Answer:
S)No solution
T)Infinite solution
U)x=0
V)No solution
W) -1.2
X) No solution
Step-by-step explanation:
For solving linear equations of one variable, take variable on one side and constants on one side.
S)[tex]\frac{-1}{3}(x) - \frac{1}{2} = \frac{-1}{3}(x) +12[/tex]
[tex]\frac{1}{2} = 12[/tex] , this is false.
thus, for what ever value of x, this is false, thus, no solution.
T)[tex]\frac{-5}{6}(x) +40 = 40 + \frac{-5}{6}(x)[/tex]
[tex]40 = 40[/tex] , this is true.
thus, for what ever value of x, this is true, thus, infinite solutions.
U)[tex](-10)(x) = 0[/tex]
Thus, x= 0.
V) [tex]-6 = 6[/tex],
this is false.
thus, for what ever value of x, this is false, thus, no solution
W)[tex]\frac{5}{6}(x) = -1[/tex]
[tex]x = \frac{-6}{5} = -1.2[/tex]
X)[tex]8x -5 = 6x+2+2x = 8x +2[/tex]
[tex]-5 = 2[/tex] , which is false.
thus, for what ever value of x, this is false, thus, no solution.
