<p>Didn't see this one in the consolidated list.</p>
<p>I know how to simplfy everything down to get rid of the dumb symbols, but then how else can it be done without guess and check?</p>
<p>ok heres how i did it...
* = the square
x<em>y = xy-y
I. a</em>b
II. (a+b)<em>b
III. a</em>(a+b)</p>
<p>plugin 1 for both a and b in all instances and you find that you get I and III are the only ones that work</p>
<p>Plugging in 1 for both a and b proves that I and III can be equal to zero.
It does not prove that II can not be zero for some a and b.</p>
<ul>
<li>= the square
x<em>y = xy - y = y(x - 1)
II. (a+b)</em>b = b(a + b - 1) can't be equal to zero because
a and are positive integers, so a>=1 and b>=1.</li>
</ul>
<p>
In this case, you don’t actually need to prove that, because there’s no “all of them” answer. Just an example of a quick way to use your options on the MC section.</p>
<p>Wow you really dug this post up :)</p>