[tex]\left \{ {{x+y=4} \atop {x^{2}-y= 2}} \right.[/tex]

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вопрос

We can solve this system of equations by substitution or elimination. Let's use substitution method here.

We have the equations:
1) x + y = 4
2) x^2 - y = 2

From equation 1, we can write y = 4 - x. Now, substitute this value of y into equation 2:

x^2 - (4 - x) = 2
x^2 - 4 + x = 2
x^2 + x - 2 = 0
(x + 2)(x - 1) = 0

So, either x + 2 = 0 or x - 1 = 0
If x + 2 = 0, then x = -2. Substituting this back into the first equation:
-2 + y = 4
y = 6
So, one solution is x = -2 and y = 6.

If x - 1 = 0, then x = 1. Substituting this back into the first equation:
1 + y = 4
y = 3
So, the other solution is x = 1 and y = 3.

Therefore, the solutions to the system of equations are x = -2, y = 6 and x = 1, y = 3.