Question

{ x2-y = 0 > { y+6 = 31

Answer 1

$\left \{ {{x^{2}- y=0} \atop {y+6=31}} \right.\\\\\left \{ {{y=25} \atop {x^{2}=y }} \right. \\\\\left \{ {{y=25} \atop {x^{2} =25}} \right.\\\\1)\left \{ {{x=-5} \atop {y=25}} \right.\\\\2)\left \{ {{x=5} \atop {y=25}} \right.$

Ответ : (- 5 ; 25) , ( 5 ; 25)

Answer 2

It appears there is a typo in the equation you provided. The ">" sign is not used correctly. I assume you meant to write a system of equations like this:

1. x^2 - y = 0
2. y + 6 = 31

If that's the case, let's solve the system of equations:

Step 1: Solve equation (2) for y y + 6 = 31 Subtract 6 from both sides: y = 31 - 6 y = 25

Step 2: Substitute the value of y (which is 25) into equation (1) x^2 - y = 0 x^2 - 25 = 0

Step 3: Solve for x x^2 = 25 Taking the square root of both sides: x = ±√25 x = ±5

So, the solutions to the system of equations are x = 5 and y = 25.