Odpowiedź :
[tex]a) \\\\\left \{ {{0=-2a+b} \atop {1=0a+b}} \right. \\\left \{ {{0=-2a+b} \atop {1=b}} \right. \\\\0=-2a+1\\2a=1 /:2\\a=\frac12\\\\y=\frac12x+1\\\\b) \\\\\left \{ {{0=2a+b} \atop {-5=0a+b}} \right. \\\left \{ {{0=2a+b} \atop {-5=b}} \right. \\\\0=2a-5\\-2a=-5 /:(-2)\\a=\frac52\\\\y=\frac52x-5\\y=\frac52x-\frac{10}2\\y=\frac{5x-10}2[/tex]
[tex]c)\\\\\left \{ {{0=4a+b} \atop {3=0a+b}} \right. \\\left \{ {{0=4a+b} \atop {3=b}} \right. \\\\0=4a+3\\-4a=3 /:(-4)\\a=-\frac34\\\\y=-\frac34x+3\\y=-\frac34x+\frac{12}4\\y=\frac{-3x+12}4\\y=\frac{-(3x-12)}4\\y=-\frac{3x-12}4[/tex]