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?

Answer 1

Answer:

A. y≥$\frac{1}{3\\}$x+3, 3x-y>2

Step-by-step explanation:

The linear equations above match the lines in terms of slope and y-intercept.

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Answer 2

The system of linear inequalities of the given points is required.

The system of inequality is

$x-3y\leq -9$

$3x-y>2$

Points of the first line

$(0,3)$

$(3,4)$

The equation of the line is

$y-3=\dfrac{4-3}{3-0}(x-0)\\\Rightarrow y-3=\dfrac{1}{3}x\\\Rightarrow y-\dfrac{1}{3}x=3\\\Rightarrow 3y-x=9\\\Rightarrow x-3y=-9$

Let us take a point above this line $(2,4)$

It is also a solid line.

$2-3\times 4=-10<-9$

So, the inequality is $x-3y\leq -9$

Points of the second line

$(0,-2)$

$(1,1)$

The equation of the second line is

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

Let us take a point towards the right of the line $(2,2)$

$3\times 2-2=4>2$

Since, the line is dashed the second inequality will be $3x-y>2$

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