t^2-k^2<6 ; t+k>4

<p>If t and k are positive integers in the inequalities above and t>k what is the t how do I solve this algebraicly answer choice1-5.</p>

<p>T=3 because common sense</p>

<p>Clearly, t must be at least 3 (if t was 2, k would be 1, t+k is less than 4).</p>

<p>If t = 4, then k is at most 3, in which 4^2 - 3^2 = 7 which is not less than 6. If t is greater than 4, it can be shown that t^2 - k^2 would have to be larger, which doesn’t work.</p>

<p>The only possible value for t is 3.</p>

<p>plug in integers, it shouldn’t take long. But you’re missing a piece of information k cannot equal t (or be greater than t).</p>

<p>Also, a slightly more algebraic solution could be, note that</p>

<p>(t-k)(t+k) < 6</p>

<p>Since t+k > 4, and t-k is a positive integer, the only possible value for t-k is 1, so t = k+1. Then just plug in random integers, t = 3.</p>

<p>@rspense thank you.</p>

<p>What I do for this question is that:</p>

<p>1) (t-k)(t+k) < 6
So (t+k) > 4
Now, (t-k) has to be 1
Following this line of thought, t+k has to be five. Why? because t+k has to be greater than 4 but can be no greater than 6. And, since all values are positive integers, 5 is the only option.</p>

<p>So, we get:</p>

<p>t-k = 1 and t+ k = 5. From this you can say 2t =6 and t=3. </p>

<p>(You can obviously just go with the guessing too, though!)</p>