Question

Система x-3y =4, xy-7y =6​

Answer 1

$\left \{ {{x-3y=4} \atop {xy-7y=6}} \right.\\\\\left \{ {{x=3y+4} \atop {(3y+4)*y-7y=6}} \right.\\\\\left \{ {{x=3y+4} \atop {3y^{2}+4y-7y-6=0 }} \right. \\\\\left \{ {{x=3y+4} \atop {3y^{2}-3y-6=0 }} \right.\\\\\left \{ {{x=3y+4} \atop {y^{2}-y-2=0 }} \right.$

$\left \{ {{x=3y+4} \atop {\left[\begin{array}{ccc}y_{1}=2 \\y_{2}=-1-teorema.Vieta \end{array}\right }} \right.\\\\\left[\begin{array}{ccc}\left \{ {{y_{1} =2} \atop {x_{1} =3*2+4}} \right. \\\left \{ {{y_{2}=-1 } \atop {x_{2} =3*(-1)+4}} \right. \end{array}\right\\\\\left[\begin{array}{ccc}\left \{ {{x_{1}=10 } \atop {y_{1} =2}} \right. \\\left \{ {{x_{2}=1 } \atop {y_{2} =-1}} \right. \end{array}\right\\\\Otvet:\boxed{(10;2),(1;-1)}$

Answer 2

To solve the system of equations:

1. x - 3y = 4
2. xy - 7y = 6

We can use substitution or elimination method. Let's solve it using the substitution method:

From equation 1, we have x = 3y + 4. Now substitute this value of x into equation 2:

(3y + 4)y - 7y = 6 3y^2 + 4y - 7y = 6 3y^2 - 3y = 6 3(y^2 - y) = 6 y^2 - y = 2

Now we have a quadratic equation. Rearrange it to standard form:

y^2 - y - 2 = 0

We can factorize the equation:

(y - 2)(y + 1) = 0

So either y - 2 = 0 or y + 1 = 0:

1. If y - 2 = 0, then y = 2. Substitute this value of y back into equation 1:

x - 3(2) = 4 x - 6 = 4 x = 4 + 6 x = 10

So one solution is x = 10 and y = 2.

2. If y + 1 = 0, then y = -1. Substitute this value of y back into equation 1:

x - 3(-1) = 4 x + 3 = 4 x = 4 - 3 x = 1

So another solution is x = 1 and y = -1.

Therefore, the system of equations has two solutions: (x = 10, y = 2) and (x = 1, y = -1).