To form a system that will have no solutions, let's graph $3 x - y > 4$ first. Then we can identify where we want the second inequality to be on the coordinate plane. Before we begin, please note that there are infinitely many inequalities that would satisfy the required conditions. Here we will look at only one scenario.

Graphing $3 x - y > 4$

Let's find the equation of the boundary line first. To do so, we can exchange the inequality symbol with an equality sign and rewrite it in the slope-intercept form.

3 x - y = 4 \Leftrightarrow y = 3 x - 4

Since the inequality sign is strict, the boundary will be a dashed line. Now we can graph it!

Next we will test the point

(0, 0)

to check which area to shade. We must substitute it into the original inequality.

3 x - y > 4

SubstituteII

$x = 0$ , $y = 0$

3 (0) - 0 > ? 4

Multiply

0 - 0 > ? 4

SubTerm

Subtract term

0 ≯ 4 \times

We got a false statement, so the point

(0, 0)

is not a part of the solution set. We will therefore shade the half-plane without the point

(0, 0) .

Second Inequality

The second inequality must have a solution set somewhere in the half-plane so that it does not overlap with the solution of $3 x - y > 4 .$ We can tell that the boundary line has to be parallel to $y = 3 x - 4,$ or else the lines will eventually intersect and the system will have a solution. Therefore, we need a line with the slope of $3 .$ Let's take $y = 3 x + 2 .$

As we can see, we must shade the half plane without the solution of $3 x - y > 4 .$

This system of inequalities has no solution, which is what we wanted! Now we just need to write down this inequality. We already know that the boundary line is

y = 3 x + 2 .

Since the solution set is above the boundary and the line is solid, we can identify the inequality that represents this region.

y \geq 3 x + 2

From the graph we can tell that the point

(0, 0)

is not a part of the solution set. We can check if our inequality is correct by substituting this point into

y \geq 3 x + 2

and simplifying.

y \geq 3 x + 2

SubstituteII

$x = 0$ , $y = 0$

0 \geq ? 3 (0) + 2

Multiply

0 \geq ? 0 + 2

AddTerms

Add terms

0 ≱ 2 \times

We got a false statement, which is what we wanted! This means that the point

(0, 0)

is not included in the solution set of either inequality and therefore in the system as a whole. Note that this is just one solution out of infinitely many.

Graphing 3x−y>4

Second Inequality

Graphing $3 x - y > 4$

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