zad1.Rozwiąz nierłównośc
a) 3 – y < 5 + y -2y =
b) 5 (2x + 3) > 2 (x – 3) + x =
c) -3 (z – 2) – 2 ≤ z – 5 – 4x =
d) 7y – 2 (y – 4) ≥ 6 – (2 – y) =
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a) 3-y<5+y-2y
3<5+2y-2y
3-5<0
-2<0
b)5 (2x + 3) > 2 (x – 3) + x =
10 x + 15> 2x -6
10 x - 2x > -6 - 15
8x> - 21
x > 21/8
c) -3 (z – 2) – 2 ≤ z – 5 – 4z =
-3z + 6 - 2 < z - 5 - 4z
-3z +z + 4z < -5 - 6 + 2
2z
d) 7y – 2 (y – 4) ≥ 6 – (2 – y) =
7y -2y +4 > 6-2 + y