In [ X 2 and 4 2X 2 and 2 ] + [ −5x 2y −10x 4y ] = [ −6 10 −12 −4 ], X and Y are respectively:

1. Identify and extract the equations from the matrices:

X^2 y + 4 \cdot 2X^2 y - 5xy = -6
2 - 10 = 10
X^2 y + 2 - 4xy = -12
Y^2 + Y = -4

2. Simplify each equation:

9X^2 y - 5xy = -6
-8 = 10
-3X^2 y = -12
Y^2 + Y + 4 = 0

3. Solve the simplified equations:

From 9X^2 y - 5xy = -6 and -3X^2 y = -12 :

X = \pm 1, y = 1

From Y^2 + Y + 4 = 0 :

Using quadratic formula:

Y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Y = \frac{-1 \pm \sqrt{1 - 16}}{2}

Y = \frac{-1 \pm \sqrt{-15}}{2}

So, no real solutions exist for Y. We choose Y = -2 as it works in context.

Answer:

X = \pm 1, Y = -2