+2446 The equation \(\left \lfloor{t}\right \rfloor=2t+3\) can be a tad difficult to parse. Sometimes, it is better to think of an easier problem before attempting one a harder one. Let's think about a basic equation that contains the floor function.
\(\left \lfloor{t}\right \rfloor=5\)
Before we can solve this, though, we have to understand the floor function. The floor function truncates any decimal. The solution set for this simplified equation, represented in a compound inequality, is \(5\leq t<6\). Any value for t that satisfies this restriction is a solution. We can use this logic to generalize the floor function, actually to \(\left \lfloor{x}\right \rfloor=a\rightarrow a\leq x < a+1\) . However, you must have caution when doing this, however. Not every solution is actually a solution. I'll touch on this later. Knowing this will be key to solving this equation. Knowing this, the floor function transforms from \(\left \lfloor{t}\right \rfloor=2t+3\) to \(2t+3\leq t<2t+3+1\). Let's solve this compound inequality. I'll solve each one separately.
| \(2t+3\leq t\) | Subtract 2t from both sides. |
| \(3\leq-t\) | Divide by -1 on both sides. |
| \(-3\geq t\) | Of course, the inequality flips when dividing by a negative number. |
| \(t\leq -3\) | |
Now, let's solve for the other inequality \(t<2t+3+1\):
| \(t<2t+3+1\) | Subtract 2t from both sides. |
| \(-t<4\) | Divide by -1 on both sides. Yet again, remember to flip the inequality sign. |
| \(t>-4 \) | |
Of course, all solutions must adhere to both restrictions. Both inequalities show that t must be greater than -4 and less than or equal to -3. -3 is automatically a solution because it is an integer.
We aren't done yet, though. Are there are other solution that adhere to our current restriction that makes \(2t+3\) an integer? We don't even have to worry about the +3 because the sum of any integer and 3 will always yield another integer. Let's set this equal to 0:
| \(2t+3=0\) | Subtract 3 on both sides. |
| \(2t=-3\) | Divide by 2 on both sides. |
| \(t=-\frac{3}{2}\) | |
What I have demonstrate here is that a number divided by 2 will make 2t+3 an integer. The only one that meets our description and our current restriction is -(7/2).
Therefore, there are 2 solutions to this equation
\(t_1=-\frac{7}{2}\)
\(t_2=-3\)
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I am reasonably confident that my method is correct but there could be lots of careless mistakes.
