👤

Zapisz równanie w postaci kierunkowej.

a) [tex]\sqrt{2y}[/tex] - 3[tex]\sqrt{6}[/tex] = 0
b) 7x + 8y - 2 = 0


Odpowiedź :

[tex]Zadanie\\\\y=ax+b\\\\a)\\\\\sqrt{2}y-3\sqrt{6}=0\ \ \ \mid+3\sqrt{6}\\\\\sqrt{2}y= 3\sqrt{6}\ \ \ \mid:\sqrt{2}\\\\y=3\sqrt{3}\\\\b)\\\\7x+8y-2=0\ \ \ \mid-7x,\ +2\\\\8y=-7x+2\ \ \ \mid:8\\\\y=-\frac{7}{8}x+\frac{1}{4}[/tex]