Question

Решите систему

Answer 1

$\left \{ {{4x-3y=-3} \atop {-10x-6y=3}} \right.$
Решим систему методом добавления
$\left \{ {{4x-3y=-3}*(-2) \atop {-10x-6y=3}} \right.$
$\left \{ {{-8x+6y=6} + \atop {-10x-6y=3}} \right.$
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-18х=9
х=9:(-18)
х=-0,5
Подставим значение х в 1-е уравнение
4*(-0,5)-3у=-3
-2-3у=-3
-3у=-3+2
-3у=-1
у=-1:(-3)
у=1/3
(-0,5; 1/3)
Ответ (-0,5; 1/3)

Answer 2

System of Equations

To solve the given system of equations, we need to find the values of the variables that satisfy both equations. Let's analyze the provided equations and find the solution.

Equation 1: 2x - y = 7Equation 2: x^2 - xy = 12 To solve this system, we can use substitution or elimination methods. Let's use the substitution method.

From Equation 1, we can express y in terms of x: y = 2x - 7

Substituting this value of y into Equation 2, we get: x^2 - x(2x - 7) = 12

Simplifying the equation: x^2 - 2x^2 + 7x = 12 -x^2 + 7x = 12 x^2 - 7x + 12 = 0

Now, we can solve this quadratic equation to find the values of x.

Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 1, b = -7, and c = 12.

Substituting the values into the formula: x = (-(-7) ± √((-7)^2 - 4(1)(12))) / (2(1)) x = (7 ± √(49 - 48)) / 2 x = (7 ± √1) / 2 x = (7 ± 1) / 2

So, we have two possible values for x: x = (7 + 1) / 2 = 8 / 2 = 4 x = (7 - 1) / 2 = 6 / 2 = 3

Now, let's substitute these values of x back into Equation 1 to find the corresponding values of y.

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

For x = 3: 2(3) - y = 7 6 - y = 7 y = 6 - 7 y = -1

Therefore, the solution to the given system of equations is: (x, y) = (4, 1) and (x, y) = (3, -1).