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:

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

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 :

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

Answer 2

On a coordinate plane, when two straight line are drawn the, the slope of the line will be the derivative and the values of the points on the line will be given by the shaded region.

The system of linear inequalities is represented by the graph is:

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

$3x-y> 2$

Given information:

Two straight lines are given,

The first solid line has a positive slope.

The points (0,3) and (3,4)

Now, the equation of line passes through points (a,b) and (c,d) is given by:

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

Now ,

On putting the values in above equation:

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

Also, the shaded region id above this line

Hence, the inequality represents the graph is $y\geq \frac{1}{3}x+3$

Now for the second point:

The equation can be written as:

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

As, the shaded region is right to the line

Hence, inequality represents the second graph is $3x-y> 2$

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

• $y\geq \frac{1}{3}x+3$

• $3x-y> 2$

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