pounds of beets on the scale. To express this in a more familiar format, \( \frac{7}{4} \) is equivalent to \( 1 \frac{3}{4} \) pounds. Happy baking!