Question

Which system of linear inequalities is represented by the graph? y > x + 3 and 3x – y > 2 y > x + 3 and 3x – y > 2 y > x + 3 and 3x + y > 2 y > x + 3 and 2x – y > 2

Answer 1

Step 1: Find equations of the lines
Red, solid line:
Find slope: 1square in y / 3 squares in x => slope = 1/3
y-intercept : (0,3)
Equation of the line: one of the following:
y=x/3+3   .................(1a)
3y=x+9    .................(1b)
x-3y+9=0 .................(1c)
Black, dotted line:
Slope = 3 squares in y / 1 square in x => slope = 3
y-intercept (0,-2)
Equation of the line: one of the following:
y=3x-2.....................(2a)
3x-y=2..................(2b)

2. Find location of region, i.e. greater or less than
Red region:
The region is above the solid line, i.e. we look for y>= .....
From equation (1a), we change the equality to >=, or
y >= x/3+3   [>= because line is solid]
Black region:
The region is to the right of the dotted line, we look for x > ...
From equation (2b),, we change equality to >, or
3x-y > 2   [ > because line is dotted ]


Note: All the given answers have common typos, namely
the first inequality should read y > x/3+3   or y>=x/3+3
the second inequality should read 3x-y > 2 or 3x-y >=2
The way question is presented, all the answer choices are the same (and incorrect).

@lazalewagoner  you still have time to correct the answer choices.

Answer 2

The lines represent the inequalities $\boxed{y>\frac{1}{3}x + 3{\text{ and }}3x - y > 2.}$ .

Further explanation:

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

 $\boxed{y = mx + c}$

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

 $\boxed{m=\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}}$

Given:

The inequalities are as follows.

a. $y > x + 3{\text{ and }}3x - y > 2$  

b.$y > x + 3{\text{ and }}3x - y > 2$  

c.$y > x + 3{\text{ and }}3x + y > 2$  

d.$y > x + 3{\text{ and 2}}x - y > 2$

Explanation:

The blue line intersects y-axis at $(0, - 2)$, therefore the y-intercept is  .

The blue line intersect the points that are $\left( {1,1} \right)$ and $\left( {0, - 2}\right)$.

The slope of the line can be obtained as follows.

$\begin{aligned}m&=\frac{{1 -\left({- 2} \right)}}{{1 - 0}}\\&=\frac{{1 + 2}}{1}\\&=3\\\end{aligned}$

The slope of the line is $m = 3$.

Now check whether the inequality included origin or not.

Substitute $\left( {0,0}\right)$ in equation $y < 3x - 2$.

$\begin{aligned}\left( 0 \right) &< 3\left( 0 \right)- 2\\0 &< - 2\\\end{aligned}$

0 is not less than which mean that the inequality doesn’t include origin.

Therefore, the blue line is $y < 3x--2{\text{ or }}3x - y > 2$.

The orange line intersects y-axis at  $\left( {0,3}\right)$, therefore the y-intercept is 3.

The orange line intersect the points that are $\left( { - 3,2}\right)$ and $\left( {0,3}\right)$.

The slope of the line can be obtained as follows.

 $\begin{aligned}m&=\frac{{3 - 2}}{{0 - \left( { - 3}\right)}}\\&=\frac{1}{3}\\\end{aligned}$

The slope of the line is  $m = \dfrac{1}{3}$.

Now check whether the inequality included origin or not.

Substitute $\left( {0,0}\right)$ in equation $y > \dfrac{1}{3}\left( x \right) + 3$.

$\begin{aligned}\left(0 \right)&>\frac{1}{3}\left(0\right) + 3\\0&>3\\\end{aligned}$

0 is not greater than 3 which mean that the inequality doesn’t include origin.

Therefore, the orange line is $y > \dfrac{1}{3}x + 3$.

The lines represent the inequalities $\boxed{y > \frac{1}{3}x + 3{\text{ and }}3x - y > 2.}$.

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function brainly.com/question/3412497

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.