Solving Complex Number Equation

The problem requires finding the real values of $x$ and $y$ that satisfy the given complex equation $5x+2iy-ix+7y=2+3i$. We need to equate the real and imaginary parts of the equation.

Separating Real and Imaginary Parts

First, group the real terms (those without $i$) and the imaginary terms (those with $i$) on the left side of the equation:

$(5x + 7y) + i(2y - x) = 2 + 3i$

Forming Linear Equations

For two complex numbers to be equal, their real parts must be equal, and their imaginary parts must be equal. This gives us a system of two linear equations:

1. Real part equality: $5x + 7y = 2$
2. Imaginary part equality: $-x + 2y = 3$

Solving the System of Equations

We can solve this system using substitution or elimination. Let's use substitution. From the second equation, we can express $x$ in terms of $y$:

$x = 2y - 3$

Now, substitute this expression for $x$ into the first equation:

$5(2y - 3) + 7y = 2$

Simplify and solve for $y$:

$10y - 15 + 7y = 2$

$17y - 15 = 2$

$17y = 17$

$y = 1$

Substitute the value $y=1$ back into the expression for $x$:

$x = 2(1) - 3$

$x = 2 - 3$

$x = -1$

Final Values of (x,y)

The values obtained are $x = -1$ and $y = 1$. Therefore, the pair $(x,y)$ is $(-1,1)$.

This corresponds to Option A.