<p>If 2x-5, x+1, and 3x-8 are all intergers, and x+1 is the median of these intergers, which of the following could be a value for x?
a. 5
b.7
c. 9
d.10
e.11</p>
<p>Do you just plug in or is there another way? I got the answer by plugging in I just want to know an easier way.</p>
<p>I think plugging them in is the fastest way. You could set up an inequality and manipulate the values, but I don't think that would help much.</p>
<p>Yeah there is an easy way to find it. If x+1 is the median, that means it falls in between the two: 2x-5<x+1 and x+1<3x-8</p>
<p>2x-5<x+1
x<6</p>
<p>So only (a) works, and you're done. But is 2x-5 necessarily the smaller one? What if it's the bigger one? Assume it's bigger and see what x would need to be for that to be true. </p>
<p>2x-5>3x-8
3>x
x<3</p>
<p>So 2x-5 is bigger only when x is less than 3, which is never true of the answers. In other words, for x>3, 2x-5 will be smaller. </p>
<p>If you want to be tidy, you can check the other inequality. </p>
<p>x+1<3x-8
9<2x
x>9/2</p>
<p>The conditions are x>4.5 and x<6. x=5 works and the other answers do not.</p>
<p>Plugging in here is probably best. You could however graph all three lines and see when the y values are integers with y=x+1 as the middle value...</p>