Answer:
Problem 1:
a. x=2
b. x=3
c. x=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]
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.
