Browse · MATH Print → jmc prealgebra senior Problem Simplify 2k−3+3k+1+23k+1. Solution — click to reveal Notice that the two fractions have the same denominator, so we can add them. Addition is commutative, so we can rearrange the terms to get 2k−3+23k+1+3k+1=24k−2+3k+1=2k−1+3k+1=5k. Final answer 5k ← Previous problem Next problem →