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Graph the given inequality. Add another inequality whose solution set does not intersect the solution set of the given one.
Create any inequality so that the system has a solution.
Example Solution: y ≤ 2x-1
Example Solution: y ≥ 2x-1
We will start by graphing the given inequality. Then, we can write a second inequality that does not have any common solutions with the given one.
To begin, we will isolate y. This will make graphing the boundary line easier.
LHS+4x>RHS+4x
.LHS /2.>.RHS /2.
Write as a sum of fractions
a* b/c=a/c* b
Calculate quotient
Let's write the boundary line. To do so, we substitute the inequality sign with an equals sign. ccc Inequality & & Boundary Line y > 2x+3 & & y = 2x+ 3
To decide which side to shade, we can use a test point that is not on the line. If the inequality holds true when substituting the coordinates of this point, we will shade the region that contains it. Otherwise, we will shade the opposite region. For simplicity, let's use (- 2,2).
x= - 2, y= 2
a(- b)=- a * b
Add terms
The inequality holds true. This means we should shade the side of the boundary line that contains the point.
For our second inequality not to overlap with the one above, the boundary lines must be parallel. Otherwise, the lines will inevitably intersect and the solution sets will overlap. Moreover, to also avoid overlapping solution sets, the y-intercept of our line must be less than 3. Let's arbitrarily choose a y-intercept of - 1. y=2x-1 Let's draw this line on the same coordinate plane.
To shade the region below the second boundary line, we have to determine the inequality sign. To do so, we will use the point (2,0). When substituted into the inequality, it must produce a true statement. Note that we do not really care if the line is dashed or solid. For simplicity, we will leave it solid and the inequality will be not strict.
x= 2, y= 0
Multiply
Subtract term
The inequality sign is less than or equal to. Therefore, our inequality is y≤ 2x-1. Let's shade the region!
Finally, we form our system of inequalities by combining the inequalities. y>2x+3 y ≤ 2x-1
If any system of inequalities has a solution, it does in fact, have infinitely many solutions. This is because any region on a coordinate plane contains infinitely many points. In Part A, we created a system with no solution.
y>2x+3 & (I) y ≤ 2x-1 & (II)
By flipping the inequality sign of Inequality (II), we will make it point the opposite direction. Therefore, we will get overlapping regions.
Let's finally cut away the non-overlapping region.
The shaded region, which is the solution to the system, contains infinitely many points. y>2x+3 y≥ 2x-1 The system above has infinitely many solutions.