Question

Решите методом сложения систему уравнений :

Answer 1

{7x+3y=43
[4x-3y=67
прибавим
11x=110
x=110:11
x=10
70+3y=43
3y=43-70
3y=-27
y=-27:3
y=-9
(10;-9)

Answer 2

Solving the System of Equations using the Method of Addition:

To solve the system of equations, we will use the method of addition. The given system of equations is:

Equation 1: \(x^2 + y^2 = 20\) (1) Equation 2: \(3x + y = 2\) (2)

To eliminate one variable, we can multiply Equation 2 by 3 and subtract it from Equation 1:

\(3(x^2 + y^2) - (3x + y) = 3(20) - 2\)

Simplifying the equation gives us:

\(3x^2 + 3y^2 - 3x - y = 60 - 2\)

\(3x^2 + 3y^2 - 3x - y = 58\) (3)

Now, let's solve the resulting equation for one variable. We can rearrange Equation 2 to solve for y:

\(y = 2 - 3x\) (4)

Substituting Equation 4 into Equation 3, we get:

\(3x^2 + 3(2 - 3x)^2 - 3x - (2 - 3x) = 58\)

Simplifying the equation gives us:

\(3x^2 + 3(4 - 12x + 9x^2) - 3x - 2 + 3x = 58\)

\(3x^2 + 12 - 36x + 27x^2 - 3x - 2 + 3x = 58\)

Combining like terms, we have:

\(30x^2 - 36x + 10 = 58\)

Rearranging the equation, we get:

\(30x^2 - 36x - 48 = 0\) (5)

Now, we can solve Equation 5 for x using the quadratic formula:

\(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\)

For Equation 5, the values of a, b, and c are:

\(a = 30\) \(b = -36\) \(c = -48\)

Substituting these values into the quadratic formula, we get:

\(x = \frac{{-(-36) \pm \sqrt{{(-36)^2 - 4(30)(-48)}}}}{{2(30)}}\)

Simplifying the equation gives us:

\(x = \frac{{36 \pm \sqrt{{1296 + 5760}}}}{{60}}\)

\(x = \frac{{36 \pm \sqrt{{7056}}}}{{60}}\)

\(x = \frac{{36 \pm 84}}{{60}}\)

Simplifying further, we have:

\(x_1 = \frac{{36 + 84}}{{60}} = \frac{{120}}{{60}} = 2\)

\(x_2 = \frac{{36 - 84}}{{60}} = \frac{{-48}}{{60}} = -\frac{{4}}{{5}}\)

Now that we have the values of x, we can substitute them back into Equation 4 to find the corresponding values of y:

For \(x_1 = 2\): \(y_1 = 2 - 3(2) = 2 - 6 = -4\)

For \(x_2 = -\frac{{4}}{{5}}\): \(y_2 = 2 - 3\left(-\frac{{4}}{{5}}\right) = 2 + \frac{{12}}{{5}} = \frac{{22}}{{5}}\)

Therefore, the solutions to the system of equations are: \(x_1 = 2, y_1 = -4\) \(x_2 = -\frac{{4}}{{5}}, y_2 = \frac{{22}}{{5}}\)

Solving the Second System of Equations:

The second system of equations is:

Equation 1: \(xy - 3y^2 = -24\) (6) Equation 2: \(xy + 2y^2 = 21\) (7)

To solve this system, we can use the method of substitution. Let's solve Equation 1 for x:

\(x = \frac{{-24 + 3y^2}}{{y}}\) (8)

Substituting Equation 8 into Equation 2, we get:

\(\left(\frac{{-24 + 3y^2}}{{y}}\right)y + 2y^2 = 21\)

Simplifying the equation gives us:

\(-24 + 3y^2 + 2y^3 = 21y\)

Rearranging the equation, we have:

\(2y^3 + 3y^2 - 21y + 24 = 0\) (9)

To solve Equation 9, we can try different values of y and check if they satisfy the equation. By inspection, we can see that y = -3 is a solution. Dividing Equation 9 by (y + 3), we get:

\((y + 3)(2y^2 - 3y + 8) = 0\)

The quadratic equation \(2y^2 - 3y + 8 = 0\) does not have real solutions. Therefore, the only solution for y is y = -3.

Substituting y = -3 into Equation 8, we can find the corresponding value of x:

\(x = \frac{{-24 + 3(-3)^2}}{{-3}}\)

Simplifying the equation gives us:

\(x = \frac{{-24 + 27}}{{-3}} = \frac{{3}}{{-3}} = -1\)

Therefore, the solution to the second system of equations is: \(x = -1, y = -3\)

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