Respuesta :

craigismadsus
Sure, let's solve the system of equations:

y = x² + 4x - 2
y = 6x - 3

To find the values of x and y that satisfy both equations, we can set them equal to each other:

x² + 4x - 2 = 6x - 3

Now, let's simplify and solve for x:

x² + 4x - 6x = -3 + 2
x² - 2x = -1

To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 1, b = -2, and c = -1. Substituting these values into the formula:

x = (-(-2) ± √((-2)² - 4(1)(-1))) / (2(1))
x = (2 ± √(4 + 4)) / 2
x = (2 ± √8) / 2
x = (2 ± 2√2) / 2

Now, let's simplify further:

x = 1 ± √2

So we have two possible values for x: x = 1 + √2 and x = 1 - √2.

To find the corresponding y-values, we can substitute these x-values into either of the original equations. Let's use the second equation y = 6x - 3:

For x = 1 + √2:
y = 6(1 + √2) - 3
y = 6 + 6√2 - 3
y = 3 + 6√2

For x = 1 - √2:
y = 6(1 - √2) - 3
y = 6 - 6√2 - 3
y = 3 - 6√2

So the solutions to the system of equations are:
(x, y) = (1 + √2, 3 + 6√2) and (x, y) = (1 - √2, 3 - 6√2).

Since you mentioned "during before 12," I assume you're referring to the time. However, the equations you provided are not related.
ayenoe13

Answer:

Step-by-step explanation:

To solve the system of equations:

{y=x2+4x−2y=6x−3​

We can substitute the second equation into the first equation to get:

6x−3=x2+4x−2

Rearranging the terms, we get:

x2−2x+1=0

This is a perfect square trinomial, which can be factored as:

(x−1)2=0

Therefore, the only solution to the system of equations is:

{x=1y=3​

Thus, the solution to the system of equations is x = 1 and y = 3.