Given: \(4x^2 + 4y^2 - 4xy - 6y + 3 = 0\)
Rearrange the terms: \[ 4x^2 - 4xy + 4y^2 - 6y + 3 = 0 \] We can complete the square or rewrite the expression. Let's try grouping terms:
\[ 4x^2 - 4xy + 4y^2 - 6y + 3 \] Notice that: \[ 4x^2 - 4xy + 4y^2 = 4(x - y)^2 \] Substitute this back into the equation: \[ 4(x - y)^2 - 6y + 3 = 0 \] Now, let's rewrite the equation as: \[ 4(x - y)^2 = 6y - 3 \]
Let's try a substitution. Let \(x = y + a\). This implies \(x - y = a\). Substituting this into the equation: \[ 4a^2 = 6y - 3 \] So, \[ a^2 = \frac{6y - 3}{4} \]
We want to find the value of \(4x + 5y\). Using the substitution \(x = y + a\): \[ 4x + 5y = 4(y + a) + 5y = 4y + 4a + 5y = 9y + 4a \] Let \(z = 4x + 5y\). So, \(z = 9y + 4a\). Substitute \(a = \sqrt{\frac{6y - 3}{4}}\) into the expression for \(z\).
Let's try a specific value for \(y\) to find a rational result. If \(y = 1\): The original equation becomes: \[ 4x^2 + 4(1)^2 - 4x(1) - 6(1) + 3 = 0 \] \[ 4x^2 + 4 - 4x - 6 + 3 = 0 \] \[ 4x^2 - 4x + 1 = 0 \] This is a perfect square: \[ (2x - 1)^2 = 0 \] Solving for \(x\), we get \(2x - 1 = 0\), so \(x = \frac{1}{2}\).
Now, calculate \(4x + 5y\) with \(x = \frac{1}{2}\) and \(y = 1\): \[ 4x + 5y = 4 \left( \frac{1}{2} \right) + 5(1) = 2 + 5 = \boxed{7} \]Nbsp;
∴ The value of \(4x + 5y\) is \(\boxed{7}\)