<p>Hi, I was correcting my math section, and this 4 exercises are the ones I made a mistake and can't understand the reason of the answer. So the idea is that you, math genius, explain me how to get to the result shown.</p>
<p>Ok, thanks in advance, and I really hope you can explain these answers to me :)
bye</p>
<ol>
<li>Do you understand the concept of absolute value? You can make this into two equations, and restate the problem:</li>
</ol>
<p>Find a possible value such that </p>
<p>6 < x - 3 < 7 OR -7 < x-3 < -6</p>
<p>But x has to be less than 0, and the only solutions to the first equation are positive, so solve the latter one by first adding 3 to all sides:</p>
<p>-7 + 3 < x - 3 + 3 < -6 + 3 ==> -4 < x < -3</p>
<ol>
<li>y can be any integer between 7 and 16; i.e., 8,9,10,11,12,13,14, or 15</li>
</ol>
<p>So plug and chug:</p>
<p>x/8 = 2/5 => x = 2/5<em>8 = 16/5, which is not an integer
x/9 = 2/5 => x = 2/5</em>9 = 18/5, which is not an integer
x/10 = 2/5 => x = 2/5<em>10 = 4, which is an integer
x/11 = 2/5 => x = 2/5</em>11 = 22/5, not an integer
x/12 = 2/5 => x = 2/5<em>12 = 24/5, not an integer
x/13 = 2/5 => x = 2/5</em>13 = 26/5, not an integer
x/14 = 2/5 => x = 2/5<em>14 = 28/5, not an integer
x/15 = 2/5 => x = 2/5</em>15 = 6, which is an integer</p>
<p>It's obvious that y has to be a multiple of 5 for this to work, but since there aren't many values, we can just calculate them all through if you don't have that sort of insight. Counting up, we have 2 values of x which are integers.</p>
<ol>
<li>We have to find an integer such that 1/x + x is an integer for any integer x. It's obvious that only comes out to an integer if x = -1 or 1, so plug them in, and the two possible values for t are -2 and 2. If you don't believe me that these are the only values for x that work, remember that an integer + a non-integer is a non-integer.</li>
</ol>
<h1>1: I used some algebra and trial and error:</h1>
<p>YOu know that j and n must be two digits from each other since they are in consecutive order and separated by one digit less than n and 1 greater than n.</p>
<p>then, start with 1 and multiply by a number 2 greater, aka 3. 1<em>3=3, which doesn't end in a 9. continue: 3</em>5=15. no again. (adn u can't use two even numbers because even*even=end in even digits).</p>
<p>then, a few more steps later, you get 9*11=99, which fits the criteria of ending in 9.</p>
<p>k, one less than n (11) and one more than 9 = 10, whose one digit is 0.</p>
<h1>3:</h1>
<p>Really, you must find the appropriate y values. since each y is paired with an x, you are also finding the number of x's. Since these are integers, y and x must be multiples of 2 and 5. the only multiples of 5 (y) between 7 and 16 are 10 and 5, so there are two y's and thus only two corresponding x's (4, when y=10, and 6 when y=15)</p>
<h1>4: It's easiest to plug in values for x, and not to use the choices provided.</h1>
<p>then you can see that x + 1/x = t (just substitute the function, the symbol, with x + 1/x).</p>
<p>now, plug in a very easy number to understand the function simply. if x= 1, t= 1+ 1/1, or two. however ,this number is positive and all the choices are negative. therefore, you must use negative x values.</p>
<p>the easiest to plug in is -1. -1 + -1/1 = t ==> -1+-1= -2, choice d</p>
<h1>4</h1>
<p>think of it as f(x) rather than the triangle thing.</p>
<p>you're given that </p>
<p>f(x) = x + (1/x)</p>
<p>where x is an integer</p>
<p>so if </p>
<p>f(x) = t</p>
<p>where t is also an integer</p>
<p>essentially, what you have to do is make t = x + (1/x) where x is an integer</p>
<p>t can only equal -2 because you can't plug in any integer for x and have it equal any other choice.</p>
<p>ie. -1 + (1/-1) = -2
or x + (1/x) = t
t = -2</p>
<p>Thanks a lot to the three of you! I posted this less than an hour ago.</p>
<p>I really appreciate it, thanks again!</p>
<p>PS : Yes, I understood all of them. :D</p>
<p>For the absolute value one, when you got the second equation, did you just multiply through by negative one?</p>
