Let's define:

• $b$ as the number of boys at the party
• $g$ as the number of girls at the party

First Round:

• 4 boys decide not to play, so the number of boys playing is $b - 4$.
• The number of girls playing is $g$.
• Given that there were twice as many girls playing as boys, we can write:

$g = 2(b - 4)$

Second Round:

• 2 more boys and 2 girls decide not to play.
• The number of boys now playing is $(b - 4) - 2 = b - 6$.
• The number of girls now playing is $g - 2$.
• Given that there were twice as many boys playing as girls, we can write:

$b - 6 = 2(g - 2)$

Solving the System of Equations:

We have the two equations:

1. $g = 2(b - 4)$ → $g = 2b - 8$
2. $b - 6 = 2(g - 2)$

Substituting $g = 2b - 8$ into the second equation:

$b - 6 = 2((2b - 8) - 2)$

$b - 6 = 2(2b - 10)$

$b - 6 = 4b - 20$

$b - 4b = -20 + 6$

$-3b = -14$

$b = 14$

Now, substituting $b = 14$ into $g = 2b - 8$:

$g = 2(14) - 8 = 28 - 8 = 20$

Answer:

The total number of boys and girls at the party is $14 + 20 = 34$.

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Would you like more details or another approach?

Related Questions:

1. What if only 3 boys had not played in the first round instead of 4?
2. How would the answer change if instead of 2, 3 boys and 3 girls had not played in the second round?
3. Can we generalize a formula for such problems involving ratios of boys and girls?
4. What happens if we swap the given conditions in the two rounds?
5. How can we verify our solution using substitution?

Tip:

Always check your solution by plugging values back into the original conditions!