Which system of linear inequalities is represented by the graph?


y > x – 2 and y < x + 1

y < x – 2 and y > x + 1

y < x – 2 and y > x + 1

y > x – 2 and y < x + 1

Which system of linear inequalities is represented by the graph y gt x 2 and y lt x 1 y lt x 2 and y gt x 1 y lt x 2 and y gt x 1 y gt x 2 and y lt x 1 class=

Respuesta :

Answer:

Option D-  y > x – 2 and y < x + 1

Step-by-step explanation:

To determine : Which system of linear inequalities is represented by the graph?

Solution : First we examine the graph

The shaded region in purple is drawn using two line

The first line above the origin passing through the points (0,1) and (-1,0)

The equation of line passing through two points is [tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]

[tex]y-1=\frac{0-1}{-1-0}(x-0)[/tex]

[tex]y-1=x[/tex]

So, the equation of that line [tex]y = x + 1[/tex]

The second line below the origin passing through the points (0,-2) and (2,0)

The equation of line passing through two points is [tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]

[tex]y-(-2)=\frac{0-(-2)}{2-0}(x-0)[/tex]

[tex]y+2=x[/tex]

So, the equation of that line [tex]y=x-2[/tex]

Now, to check the system of inequality substitute point (x,y)=(0,0)

First equation  [tex]y = x + 1[/tex]

[tex]0<1[/tex] means  [tex]y< x + 1[/tex]

Second equation  [tex]y = x-2[/tex]

[tex]0>-2[/tex] means [tex]y> x-2[/tex]

Therefore, Option D is correct.

The linear inequalities of the system is y > x-2 and y < x + 1

The lines represent the inequalities [tex]\boxed{y>x-2}{\text{ and }}\boxed{y<x+1}[/tex].

Further explanation:

The linear equation with slope m and intercept c is given as follows.

[tex]\boxed{y=mx+c}[/tex]

The formula for slope of line with points[tex]\left({{x_1},{y_1}}\right)[/tex] and [tex]\left({{x_2},{y_2}}\right)[/tex] can be expressed as,

[tex]\boxed{m=\frac{{{y_2}-{y_1}}}{{{x_2}-{x_1}}}}[/tex]

Given:

The inequalities are as follows.

a. [tex]y>x-2{\text{ and }}y<x+1[/tex]

b. [tex]y<x-2{\text{ and }}y>x+1[/tex]

c. [tex]y<x-2{\text{ and }}y>x+1[/tex]

d. [tex]y>x-2{\text{ and }}y<x+1[/tex]

Explanation:

The blue line intersects y-axis at [tex]\left({0,1}\right)[/tex], therefore the y-intercept is 1.

The blue line intersect the points that are [tex]\left({-1,0}\right)[/tex] and [tex]\left({0,1}\right)[/tex].

The slope of the line can be obtained as follows.

[tex]\begin{aligned}m&=\frac{{1-0}}{{0-\left({-1}\right)}}\\&=\frac{1}{1}\\&=1\\\end{aligned}[/tex]

The slope of the line is m = 1.

Now check whether the inequality included origin or not.

Substitute [tex]\left({0,0}\right)[/tex] in equation y=x+1.

[tex]\begin{aligned}0&>1\left(0\right)+1\hfill\\0&>1\hfill\\\end{aligned}[/tex]

0 is not greater than 1 which means that the inequality doesn’t include origin.

Therefore, the blue line is y < x + 1.

The orange line intersects y-axis at [tex]\left({0,-2}\right)[/tex], therefore the y-intercept is -2.

The orange line intersect the points that are [tex]\left({2,0}\right)[/tex] and [tex]\left({0,-2}\right)[/tex].

The slope of the line can be obtained as follows.

[tex]\begin{aligned}m&=\frac{{-2-0}}{{0-\left(2\right)}}\\&=\frac{2}{2}\\&=1\\\end{aligned}[/tex]

The slope of the line is m = 1.

Now check whether the inequality included origin or not.

Substitute [tex]\left({0,0}\right)[/tex] in equation y=x-2.

[tex]\begin{aligned}0&>1\left(0\right)-2\hfill\\0&>-2\hfill\\\end{aligned}[/tex]

0 is greater than -2 which means that the inequality include origin.

Therefore, the orange line is y > x - 2.

Option (a) is correct as it satisfy the inequalities of the graph.

Option (b) is not correct as it satisfy the inequalities of the graph.

Option (c) is not correct as it satisfy the inequalities of the graph.

Option (d) is is not correct as it satisfy the inequalities of the graph.

Hence, [tex]\boxed{{\text{Option (a)}}}[/tex] is correct.

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Linear inequalities

Keywords: numbers, slope, slope intercept, inequality, equation, linear inequality, shaded region, y-intercept, graph, representation, origin.