Question

Решите систему уравнений с параметром. Найдите все значения параметра а, при каждом из которых следующая система уравнений имеет хотя бы одно решение. {1/x-2/y=4a {2/x-6/y=3+4a Только НЕ графическим методом

Answer 1

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Answer 2

Solving the System of Equations with a Parameter

To solve the system of equations with a parameter and find all the values of the parameter 'a' for which the system has at least one solution, we will use algebraic methods rather than the graphical method.

The given system of equations is:

1/x - 2/y = 4a ---(1) 2/x - 6/y = 3 + 4a ---(2)

To solve this system, we will eliminate one variable and express it in terms of the other variable. Let's start by eliminating 'x'.

Eliminating 'x'

To eliminate 'x', we can multiply equation (1) by 2 and equation (2) by 1, and then subtract equation (2) from equation (1). This will eliminate 'x' and allow us to solve for 'y'.

2 * (1/x - 2/y) = 2 * (4a) 1 * (2/x - 6/y) = 1 * (3 + 4a)

Simplifying the equations:

2/x - 4/y = 8a ---(3) 2/x - 6/y = 3 + 4a ---(4)

Subtracting equation (4) from equation (3):

(2/x - 4/y) - (2/x - 6/y) = (8a) - (3 + 4a)

Simplifying the equation:

-2/y + 2/y = 8a - 3 - 4a

The 'x' terms cancel out, and we are left with:

-2/y + 2/y = 8a - 3 - 4a

Simplifying further:

0 = 4a - 3 - 4a

0 = -3

This equation is inconsistent and has no solution. Therefore, there are no values of 'a' for which the system of equations has at least one solution.

Conclusion

After solving the system of equations, we found that there are no values of the parameter 'a' for which the given system of equations has at least one solution.