Answer:
[tex]y<2x-1[/tex]
[tex]y<\frac{1}{4}x+\frac{5}2[/tex]
Skills needed: Linear Inequalities
Step-by-step explanation:
1) Now, we need to understand the idea of linear inequalities. Linear inequalities have usually a region of points that make up the solution. Check the images I posted for reference.
There are usually 4 symbols used for inequalities: [tex]<, >, \geq, \leq[/tex]
Important note: If it includes [tex]\geq[/tex] or [tex]\leq[/tex], then that means the solution includes points on the line.
2) In this situation, we have two lines:
[tex]y=2x-1[/tex] (Which is the line that is more steep + Contains BOTH points A and D) and [tex]y=\frac{1}4x+\frac{5}{2}[/tex] (Which is the line that is less steep + Contains ONLY point A as part of it).
We need to make 2 inequalities so Point C is the only solution.
---> This means we cannot use [tex]\geq[/tex] and [tex]\leq[/tex] as those will include points on the line (Points a and d) as solutions.
3) So [tex]y>2x-1[/tex] or [tex]y < 2x-1[/tex]
AND [tex]y>\frac{1}4x+\frac{5}2[/tex] or [tex]y < \frac{1}4x+\frac{5}{2}[/tex]
----> For the inequality involving the line [tex]2x+1[/tex], we have two possibilities:
The y-value is greater than twice the x-value plus one
The y-value is less than twice the x-value plus one
----> Let's use Point C to see which one it will fall under. The coordinates of point C are: (3, 1) --> We plug in these values for y and x
1 is the y-value, 3 is the x-value.
[tex]1 = 2(3)-1 \\ 1=6-1 \\ 1< 5[/tex]
This means that: [tex]y<2x-1[/tex] for Point C to be a solution
4) The same rules above apply for the next line (2 possibilities):
The y-value is greater than one-fourths the x-value plus five-halves OR
The y-value is less than one-fourths the x-value plus five-halves
---> Let's plus (3, 1) in again
[tex]1 = \frac{1}{4}*3+\frac{5}{2} \\ 1=\frac{3}{4}+\frac{5}2 \\ 1 < \frac{13}4[/tex]
This means that: [tex]y<\frac{1}4x+\frac{5}2[/tex] for Point C to be a solution
