Question

Розв'язати графічно систему рівняння {xy= -12 {x+y=1 Смотреть ответ D281993 D281993 Ответ:(х1у1)= 4,-3(х2у2)=-3,4Объяснение:{xy= -12{x+y=1{xy= -12{x=1-у(1-у)у=-12{y= -3{y=4х=1-(-3)х=1-4х=4х=-3(х1у1)= 4,-3(х2у2)=-3,4{4*(-3)=-12{4=(-3)=11 {-3*4=-12{-3+4=12 {-12=-12{1=1{-12=-12{1=1(х1у1)= 4,-3(х2у2)=-3,4 Новые вопросы в Алгебра Нарисуйте график некоторой функции, возрастающей на интервале (-∞; 2) и убывающей на интервале (2; +∞).​ 10^√p^30×q^40 якщо меньше Або дорівнює 0 Даю 100 баллов! Скоротіть дріб: [tex]\frac{12a}{4a^{2}+a }[/tex]

Answer 1

Graphical Solution of the System of Equations

To solve the system of equations graphically, we will plot the graphs of the two equations and find the points of intersection. Let's start by graphing the equations:

Equation 1: xy = -12 Equation 2: x + y = 1

To graph the first equation, we can rearrange it to solve for y: y = -12/x

Now, let's plot the graphs:

- For the first equation, we can choose different values of x and calculate the corresponding values of y using the equation y = -12/x. For example, when x = -1, y = 12; when x = 1, y = -12; when x = 2, y = -6; and so on. Plot these points on the graph.

- For the second equation, we can rearrange it to solve for y: y = 1 - x

Now, let's plot the graph of the second equation:

- When x = 0, y = 1; when x = 1, y = 0; when x = 2, y = -1; and so on. Plot these points on the graph.

By plotting the graphs of both equations, we can find the points of intersection, which represent the solutions to the system of equations.

Solution to the System of Equations

From the graph, we can see that the two graphs intersect at two points: (4, -3) and (-3, 4). These are the solutions to the system of equations.

Therefore, the solution to the system of equations {xy = -12, x + y = 1} is: - (x1, y1) = (4, -3) - (x2, y2) = (-3, 4)

Simplifying the Fraction

The fraction you provided is: [tex]\frac{12a}{4a^{2}+a+}[/tex]

To simplify this fraction, we need to factor the denominator. Let's factor the denominator:

4a^2 + a

The common factor in both terms is 'a', so we can factor it out:

a(4a + 1)

Now, let's rewrite the fraction with the factored denominator:

[tex]\frac{12a}{a(4a + 1)}[/tex]

Next, we can cancel out the common factor 'a' in the numerator and denominator:

[tex]\frac{12}{4a + 1}[/tex]

Therefore, the simplified form of the fraction [tex]\frac{12a}{4a^{2}+a+}[/tex] is [tex]\frac{12}{4a + 1}[/tex].

I hope this helps! Let me know if you have any further questions.