Question

Решите систему уравнений,используя способ подстановки {x=3-y{y2-x=39 Посмотреть ответы valeriyayezhov valeriyayezhov y^2 - (3-y)=39           y^2 - 42+y=0 y^2 + y - 42=0 y=6 y= -7 x= 3-6       x= -3 x = 3-(-7)     x=10 Ответ: (-3, 6); (10, -7) askerbrainly askerbrainly y^2-(3-y)=39 y^2-3+y=39 y^2+y=42 y^2+y-42=0 D=b^2-4ac=1+168=169 √D=13

Answer 1

To solve the system of equations using the method of substitution, we have the following equations:

1. $y^2 - (3 - y) = 39$
2. $y^2 - x = 39$

Let's proceed with the substitution method:

Step 1: Solve the first equation for $x$: $y^2 - (3 - y) = 39$

Expand the bracket: $y^2 - 3 + y = 39$

Combine like terms: $y^2 + y - 42 = 0$

Step 2: Factor the quadratic equation: $(y + 7)(y - 6) = 0$

Step 3: Solve for $y$ from the factors: Either $y + 7 = 0 \implies y = -7$ or $y - 6 = 0 \implies y = 6$.

Step 4: Substitute the values of $y$ back into the second equation to find $x$.

For $y = -7$: $x = y^2 - 39 = (-7)^2 - 39 = 49 - 39 = 10$

For $y = 6$: $x = y^2 - 39 = 6^2 - 39 = 36 - 39 = -3$

So, the solutions to the system of equations are: $x = 10, y = -7$ $x = -3, y = 6$

Hence, the correct answer is $(-3, 6)$ and $(10, -7)$.