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

Answer 1

We can rewrite the given equation as: $(2x - y)^2 + 3(y - 1)^2 = 0$

For this equation to hold true, both terms must be equal to zero: $(2x - y)^2 = 0$ and $(y - 1)^2 = 0$

Solving these equations, we get: $x = \frac{1}{2}$ and $y = 1$

Therefore, $4x + 5y = 4(\frac{1}{2}) + 5(1) = 7$

So, the value of $(4x + 5y)$ is 7.

Answer 2

We can rewrite the given equation as: $(2x - y)^2 + 3(y - 1)^2 = 0$

For this equation to hold true, both terms must be equal to zero: $(2x - y)^2 = 0$ and $(y - 1)^2 = 0$

Solving these equations, we get: $x = \frac{1}{2}$ and $y = 1$

Therefore, $4x + 5y = 4(\frac{1}{2}) + 5(1) = 7$

So, the value of $(4x + 5y)$ is 7.