Solving Expression Value with Algebraic Equations

We are given the following two equations:

• Equation 1: \( 3x + 2y = 15 \)
• Equation 2: \( xy = 6 \)

The goal is to find the value of this specific expression:

Expression to evaluate: \( \frac{3x^3}{2} + \frac{4y^3}{9} \)

Algebraic Manipulation Strategy

To find the value of the expression \( \frac{3x^3}{2} + \frac{4y^3}{9} \), we can utilize the given equations. Notice that the expression involves terms like \( x^3 \) and \( y^3 \). A common strategy is to cube one of the given equations. Let's cube Equation 1:

We use the algebraic identity \( (a+b)^3 = a^3 + b^3 + 3ab(a+b) \).

Let \( a = 3x \) and \( b = 2y \). Applying the identity to \( (3x + 2y)^3 \):

\( (3x + 2y)^3 = (3x)^3 + (2y)^3 + 3(3x)(2y)(3x + 2y) \)

From Equation 1, we know \( 3x + 2y = 15 \). Substitute this value:

\( (15)^3 = (3x)^3 + (2y)^3 + 3(3x)(2y)(15) \)

Let's simplify the terms:

• \( 15^3 = 3375 \)
• \( (3x)^3 = 27x^3 \)
• \( (2y)^3 = 8y^3 \)
• The middle term simplifies to \( 3 \times (6xy) \times 15 \), since \( (3x)(2y) = 6xy \).

Substitute these simplified terms back into the equation:

\( 3375 = 27x^3 + 8y^3 + 3(6xy)(15) \)

\( 3375 = 27x^3 + 8y^3 + 18xy \times 15 \)

Substituting Known Values

Now, substitute the value of \( xy = 6 \) from Equation 2 into the equation derived above:

\( 3375 = 27x^3 + 8y^3 + 18(6) \times 15 \)

Calculate the product:

\( 18 \times 6 \times 15 = 108 \times 15 = 1620 \)

So the equation becomes:

\( 3375 = 27x^3 + 8y^3 + 1620 \)

Calculating Intermediate Result

Rearrange the equation to isolate the sum of the cubes, \( 27x^3 + 8y^3 \):

\( 27x^3 + 8y^3 = 3375 - 1620 \)

\( 27x^3 + 8y^3 = 1755 \)

Evaluating Target Expression

We need to find the value of \( \frac{3x^3}{2} + \frac{4y^3}{9} \). Let's see how this relates to the \( 27x^3 + 8y^3 \) we just calculated.

Consider manipulating the target expression. We can factor out \( \frac{1}{18} \) from the expression:

\( \frac{3x^3}{2} + \frac{4y^3}{9} = \frac{1}{18} \left( 18 \times \frac{3x^3}{2} + 18 \times \frac{4y^3}{9} \right) \)

Simplify inside the parentheses:

\( = \frac{1}{18} \left( (9 \times 2) \times \frac{3x^3}{2} + (2 \times 9) \times \frac{4y^3}{9} \right) \)

\( = \frac{1}{18} \left( 9 \times 3x^3 + 2 \times 4y^3 \right) \)

\( = \frac{1}{18} \left( 27x^3 + 8y^3 \right) \)

This is exactly what we need! Substitute the value \( 27x^3 + 8y^3 = 1755 \):

\( \frac{1}{18} (1755) = \frac{1755}{18} \)

Final Calculation

Perform the final division to get the value:

\( \frac{1755}{18} \)

To simplify or calculate, we can divide 1755 by 18.

Dividing 1755 by 18 gives 97.5.

Alternatively, simplify the fraction. Both numerator and denominator are divisible by 9:

\( \frac{1755 \div 9}{18 \div 9} = \frac{195}{2} \)

The value \( \frac{195}{2} \) matches the first option provided.