All three decompositions correctly sum to $$\frac{8}{10}$$ and represent valid ways to break apart the fraction. Decomposition 1: $$\frac{5}{10} + \frac{3}{10} = \frac{8}{10}$$. Decomposition 2: $$\frac{4}{10} + \frac{4}{10} = \frac{8}{10}$$. Decomposition 3: $$\frac{2}{10} + \frac{2}{10} + \frac{4}{10} = \frac{8}{10}$$. Choices A, C, and D incorrectly suggest that only one decomposition method is valid.

When you see fractions with the same denominator being added together, you can add just the numerators while keeping the denominator the same. This equation asks you to find three different positive whole numbers that add up to 9.

Let's work through this step by step. Since all fractions have 14 in the denominator, you can rewrite the equation as: $$\frac{a + b + c}{14} = \frac{9}{14}$$. This means $$a + b + c = 9$$.

Now you need to find which set of three different positive whole numbers adds up to 9. Let's check each option:

For choice D: $$a = 2, b = 3, c = 4$$. Adding these: $$2 + 3 + 4 = 9$$. Perfect! These are three different positive whole numbers that sum to 9.

Choice A gives us $$2 + 2 + 5 = 9$$, but this violates the requirement that $$a$$, $$b$$, and $$c$$ must be different numbers since both $$a$$ and $$b$$ equal 2.

Choice B gives us $$1 + 3 + 6 = 10$$, which is too large. This would make the right side equal $$\frac{10}{14}$$, not $$\frac{9}{14}$$.

Choice C gives us $$3 + 3 + 3 = 9$$, which has the right sum but fails because all three values are the same, violating the "different" requirement.

When solving fraction addition problems like this, always check two things: do the numerators add up correctly, and do all values meet the given conditions? Here, both the sum and the "different numbers" requirement matter equally.

Choice A correctly decomposes $$\frac{6}{8}$$ into three equal addends: $$\frac{2}{8} + \frac{2}{8} + \frac{2}{8} = \frac{6}{8}$$. This shows a different grouping pattern than the given decomposition (equal parts vs. unequal parts). Choice B sums to $$\frac{7}{8}$$. Choice C sums to $$\frac{6}{8}$$ and is a valid decomposition. Choice D sums to $$\frac{7}{8}$$.

To show $$\frac{10}{12} = \frac{3}{12} + \frac{3}{12} + \frac{4}{12}$$, Marcus needs to circle exactly three groups containing 3, 3, and 4 shaded parts respectively. Choice B creates only two groups instead of three. Choice C creates four groups instead of three. Choice D creates five groups and doesn't match the required group sizes.

Choice C is incorrect because $$\frac{4}{15} + \frac{4}{15} + \frac{4}{15} = \frac{12}{15}$$, which is greater than $$\frac{11}{15}$$. Choices A, B, and D all correctly sum to $$\frac{11}{15}$$: Choice A gives $$\frac{5+3+3}{15} = \frac{11}{15}$$, Choice B gives $$\frac{6+3+2}{15} = \frac{11}{15}$$, and Choice D gives $$\frac{7+2+2}{15} = \frac{11}{15}$$.

Choice A correctly decomposes $$\frac{5}{6}$$ into four addends: $$\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{2}{6} = \frac{5}{6}$$. Choice B sums to $$\frac{4}{6}$$, not $$\frac{5}{6}$$. Choice C sums to $$\frac{6}{6}$$, which is greater than $$\frac{5}{6}$$. Choice D sums to $$\frac{5}{6}$$ but includes $$\frac{0}{6}$$, which is not a meaningful part in a decomposition.

When you see a fraction decomposition problem, you're working with breaking apart fractions into smaller pieces that add up to the original amount. Think of it like breaking a chocolate bar into different sized pieces.

Ava starts with $$\frac{8}{16}$$ and wants to split it into two equal parts. To find equal parts, you divide: $$\frac{8}{16} ÷ 2 = \frac{4}{16}$$. So $$\frac{8}{16} = \frac{4}{16} + \frac{4}{16}$$.

Now you can substitute this back into the original equation. Instead of $$\frac{12}{16} = \frac{8}{16} + \frac{4}{16}$$, you get $$\frac{12}{16} = \frac{4}{16} + \frac{4}{16} + \frac{4}{16}$$. You can verify this works: $$\frac{4}{16} + \frac{4}{16} + \frac{4}{16} = \frac{12}{16}$$ ✓

Answer choice A gives $$\frac{2}{16} + \frac{6}{16} + \frac{4}{16}$$, but $$\frac{2}{16}$$ and $$\frac{6}{16}$$ aren't equal parts, so this doesn't follow the directions.

Answer choice B has the same problem as A—$$\frac{6}{16}$$ and $$\frac{2}{16}$$ aren't equal, just written in different order.

Answer choice C shows $$\frac{3}{16} + \frac{5}{16} + \frac{4}{16}$$, where $$\frac{3}{16}$$ and $$\frac{5}{16}$$ also aren't equal parts.

Only answer choice D correctly splits $$\frac{8}{16}$$ into two equal parts of $$\frac{4}{16}$$ each.

Study tip: When decomposing fractions into equal parts, divide the numerator by the number of parts you want, keeping the denominator the same. Always check that your pieces add back to the original fraction.

