Which system of linear inequalities is represented by the graph?

 ax – 3y > 6 and y > 2x + 4 bx + 3y > 6 and y > 2x + 4 cx + 3y > 6 and y > 2x – 4 dx – 3y > 6 and y > 2x – 4

Which system of linear inequalities is represented by the graph ax 3y gt 6 and y gt 2x 4 bx 3y gt 6 and y gt 2x 4 cx 3y gt 6 and y gt 2x 4 dx 3y gt 6 and y gt 2 class=

Respuesta :

The correct answer is:

bx + 3y > 6 and y > 2x + 4

Explanation:

Looking at the second inequality, the y-intercept is 4 and the slope is 2.  This means the graph of the line crosses the y-axis at (0, 4) and the line goes up 2 and over 1.  Since it is greater than, this means the graph is shaded above it.  Comparing this to the graph, the line for the blue part crosses the y-axis at (0, 4) and goes up 2 and over 1.  The graph is also shaded above the line.

For the first inequality, bx+3y > 6, we want to isolate y.  To do this, we subtract bx from each side:

bx+3y-bx > 6-bx

3y > 6-bx

Divide both sides by 3:

3y/3 > 6/3 - bx/3

y > 2 - (b/3)x

This means the line for this will have a y-intercept of 2 and decrease 1 while going over 3.  The orange section does this.  Additionally, since it is greater than, the graph should be shaded above the line.  This one is, so this is the correct answer.

The lines represent the inequalities [tex]\boxed{bx + 3y > 6} and \boxed{y > 2x + 4}[/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.

1.[tex]ax - 3y > 6{\text{ and }}y > 2x + 4[/tex]

2. [tex]bx + 3y > 6{\text{ and }}y > 2x + 4[/tex]

3.[tex]cx + 3y > 6{\text{ and }}y > 2x - 4[/tex]

4. [tex]dx - 3y > 6{\text{ and }}y > 2x - 4[/tex]

Explanation:

The blue line intersects y-axis at[tex]\left( {0,4} \right)[/tex], therefore the y-intercept is 4.

The blue line intersect the points that are [tex]\left( {-2,0}\right)[/tex] and [tex]\left( {0,4}\right)[/tex].

The slope of the line can be obtained as follows.

[tex]\begin{aligned}m&=\frac{{4 - 0}}{{0 - \left({ - 2}\right)}}\\&=\frac{4}{2}\\&=2\\\end{aligned}[/tex]

The slope of the line is m = 2.

Now check whether the inequality included origin or not.

Substitute [tex]\left( {0,0}\right)[/tex] in equation [tex]y=2x+4.[/tex]

[tex]\begin{aligned}0 &> 2\left( 0 \right) + 4 \hfill\\0 &> 4 \hfill\\\end{aligned}[/tex]

0 is not greater than 4 which mean that the inequality doesn’t include origin.

Therefore, the blue line is y > 2x + 4.

Solve inequality ax-3y > 6 to obtain the standard form of inequality.

[tex]\begin{aligned}ax - 3y &> 6 \hfill\\ax - 6 &> 3y \hfill\\y &< \frac{{ax - 6}}{3}\hfill \\y &< \frac{a}{3}x - 2\hfill\\\end{aligned}[/tex]

The y intercept of the equation is -2 and the y-intercept of the line is 2.

Therefore, the inequality doesn’t satisfy the line.

Solve inequality bx + 3y > 6 to obtain the standard form of inequality.

[tex]\begin{aligned}bx + 3y &> 6 \hfill \\3y &> 6 - bx \hfill\\y &< \frac{{6 - bx}}{3} \hfill\\y &< - \frac{{bx}}{3} + 2 \hfill\\\end{aligned}[/tex]

The y intercept of the equation is 2 and the y-intercept of the line is 2.

Therefore, the inequality bx + 3y > 6 satisfies the line.

Option 1 is not correct as it doesn’t satisfy the inequalities of the graph.

Option 2 is correct as the inequalities satisfy the graph.

Option 3 is not correct as the y-intercept is not 4.

Option 4 is not correct as the y-intercept is not 4.

Hence, [tex]\boxed{{\text{Option 2}}}[/tex] is correct

Learn more:

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

3. Learn more about range and domain of the function https://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.

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