Question

Решить систему методом подстановки х+у=4 у+ху=6

Answer 1

$\left \{ {{x+y=4} \atop {y+xy=6}} \right. \; \; \left \{ {{x=4-y} \atop {y+y(4-y)=6}} \right. \; \; \left \{ {{x=4-y} \atop {y+4y-y^2=6}} \right. \; \; \left \{ {{x=4-y} \atop {y^2-5y+6=0}} \right. \\\\ \left \{ {{x=4-y} \atop {(y-2)(y-3)=0}} \right. \; \; \left \{ {{x_1=2\; ,\; x_2=1} \atop {y_1=2\; ,\; y_2=3}} \right. \\\\Otvet:\; \; (2,2)\; ,\; (1,3)\; .$

Answer 2

System of Equations: x + y = 4, y + xy = 6

To solve the system of equations x + y = 4 and y + xy = 6 using the method of substitution, we will solve one equation for one variable and substitute it into the other equation.

Let's solve the first equation, x + y = 4, for x: - Subtract y from both sides: x = 4 - y. Now, substitute this value of x into the second equation, y + xy = 6: - Replace x with 4 - y: y + (4 - y)y = 6. - Simplify the equation: y + 4y - y^2 = 6. - Combine like terms: 5y - y^2 = 6.

We now have a quadratic equation in terms of y. Let's solve it to find the values of y:

- Rearrange the equation: y^2 - 5y + 6 = 0. - Factor the quadratic equation: (y - 2)(y - 3) = 0. - Set each factor equal to zero and solve for y: - y - 2 = 0 → y = 2. - y - 3 = 0 → y = 3.

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

- For y = 2: x = 4 - 2 → x = 2. - For y = 3: x = 4 - 3 → x = 1.

Therefore, the solution to the system of equations x + y = 4 and y + xy = 6 is (x, y) = (2, 2) and (1, 3).

Please let me know if you need any further clarification or assistance!