Final answer:
After correcting the calculations, the equation 6C2 + 6C3 equals 7C3 is indeed true, resulting in 35 on both sides. The use of least common denominator is not relevant here as this combinatorial problem does not involve fraction addition.
Step-by-step explanation:
The equation 6C2 + 6C3 = 7C3 involves combinations where 6C2 refers to the number of ways to choose 2 items out of 6, and similarly for 6C3 and 7C3. Let's calculate these combinations to evaluate if the equation is true.
6C2 is equivalent to 6! / (2! * (6-2)!) which simplifies to 6 * 5 / (2 * 1) or 15.
6C3 is equivalent to 6! / (3! * (6-3)!) which simplifies to 6 * 5 * 4 / (3 * 2 * 1) or 20.
Thus, 6C2 + 6C3 equals 15 + 20, which is 35.
7C3 is equivalent to 7! / (3! * (7-3)!) which simplifies to 7 * 6 * 5 / (3 * 2 * 1) or 35.
Therefore, the original equation 6C2 + 6C3 equals to 7C3, making 35 = 35 true after the correct calculation. Jayden and Alexa's calculations contained mistakes, but when calculated correctly, the equation holds.
When dealing with combinations or combinatorics, it's important to correctly apply the formula for combinations and perform arithmetic simplifications accurately. The use of least common denominator (LCD) that Alexa referred to is not related to this type of problem, as that approach applies to fraction addition, not combinatorial problems.