<ol>
<li>If the absolute value of ax-1 is less than or equal to 1, where a is a positive even integer, which on the following CANNOT be a value of x^2?</li>
</ol>
<p>a. 0
b. 1/4
c. 1/2
d. 1
e. 4</p>
<ol>
<li>If x/y=k and k>0, what is x+y/x in terms of k?</li>
</ol>
<p>it's easy to do assigning values, but how do you do it algebraically?
the answer is k+1/k</p>
<p>For (1) write the inequality as:</p>
<p><a href=“2nx-1”>1</a> is greater or equal to -1
and
<a href=“2nx-1”>2</a> is less than or equal to 1. </p>
<p>Where n is a positive integer. The reason for “2n” in the inequalities is because of the requirement “positive even integer”.</p>
<p>The first inequality has the solution that x is greater or equal to 0.</p>
<p>The second inequality has the solution that x is less than or equal to 1/n.</p>
<p>So x is greater or equal to 0 and less than or equal to 1/n. So x can never be greater than 1, and therefore x^2 can’t be greater than 1.</p>
<p>Choice (e) is not possible.</p>
<p>There is a likely typo in your transcription of question (2).</p>
<p>If x/y=k and k>0, what is (x+y)/x in terms of k?</p>
<p>sorry about that!</p>
<p>(x+y)/x=x/x+y/x=1+1/(x/y)=1+1/k</p>
<p>RandomHSer is correct, but it’s better to put parentheses to remove ambiguity:
(x+y)/x = (x/x) + (y/x) = 1 + (1/(x+y)) = 1 + (1+k).</p>
<p>Or in LaTeX:
$\frac{x+y}{x} = \frac{x}{x} + \frac{y}{x} = 1 + \frac{1}{\frac{x}{y}} = 1 + \frac{1}{k}$</p>