<p>t^2 - k^2 < 6
t + k > 4
If t and k are positive integers in the inequalities above and t > k what is the value of t?</p>
<p>This is an SAT question. How do I solve it?
Also, is there a way to solve this algebraically, instead of guessing?</p>
<p>(t+k)(t-k)<6
t+k>4
t>k
t-k>0
t+k can be 5
t-k will be 1
so t=3</p>
<p>It’s testing if you can play recognize a difference of perfect squares and then solve a simple 2 simultaneous equation problem. t^2 - k^2 = (t+k)(t-k). Because of the restrictions, t and k are positive integers AND t+k>4, you notice that t+k could be 5 and then t-k must be 1. Hmm… t+k can’t be 6,7,8, etc. This is the only combination that works! So t+k=5 and t-k=1… 2t=6, so t=3 (and k=2).</p>
<p>Don’t be afraid to use your pencil to play around with whatever they give you. Sometimes on the trickier problems you won’t see the solution from the very start, but you can “see” the solution once you start writing down stuff. Many problems aren’t strictly algebra, but involve some sort of insight which leads to a shortcut that allows you to solve the problem within 90 seconds.</p>
<p>Help? I’ve read the responses and I don’t get it at all.</p>
<p>@BassGuitar:</p>
<p>We’re given three conditions:
t^2 - k^2 < 6
t + k > 4
t > k</p>
<p>The key thing to notice in this example is t^2 - k^2. Once you practice more problems involving the difference of squares, you’ll begin to automatically factor out t^2 - k^2 into (t + k)(t - k). So, we know that (t + k)(t - k) is less than 6. We are also given that t + k is greater than 4. Automatically, you should think of the number 5. This is because if t + k = 6 or higher, t^2 - k^2 can’t be less than 6.</p>
<p>So, let’s have t + k = 5 and (t + k)(t - k) = 5. That means t - k = 1.</p>
<p>Given the equations:
t + k = 5
t - k = 1
We get 2t = 6 when we add the two equations together. That means t = 3.</p>
<p>Double check the solution t = 3, k = 2 and you can see that all three conditions are satisfied.</p>
<p>Hope this helps! :)</p>
<p>I don’t think of this is a factoring problem, though it can certainly be done that way. My favorite method for this one is definitely playing around with numbers. It would be hard to spend more than 30 seconds before stumbling on the right answer.</p>
<p>Another way: look at the squares of integers: 0, 1, 4, 9, 16, 25…</p>
<p>There are only a couple ways to have the difference be less than 6. And only one pair like that adds up to more than 5.</p>