2x+y=7
log3(x^2+4x-3) - log3y=1
y = 7 - 2x
log3(x^2+4x-3) - log3(7 - 2x) = 1
log3[(x^2+4x-3) /(7 - 2x)] = 1
(x^2+4x-3) /(7 - 2x) = 3
[(x^2 + 4x - 3 - 21 + 6x)] / (7 - 2x) = 0
x^2 + 10x - 24 = 0
7 - 2x ≠0, x ≠ 3,5
x1 = - 12
x2 = 2
y1 = 7 - 2*(-12)
y1 = 31
y2 = 7 - 2*2
y2 = 3
Ответ: (-12;31) (2;3)