If $$x$$ and $$y$$ are real numbers such that $$4x^2 + 4y^2 - 4xy - 6y + 3 = 0,$$ then the value of $$(4x + 5y)$$ is

Question

If $$x$$ and $$y$$ are real numbers such that $$4x^2 + 4y^2 - 4xy - 6y + 3 = 0,$$  then the value of $$(4x + 5y)$$ is

Accepted Answer

We are tasked with solving the equation:  
$$
4x^2 + 4y^2 - 4xy - 6y + 3 = 0,
$$  
and finding the value of $$ 4x + 5y $$.

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#### **Step 1: Rearrange the equation**
Group terms to identify perfect squares:  
$$
4x^2 + 4y^2 - 4xy - 6y + 3 = [(2x)^2 + y^2 - 2(2x)(y)] + [3y^2 - 6y + 3].
$$

Simplify each group:  
1. The first group becomes:  
$$
(2x - y)^2.
$$  
2. The second group becomes:  
$$
3(y - 1)^2.
$$

Thus, the equation simplifies to:  
$$
(2x - y)^2 + 3(y - 1)^2 = 0.
$$

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#### **Step 2: Solve for $$ x $$ and $$ y $$**  
Since the sum of two perfect squares equals zero, **each square must individually equal zero**. Therefore:  
1. $$ (2x - y)^2 = 0 $$, which implies:  
$$
2x = y. 	ag{1}
$$  
2. $$ 3(y - 1)^2 = 0 $$, which implies:  
$$
y = 1. 	ag{2}
$$

Substitute $$ y = 1 $$ into Equation (1):  
$$
2x = 1,
$$  
which gives:  
$$
x = \frac{1}{2}. 	ag{3}
$$

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#### **Step 3: Calculate $$ 4x + 5y $$**  
Substitute $$ x = \frac{1}{2} $$ and $$ y = 1 $$ into $$ 4x + 5y $$:  
$$
4x + 5y = 4\left(\frac{1}{2}ight) + 5(1),
$$  
$$
4x + 5y = 2 + 5 = 7.
$$

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### **Final Answer:** The value of $$ 4x + 5y $$ is **7**.