Graph the inequality y< 3x - 2 on a plane

Ethen Blackwell
2022-07-31
Answered

Graph the inequality y< 3x - 2 on a plane

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hottchevymanzm

Answered 2022-08-01
Author has **15** answers

The equation of the boundary line is,y = 3x – 2

The equation is in the slope–intercept form.

Identify the slope and y–intercept.

Slope = 3

y–intercept = –2

Draw the x– andy–axes.

Choose an interval of 1 for both theaxes.

Now, label and scale the axes

The y–intercept is–2.

So, plot the point (0, –2) on thegraph.

Consider the slope as a ratio 3/1.

Use the ratio to find another point on the graph.

The numerator of the slope is positive.

So, move 3 units up from the point (0, –2).

Now, move 1 unit to the right.

We get the point (1, –1).

Graph the equation by drawing a line through the points.

Use a solid line to show that the points on the line aresolutions of y 3x – 2.

The solutions of y 3x – 2 must lie on oneside of the boundary line.

Test a point that is not on the line, such as (0, 0).

0 3(0) – 2

0 0 – 2

0 –2

Since 0 –2 is false, the half–plane containing(0, 0) does not contain the solutions.

So, the half–plane that does not contain (0, 0) is thegraph of the solution.

Shade it to show that every point in that half–plane is asolution.

The equation is in the slope–intercept form.

Identify the slope and y–intercept.

Slope = 3

y–intercept = –2

Draw the x– andy–axes.

Choose an interval of 1 for both theaxes.

Now, label and scale the axes

The y–intercept is–2.

So, plot the point (0, –2) on thegraph.

Consider the slope as a ratio 3/1.

Use the ratio to find another point on the graph.

The numerator of the slope is positive.

So, move 3 units up from the point (0, –2).

Now, move 1 unit to the right.

We get the point (1, –1).

Graph the equation by drawing a line through the points.

Use a solid line to show that the points on the line aresolutions of y 3x – 2.

The solutions of y 3x – 2 must lie on oneside of the boundary line.

Test a point that is not on the line, such as (0, 0).

0 3(0) – 2

0 0 – 2

0 –2

Since 0 –2 is false, the half–plane containing(0, 0) does not contain the solutions.

So, the half–plane that does not contain (0, 0) is thegraph of the solution.

Shade it to show that every point in that half–plane is asolution.

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