Answer:
The y-coordinate of the solution is, 10
Explanation:
Given the system of equation:
....[1]
....[2]
Equate the equation [1] and [2] we have;

Add 2x to both sides of an equation:

Subtract 4 from both sides we have;

or

Using perfect square:

⇒We can write the equation as:

then;

⇒

Subtract 3 from both sides we have;
x = -3
Substitute value of x in [1] we have;
⇒
Solution for the given system of equation = (-3, 10)
Therefore, the y-coordinate of the solution is, 10