Question

On a coordinate plane, 2 straight lines are shown. The first solid line has a positive slope and goes through (0, 3) and (3, 4). Everything above the line is shaded. The second dashed line has a positive slope and goes through (0, negative 2) and (1, 1). Everything to the right of the line is shaded. Which system of linear inequalities is represented by the graph? y > One-thirdx + 3 and 3x – y > 2 y > One-halfx + 3 and 3x – y > 2 y > One-thirdx + 3 and 3x + y > 2 y > One-thirdx + 3 and 2x – y > 2

Answer 1

Answer:

$y > \frac{1}{3}x + 3$

$3x - y > 2$

Step-by-step explanation:

The general equation of a straight line is given by:

$y = ax + b$

Being given two points, we can replace x and y, solve the system and find the values for a and b.

The first line:

The problem states that it goes through $(0,3)$

So, when $x = 0, y = 3$. Replacing in the equation

$y = ax + b$

$3 = a(0) + b$

$b = 3$

For now, the equation of the line is:

$y = ax + 3$

The line also goes through $(3,4)$. So, when $x = 3, y = 4$. Replacing in the equation:

$y = ax + 3$

$4 = 3a + 3$

$3a = 1$

$a = \frac{1}{3}$

So, the equation of this line is:

$y = \frac{1}{3}x + 3$

Everything above the line is shaded.  So, it is represented by the following inequality:

$y > \frac{1}{3}x + 3$

The second line:

The problem states that it goes through $(0,-2)$

So, when $x = 0, y = -2$. Replacing in the equation

$y = ax + b$

$-2 = a(0) + b$

$b = -2$

The line also goes through $(1,1)$. So, when $x = 1, y = 1$. Replacing in the equation:

$y = ax - 2$

$1 = a - 2$

$a = 3$

So, the equation of the line is:

$y = 3x - 2$

Everything to the right of the line is shaded:

We have to write this in function of x:

$3x = y + 2$

$3x - y = 2$

To the right is:

$3x - y > 2$

Answer 2

Answer:

$y\geq\dfrac{1}{3}x+3\\\\3x-y>2$

Step-by-step explanation:

Equation of line passes through points (a,b) and (c,d) is given by :-

$(y-b)=\dfrac{d-b}{c-a}(x-a)$

Given :   The first solid line (≤ or ≥) has a positive slope and goes through (0, 3) and (3, 4).

$(y-3)=\dfrac{4-3}{3-0}(x-0)\\\\\Rightarrow\ y-3=\dfrac{1}{3}x\\\\\Rightarrow\ y-\dfrac{1}{3}x=3$

Also, Everything above the line is shaded .

i.e. the inequality represents the first graph: $y\geq\dfrac{1}{3}x+3$

The second dashed line (< or >) has a positive slope and goes through (0, - 2) and (1, 1).

$(y-(-2))=\dfrac{1-(-2)}{1-0}(x-(0))\\\\\Rightarrow\ y+2=\dfrac{1+2}{1}x\\\\\Rightarrow\ 3x-y=2$

Everything to the right of the line is shaded.

i.e. inequality represents the second graph : $3x-y>2$

Hence, the system of linear inequalities is represented by the graph :

$y\geq\dfrac{1}{3}x+3\\\\3x-y>2$