Answer :
To solve this problem using the substitution method, let's follow a detailed, step-by-step approach.
1. Define Variables:
- Let [tex]\( a \)[/tex] represent the number of buckets of apples sold.
- Let [tex]\( p \)[/tex] represent the number of buckets of peaches sold.
- Let [tex]\( s \)[/tex] represent the number of buckets of strawberries sold.
2. Set Up the Revenue Equation:
The total revenue is given by:
[tex]\[ 4a + 6p + 9s = 229 \][/tex]
3. Set Up the Relationships:
- The farm sold half as many strawberry buckets as peach buckets:
[tex]\[ s = \frac{p}{2} \][/tex]
- The farm sold 8 fewer apple buckets than peach buckets:
[tex]\[ a = p - 8 \][/tex]
4. Substitute Relationships into the Revenue Equation:
Substitute [tex]\( s = \frac{p}{2} \)[/tex] and [tex]\( a = p - 8 \)[/tex] into the revenue equation [tex]\( 4a + 6p + 9s = 229 \)[/tex]:
[tex]\[ 4(p - 8) + 6p + 9\left(\frac{p}{2}\right) = 229 \][/tex]
5. Simplify and Solve the Equation:
- First, distribute and combine like terms:
[tex]\[ 4(p - 8) + 6p + \frac{9p}{2} = 229 \][/tex]
[tex]\[ 4p - 32 + 6p + \frac{9p}{2} = 229 \][/tex]
- Combine the constant terms:
[tex]\[ 10p - 32 + \frac{9p}{2} = 229 \][/tex]
- To combine the terms with [tex]\( p \)[/tex], get a common denominator (2) for the fractions:
[tex]\[ \frac{20p}{2} - \frac{64}{2} + \frac{9p}{2} = 229 \][/tex]
[tex]\[ \frac{29p}{2} - 32 = 229 \][/tex]
- Clear the fraction by multiplying everything by 2:
[tex]\[ 29p - 64 = 458 \][/tex]
- Solve for [tex]\( p \)[/tex]:
[tex]\[ 29p = 522 \][/tex]
[tex]\[ p = 18 \][/tex]
6. Find [tex]\( a \)[/tex] and [tex]\( s \)[/tex] Using the Values of [tex]\( p \)[/tex]:
- Substitute [tex]\( p = 18 \)[/tex] back into the equations for [tex]\( a \)[/tex] and [tex]\( s \)[/tex]:
[tex]\[ a = p - 8 = 18 - 8 = 10 \][/tex]
[tex]\[ s = \frac{p}{2} = \frac{18}{2} = 9 \][/tex]
7. Confirm the Solution:
Verify the values [tex]\( a = 10 \)[/tex], [tex]\( p = 18 \)[/tex], and [tex]\( s = 9 \)[/tex] in the original revenue equation:
[tex]\[ 4a + 6p + 9s = 4(10) + 6(18) + 9(9) = 40 + 108 + 81 = 229 \][/tex]
The original equation is satisfied, confirming that the solution is correct.
Therefore, the number of buckets sold by the farm stand is:
[tex]\[ \boxed{C. \; a=10 ; p=18 ; s=9} \][/tex]
1. Define Variables:
- Let [tex]\( a \)[/tex] represent the number of buckets of apples sold.
- Let [tex]\( p \)[/tex] represent the number of buckets of peaches sold.
- Let [tex]\( s \)[/tex] represent the number of buckets of strawberries sold.
2. Set Up the Revenue Equation:
The total revenue is given by:
[tex]\[ 4a + 6p + 9s = 229 \][/tex]
3. Set Up the Relationships:
- The farm sold half as many strawberry buckets as peach buckets:
[tex]\[ s = \frac{p}{2} \][/tex]
- The farm sold 8 fewer apple buckets than peach buckets:
[tex]\[ a = p - 8 \][/tex]
4. Substitute Relationships into the Revenue Equation:
Substitute [tex]\( s = \frac{p}{2} \)[/tex] and [tex]\( a = p - 8 \)[/tex] into the revenue equation [tex]\( 4a + 6p + 9s = 229 \)[/tex]:
[tex]\[ 4(p - 8) + 6p + 9\left(\frac{p}{2}\right) = 229 \][/tex]
5. Simplify and Solve the Equation:
- First, distribute and combine like terms:
[tex]\[ 4(p - 8) + 6p + \frac{9p}{2} = 229 \][/tex]
[tex]\[ 4p - 32 + 6p + \frac{9p}{2} = 229 \][/tex]
- Combine the constant terms:
[tex]\[ 10p - 32 + \frac{9p}{2} = 229 \][/tex]
- To combine the terms with [tex]\( p \)[/tex], get a common denominator (2) for the fractions:
[tex]\[ \frac{20p}{2} - \frac{64}{2} + \frac{9p}{2} = 229 \][/tex]
[tex]\[ \frac{29p}{2} - 32 = 229 \][/tex]
- Clear the fraction by multiplying everything by 2:
[tex]\[ 29p - 64 = 458 \][/tex]
- Solve for [tex]\( p \)[/tex]:
[tex]\[ 29p = 522 \][/tex]
[tex]\[ p = 18 \][/tex]
6. Find [tex]\( a \)[/tex] and [tex]\( s \)[/tex] Using the Values of [tex]\( p \)[/tex]:
- Substitute [tex]\( p = 18 \)[/tex] back into the equations for [tex]\( a \)[/tex] and [tex]\( s \)[/tex]:
[tex]\[ a = p - 8 = 18 - 8 = 10 \][/tex]
[tex]\[ s = \frac{p}{2} = \frac{18}{2} = 9 \][/tex]
7. Confirm the Solution:
Verify the values [tex]\( a = 10 \)[/tex], [tex]\( p = 18 \)[/tex], and [tex]\( s = 9 \)[/tex] in the original revenue equation:
[tex]\[ 4a + 6p + 9s = 4(10) + 6(18) + 9(9) = 40 + 108 + 81 = 229 \][/tex]
The original equation is satisfied, confirming that the solution is correct.
Therefore, the number of buckets sold by the farm stand is:
[tex]\[ \boxed{C. \; a=10 ; p=18 ; s=9} \][/tex]
