Question

Somebody give me answer will award brainliest

Answer 1

Answer:

$y$

$y$

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: $, \geq, \leq$

Important note: If it includes $\geq$ or $\leq$, then that means the solution includes points on the line.

2) In this situation, we have two lines:

$y=2x-1$ (Which is the line that is more steep + Contains BOTH points A and D) and $y=\frac{1}4x+\frac{5}{2}$ (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 $\geq$ and $\leq$ as those will include points on the line (Points a and d) as solutions.

3) So $y>2x-1$ or $y < 2x-1$

AND $y>\frac{1}4x+\frac{5}2$ or $y < \frac{1}4x+\frac{5}{2}$

----> For the inequality involving the line $2x+1$, 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.

$1 = 2(3)-1 \\ 1=6-1 \\ 1< 5$

This means that: $y$ 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

$1 = \frac{1}{4}*3+\frac{5}{2} \\ 1=\frac{3}{4}+\frac{5}2 \\ 1 < \frac{13}4$

This means that: $y$ for Point C to be a solution