Question

Решите систему уравнений

Answer 1

$\left \{ {{4x+3y=-1} \atop {2x^2-y=11}} \right.$
Из уравнения 2 выразим переменную у
$\left \{ {{4x+3y=-1} \atop {y=2x^2-11}} \right.$

$4x+3(2x^2-11)=-1 \\ 4x+6x^2-33=-1 \\ 6x^2+4x-32=0|:2 \\ 3x^2+2x-16=0 \\ D=b^2-4ac=4+192=196;\,\, \sqrt{D} =14$

$x_1= \frac{-2+14}{6}=2;\,\,\,\,\,\,y_1=2\cdot 4-11=-3$
$x_2= \frac{-2-14}{6} =- \frac{8}{3} \,\,\,\,\,\,y_2=\frac{29}{9}$

Answer 2

System of Equations

To solve the system of equations, we will use the information provided in the search results snippets. Here are a few examples of system of equations:

Example 1: 2x - y = 7 x^2 - xy = 12

Example 2: 3x - y = 3 3x - 2y = 0

Example 3: 2ab - 3a/b = 15 ab + a/b = 10

Example 4: y = x - 8 xy = -7

Example 5: 2x + y = 4 x - 2y = 5

Let's solve each of these examples step by step.

Example 1:

2x - y = 7 x^2 - xy = 12

To solve this system of equations, we can use the substitution method. We can solve the first equation for y and substitute it into the second equation:

From the first equation, we have: 2x - y = 7 y = 2x - 7

Substituting y in the second equation: x^2 - x(2x - 7) = 12 x^2 - 2x^2 + 7x = 12 -x^2 + 7x = 12 x^2 - 7x + 12 = 0

Factoring the quadratic equation: (x - 3)(x - 4) = 0

Setting each factor equal to zero: x - 3 = 0 or x - 4 = 0

Solving for x: x = 3 or x = 4

Substituting the values of x back into the first equation to find y: For x = 3: 2(3) - y = 7 6 - y = 7 y = -1

For x = 4: 2(4) - y = 7 8 - y = 7 y = 1

Therefore, the solutions to the system of equations are: (x, y) = (3, -1) and (4, 1).

Example 2:

3x - y = 3 3x - 2y = 0

To solve this system of equations, we can use the elimination method. We can multiply the first equation by 2 and subtract it from the second equation to eliminate x:

Multiplying the first equation by 2: 2(3x - y) = 2(3) 6x - 2y = 6

Subtracting the first equation from the second equation: (3x - 2y) - (6x - 2y) = 0 - 6 3x - 2y - 6x + 2y = -6 -3x = -6 x = 2

Substituting the value of x back into the first equation to find y: 3(2) - y = 3 6 - y = 3 y = 3

Therefore, the solution to the system of equations is: (x, y) = (2, 3).

Example 3:

2ab - 3a/b = 15 ab + a/b = 10

To solve this system of equations, we can use the substitution method. We can solve the second equation for a/b and substitute it into the first equation:

From the second equation, we have: ab + a/b = 10 a/b = 10 - ab

Substituting a/b in the first equation: 2ab - 3(10 - ab) = 15 2ab - 30 + 3ab = 15 5ab - 30 = 15 5ab = 45 ab = 9

Substituting the value of ab back into the second equation to find a/b: a/b = 10 - 9 a/b = 1

Therefore, the solution to the system of equations is: (a/b, ab) = (1, 9).

Example 4:

y = x - 8 xy = -7

To solve this system of equations, we can substitute the value of y from the first equation into the second equation:

Substituting y in the second equation: x(x - 8) = -7 x^2 - 8x = -7 x^2 - 8x + 7 = 0

Factoring the quadratic equation: (x - 7)(x - 1) = 0

Setting each factor equal to zero: x - 7 = 0 or x - 1 = 0

Solving for x: x = 7 or x = 1

Substituting the values of x back into the first equation to find y: For x = 7: y = 7 - 8 y = -1

For x = 1: y = 1 - 8 y = -7

Therefore, the solutions to the system of equations are: (x, y) = (7, -1) and (1, -7).

Example 5:

2x + y = 4 x - 2y = 5

To solve this system of equations, we can use the elimination method. We can multiply the second equation by 2 and add it to the first equation to eliminate y:

Multiplying the second equation by 2: 2(x - 2y) = 2(5) 2x - 4y = 10

Adding the second equation to the first equation: (2x + y) + (2x - 4y) = 4 + 10 4x - 3y = 14

Simplifying the equation: 4x - 3y = 14

There are infinitely many solutions to this system of equations since we have only one equation with two variables.

Therefore, the solution to the system of equations is: (x, y) = (x, 4x - 14), where x is any real number.

Please let me know if you need help with anything else!