This question tests 4th grade ability to decompose a fraction into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation (CCSS.4.NF.3.b). Decomposing a fraction means breaking it apart into a sum of smaller fractions—like taking 7/12 and writing it as 3/12 + 4/12, or 1/12 + 6/12, or 2/12 + 2/12 + 3/12. The key requirement: all fractions in the decomposition must have the SAME denominator (all twelfths, etc.), and the numerators must add up to the original numerator. There are many different ways to decompose the same fraction, as long as these rules are followed. To decompose 7/12, students must write it as a sum of fractions with denominator 12, where the numerators add to 7. Different decompositions show different ways to partition the 7 parts: 7 twelfths can be grouped as 3+4, or 1+6, or 2+2+3, etc., all representing the same total. Choice A is correct because both decompositions use denominator 12 throughout, the numerators in first decomposition add to 7 (3 + 4 = 7), the numerators in second decomposition add to 7 (1 + 6 = 7), and the two decompositions are different from each other. This demonstrates understanding that fractions can be decomposed multiple ways while maintaining same denominator and total value. Choice D represents the same decomposition repeated in reverse order, which happens when students don't understand the need for truly different decompositions or think the commutative property creates a different decomposition. To help students: Use visual models (area models, fraction bars) to show different ways to group the same total—7/12 shaded in a rectangle can be grouped as (3 red + 4 blue)/12 or (1 red + 6 blue)/12, same total but different groupings. Emphasize SAME DENOMINATOR always—when decomposing 7/12, every fraction must have denominator 12 (all twelfths); check that numerators sum correctly: 3 + 4 = 7 ✓, 1 + 6 = 7 ✓, 2 + 2 + 3 = 7 ✓; connect to addition fact families: if you know 3 + 4 = 7 and 1 + 6 = 7, you can decompose 7/12 those ways; show that order doesn't matter (3/12 + 4/12 = 4/12 + 3/12), so don't count these as two different ways; watch for: changing denominators, numerators that don't sum correctly, repeating same decomposition, and counting commutative versions as different.

This question tests 4th grade ability to decompose a fraction into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation (CCSS.4.NF.3.b). Decomposing a fraction means breaking it apart into a sum of smaller fractions—like taking 8/12 and writing it as 5/12 + 3/12, or 2/12 + 2/12 + 4/12, or 4/12 + 4/12. The key requirement: all fractions in the decomposition must have the SAME denominator (all twelfths, etc.), and the numerators must add up to the original numerator. There are many different ways to decompose the same fraction, as long as these rules are followed. To decompose 8/12, students must write it as a sum of fractions with denominator 12, where the numerators add to 8. Different decompositions show different ways to partition the 8 parts: 8 twelfths can be grouped as 5+3, or 2+2+4, or 4+4, etc., all representing the same total. Choice A is correct because both decompositions use denominator 12 throughout, the numerators in first decomposition add to 8: 5 + 3 = 8, the numerators in second decomposition add to 8: 2 + 2 + 4 = 8, and the two decompositions are different from each other (different number of addends). This demonstrates understanding that fractions can be decomposed multiple ways while maintaining same denominator and total value. Choice B represents using different denominators, which happens when students don't maintain same denominator requirement or use equivalent fractions incorrectly. To help students: Use visual models (area models, fraction bars) to show different ways to group the same total—8/12 shaded in a rectangle can be grouped as (5 red + 3 blue)/12 or (2 red + 2 green + 4 blue)/12, same total but different groupings. Emphasize SAME DENOMINATOR always—when decomposing 8/12, every fraction must have denominator 12 (all twelfths); check that numerators sum correctly: 5 + 3 = 8 ✓, 2 + 2 + 4 = 8 ✓; connect to addition fact families: if you know 5 + 3 = 8 and 2 + 2 + 4 = 8, you can decompose 8/12 those ways; show that order doesn't matter (5/12 + 3/12 = 3/12 + 5/12), so don't count these as two different ways; watch for: changing denominators, numerators that don't sum correctly, repeating same decomposition, and counting commutative versions as different.

This question tests 4th grade ability to decompose a fraction into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation (CCSS.4.NF.3.b). Decomposing a fraction means breaking it apart into a sum of smaller fractions—like taking 3/4 and writing it as 1/4 + 2/4, or 1/4 + 1/4 + 1/4. The key requirement: all fractions in the decomposition must have the SAME denominator (all fourths), and the numerators must add up to the original numerator. To decompose 3/4, students must write it as a sum of fractions with denominator 4, where the numerators add to 3. Different decompositions show different ways to partition the 3 parts: 3 fourths can be grouped as 1+2, or 1+1+1, etc., all representing the same total. Choice A is correct because both decompositions use denominator 4 throughout, the numerators in first decomposition add to 3: 1+2=3 ✓, the numerators in second decomposition add to 3: 1+1+1=3 ✓, and the two decompositions are different from each other (one uses two fractions, the other uses three). Choice B represents using different denominators (1/2 has denominator 2, not 4), which happens when students don't maintain same denominator requirement. To help students: Use visual models (area models, fraction bars) to show different ways to group the same total—3/4 shaded in a rectangle can be grouped as (1 part + 2 parts)/4 or (1 part + 1 part + 1 part)/4, same total but different groupings. Emphasize SAME DENOMINATOR always—when decomposing 3/4, every fraction must have denominator 4 (all fourths). Connect to addition fact families: if you know 1 + 2 = 3 and 1 + 1 + 1 = 3, you can decompose 3/4 those ways.