Question

X+xy+3y=3 2x-xy-y=-2

Answer 1

Из второго уравнения:

x(2-y)+(2-y)=0
(2-y)(x+1)=0. Отсюда, либо y=2, либо х =-1

При х=-1, в первом уравнении у=2
Только один ответ : (-1;2)

Answer 2

To solve the system of equations:

1. x + xy + 3y = 3
2. 2x - xy - y = -2

We can use various methods, such as substitution or elimination. Let's use the substitution method.

Step 1: Solve one of the equations for one variable in terms of the other. Let's solve equation (1) for x in terms of y: x = (3 - xy - 3y)

Step 2: Substitute the value of x from step 1 into the second equation (equation 2): 2(3 - xy - 3y) - xy - y = -2

Step 3: Simplify and solve for y: 6 - 2xy - 6y - xy - y = -2 6 - 3xy - 7y = -2 -3xy - 7y = -2 - 6 -3xy - 7y = -8

Step 4: Now, let's solve this equation for x in terms of y: -3xy = -8 + 7y x = (8 - 7y) / (-3y)

Step 5: Now that we have both x and y in terms of each other, we can find their values. Substitute the expression for x in terms of y (from step 4) into the expression for x in terms of y (from step 1):

(8 - 7y) / (-3y) + (8 - 7y) + 3y = 3

Step 6: Simplify and solve for y: [(8 - 7y) + (-3y)(8 - 7y)] / (-3y) = 3

Step 7: Continue to solve for y: [8 - 7y - 24y + 21y^2] / (-3y) = 3 [8 - 31y + 21y^2] / (-3y) = 3

Step 8: Multiply both sides by -3y to eliminate the denominator: 8 - 31y + 21y^2 = -9y

Step 9: Rearrange the equation in standard quadratic form: 21y^2 - 31y - 9y + 8 = 0 21y^2 - 40y + 8 = 0

Step 10: Solve the quadratic equation for y. You can use the quadratic formula or factoring. The quadratic equation does not factor easily, so let's use the quadratic formula:

y = [-(b) ± √(b^2 - 4ac)] / 2a

Where a = 21, b = -40, and c = 8

y = [-(40) ± √((-40)^2 - 4 * 21 * 8)] / 2 * 21 y = [40 ± √(1600 - 672)] / 42 y = [40 ± √(928)] / 42 y = [40 ± √(16 * 58)] / 42 y = [40 ± 4√58] / 42

Now, we get two possible values for y:

y₁ = (40 + 4√58) / 42 y₂ = (40 - 4√58) / 42

Step 11: Find the corresponding values of x using the expression from Step 4:

For y₁: x = (8 - 7(40 + 4√58) / (-3(40 + 4√58)) x = (8 + 280 + 28√58) / (120 + 12√58) x = (288 + 28√58) / (120 + 12√58)

For y₂: x = (8 - 7(40 - 4√58) / (-3(40 - 4√58)) x = (8 + 280 - 28√58) / (120 - 12√58) x = (288 - 28√58) / (120 - 12√58)

So the two solutions to the system of equations are:

1. x = (288 + 28√58) / (120 + 12√58), y = (40 + 4√58) / 42
2. x = (288 - 28√58) / (120 - 12√58), y = (40 - 4√58) / 42

Please note that these solutions involve square roots and may not be very simple numerical values. You can use a calculator to get approximate decimal values for these solutions.