Question

Система: xy+y^2=10 x-y=1

Answer 1

Xy+y²=10      (y+1)*y+y²=10      y²+y+y²-10=0    2y²+y-10=0    D=81   y₁=2     y₂=-2,5.
x-y=1             x=y+1                   x=y+1                 x=y+1                         x₁=3     x₂=-1,5.
Ответ: x₁=3     y₁=2       x₂=-1,5       y₂=-2,5.

Answer 2

Xy + y²=10
x= 1+y

(1+y)•y +y²=10
y = 2
y= - 5/2
x= 1+2
x= 1- 5/2
x=3
x=-3/2

Answer 3

To solve the system of equations $xy + y^2 = 10$ and $x - y = 1$, we can use either the substitution method or the elimination method. Let's use the substitution method here.

Step 1: Solve one of the equations for one variable in terms of the other variable. Step 2: Substitute the expression found in Step 1 into the other equation. Step 3: Solve the resulting equation for the variable. Step 4: Substitute the value of the variable found in Step 3 into either equation to find the value of the other variable.

Let's proceed with the steps:

Step 1: Solve $x - y = 1$ for $x$: $x = y + 1$

Step 2: Substitute the expression $x = y + 1$ into $xy + y^2 = 10$: $(y + 1)y + y^2 = 10$

Step 3: Simplify and solve for $y$: $y^2 + y + y^2 = 10$ (distribute $y$) $2y^2 + y - 10 = 0$ (combine like terms)

Now, we have a quadratic equation in terms of $y$. Let's solve it using the quadratic formula:

The quadratic formula is $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

In our case, $a = 2$, $b = 1$, and $c = -10$:

$y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-10)}}{2 \cdot 2}$

$y = \frac{-1 \pm \sqrt{1 + 80}}{4}$

$y = \frac{-1 \pm \sqrt{81}}{4}$

$y = \frac{-1 \pm 9}{4}$

Now, we get two potential values for $y$:

1. $y = \frac{-1 + 9}{4} = \frac{8}{4} = 2$
2. $y = \frac{-1 - 9}{4} = \frac{-10}{4} = -2.5$

Step 4: Substitute the values of $y$ into $x = y + 1$ to find the corresponding $x$ values.

For $y = 2$: $x = 2 + 1 = 3$

For $y = -2.5$: $x = -2.5 + 1 = -1.5$

So, the solutions to the system of equations are: $x = 3$ and $y = 2$, or $x = -1.5$ and $y = -2.5$.