Question 259691
-6 < 2-3x < 10 is your equation:


subtract 2 from all sides of the equation to get:


-8 < -3x < 8


multiply all sides of the equation by -1 to get:


8 > 3x > -8


divide all sides of the equation by 3 to get:


8/3 > x > -8/3


you have:


x > -8/3 and x < 8/3


this translates to:


x > -2.66666666... and x < 2.66666666...


unless it gets real close, I'll use x > -2.6 and x < 2.6 as the limits test.


when x = -3, x is smaller than -2.6 so x = -3 is NOT valid.
when x = -2, x > -2.6 and x < 2.6 so this is good.
when x = -1, x > -2.6 and x < 2.6 so this is good.
when x = 0, x > -2.6 and x < 2.6 so this is good.
when x = 2, x > -2.6 and x < 2.6 so this is good.


looks like selection A is the invalid one.


to confirm, plug that value of x into your original equation.


original equation is:


-6 < 2-3x < 10


substitute (-3) for x into that equation to get:


-6 < 2 - (3*(-3)) < 10 which becomes:


-6 < 2 - (-9) < 10 which becomes:


-6 < 2 + 9 < 10 which becomes:


-6 < 11 < 10.


-6 is smaller than 11 so that part is good.
11 is NOT smaller than 10 so that part is NOT good.


This confirms that selection A cannot be a possible value of x.


If you plug all the other possible values of x into the original equation, you will see that they can be possible values of x.


your answer is selection A.


you showed selection E as 2 A.


I assumed this was a typographical error so I took out the A and left the 2 to make selection E = 2.