There is an integer n such that 2n2 − 5n + 2 is prime. To prove the statement it suffices to find a value of n such that (n, 2n2
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look at the picture i sent
<h3>Step-by-step explanation:</h3>
We will use as an example the following system of inequalities:
- 3x +2y ≥ 5x -4
- y < (1/2)x +3
For each inequality, do the following:
1. If it is not this way already, arrange the inequality so that each variable appears on only one side of the inequality. Putting the inequality into any of the standard forms will do this. It will be helpful later if you arrange the inequality so that at least one variable has a positive coefficient. (You may have to multiply by -1 (and change the sense of the inequality) to do this.)
Our example inequalities are now ...
- -2x +2y ≥ -4 . . . . . or divide by 2 to get . . . . -x +y ≥ -2
- y < (1/2)x +3
2. Notice the inequality symbol, and whether it includes the "or equal to" case.
- the first inequality is ≥, so includes "or equal to"
- the second inequality is <, so <em>does not</em> include "or equal to"
3. Replace the inequality symbol with an equal sign and graph the resulting equation. If the "or equal to" case is included, the line is graphed as a solid line. If it is not, then the line is graphed as a dashed line.
- the first equation (-x +y = -2) is graphed with a solid line
- the second equation (y = (1/2)x +3) is graphed with a dashed line
4. It is not necessary, but is less confusing, to choose a variable in the inequality (after step 1) that has a <em>positive</em> coefficient. Notice whether this variable is "less than" or "greater than" the stuff on the other side of the inequality.
- y ≥ . . . . for the first inequality
- y < . . . or . . . x > . . . . for the second inequality
If it is less than, shade the portion of the graph below (for y <) or to the left (for x <) of the line you graphed in step 3.
- the second inequality has y < , so the graph will be shaded <em>below</em> the dashed line. It also has x > , so will be shaded <em>to the right</em> of the dashed line. (These are the same shading.)
If it is greater than, shade the portion of the graph above (for y >) or to the right (for x >) of the line you graphed in step 3.
- the first inequality has y ≥ , so the graph will be shaded <em>above</em> the solid line.
5. The solution set is where the shaded areas overlap. Any point on a dashed line is not in the solution set.
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The graph for the example inequalities is attached. The first one is graphed and shaded in red; the second one is graphed and shaded in blue.