Sure, let's solve the given expression step-by-step:
Given expression:
[tex]\[
\left(\frac{1}{2} x + \frac{2}{3} y\right)^2
\][/tex]
We need to expand this perfect square binomial:
Step 1: Write the binomial squared as a product of two identical binomials:
[tex]\[
\left(\frac{1}{2} x + \frac{2}{3} y\right) \left(\frac{1}{2} x + \frac{2}{3} y\right)
\][/tex]
Step 2: Use the distributive property (FOIL method) to expand the product:
[tex]\[
\left(\frac{1}{2} x + \frac{2}{3} y\right) \left(\frac{1}{2} x + \frac{2}{3} y\right) = \left(\frac{1}{2} x \cdot \frac{1}{2} x\right) + \left(\frac{1}{2} x \cdot \frac{2}{3} y\right) + \left(\frac{2}{3} y \cdot \frac{1}{2} x\right) + \left(\frac{2}{3} y \cdot \frac{2}{3} y\right)
\][/tex]
Step 3: Simplify each term:
1. First term:
[tex]\[
\left(\frac{1}{2} x \cdot \frac{1}{2} x\right) = \frac{1}{2} \cdot \frac{1}{2} \cdot x \cdot x = \frac{1}{4} x^2
\][/tex]
2. Second term:
[tex]\[
\left(\frac{1}{2} x \cdot \frac{2}{3} y\right) = \frac{1}{2} \cdot \frac{2}{3} \cdot x \cdot y = \frac{2}{6} x y = \frac{1}{3} x y
\][/tex]
3. Third term (which is the same as the second because of symmetry):
[tex]\[
\left(\frac{2}{3} y \cdot \frac{1}{2} x\right) = \frac{2}{3} \cdot \frac{1}{2} \cdot y \cdot x = \frac{2}{6} y x = \frac{1}{3} y x = \frac{1}{3} x y
\][/tex]
4. Fourth term:
[tex]\[
\left(\frac{2}{3} y \cdot \frac{2}{3} y\right) = \frac{2}{3} \cdot \frac{2}{3} \cdot y \cdot y = \frac{4}{9} y^2
\][/tex]
Step 4: Combine the simplified terms:
[tex]\[
\frac{1}{4} x^2 + \frac{1}{3} x y + \frac{1}{3} x y + \frac{4}{9} y^2
\][/tex]
Step 5: Simplify by combining like terms (middle terms):
[tex]\[
\frac{1}{4} x^2 + \frac{2}{3} x y + \frac{4}{9} y^2
\][/tex]
Step 6: Write it in numerical form:
[tex]\[
0.25 x^2 + 0.666666666666667 x y + 0.444444444444444 y^2
\][/tex]
This is the expanded form of the given expression.
