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4x^2+4xy+y^2=25
2x-y=3

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To solve this system of equations, we can first rewrite the second equation as y = 2x - 3. Then we can substitute this expression for y into the first equation:

4x^2 + 4x(2x - 3) + (2x - 3)^2 = 25
4x^2 + 8x^2 - 12x + 4x^2 - 12x + 9 = 25
16x^2 - 24x + 9 = 25
16x^2 - 24x - 16 = 0
4x^2 - 6x - 4 = 0
(2x - 4)(2x + 1) = 0

This gives us two possible solutions for x: x = 2 or x = -1.

If x = 2:
y = 2(2) - 3
y = 4 - 3
y = 1

If x = -1:
y = 2(-1) - 3
y = -2 - 3
y = -5

Therefore, the solutions to the system of equations are:
x = 2, y = 1
or
x = -1, y = -5.