Answer:Problem 1:a. x=2b. x=3c. x=1Problem 2: A multiplication equation to hold the table true:[tex]18\times3=54[/tex]A division equation to hold the table true:[tex]\frac{54}{3} =18[/tex]Step-by-step explanation:Given in problem 1:(a). The equation is [tex]\frac{x}{6} > 1[/tex]It holds true for all values of [tex]x> 6[/tex].Let us say [tex]x=12[/tex], [tex]\frac{x}{6}=\frac{12}{6} =2[/tex] which is greater than 1.(b). The equation is [tex]\frac{x}{6} < 1[/tex]It holds true for all values of [tex]x< 6[/tex].Let us say [tex]x=3[/tex][tex]\frac{x}{6} =\frac{3}{6} =\frac{1}{2}[/tex] which is less than 1.(c). The equation is [tex]\frac{x}{6} = 1[/tex]It holds true for only [tex]x= 6[/tex].Let us say [tex]x=6[/tex],[tex]\frac{x}{6} =\frac{6}{6}=1[/tex] which is equal to 1.Problem 2: A multiplication equation to hold the table true:[tex]18\times3=54[/tex]A division equation to hold the table true:[tex]\frac{54}{3} =18[/tex]Therefore these are the values which hold true to the equation in problem 1 and 2.