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Math Lesson 10.4.4 - Systems of Inequalities where one inequality is Quadratic and the other is Linear

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Welcome to our Math lesson on Systems of Inequalities where one inequality is Quadratic and the other is Linear, this is the fourth lesson of our suite of math lessons covering the topic of Systems of Inequalities, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Solving Systems of Inequalities where one inequality is Quadratic and the other is Linear

When solving systems of inequalities where not all are linear (for example, when one inequality is quadratic), we use the same approach as in systems of linear inequalities. The only difference is that one of the lines is a parabola, so the solution zone has one curved side.

Another thing to point out in such systems is that if we are interested to find any minimum or maximum point as in the previous examples, we must solve the corresponding system of equations only by substitution method. Thus, one of the variables in the linear equation is expressed in terms of the other variable and we substitute it in the quadratic equation. Let's consider an example to clarify this point, where theory and exercise are combined to have a better understanding.

Example 5

Solve the following system of inequalities.

3x - 2y > 1y ≥ x2-2x + 1

Solution 5

First, we identify the direction of the solution set for the linear inequality. We do this by writing the first inequality in the form y (?) mx + n. Thus,

3x - 2y > 1
-2y> - 3x + 1
-2y - 2 > -3x/-2 + 1/-2
y < 3x/2 - 1/2

Hence, the solution zone of this inequality alone lies under the graph's line y = 3x/2 - 1/2.

On the other hand, the quadratic equation is already in its regular form y (?) ax2 + bx + c, where a = 1, b = -2 and c = 1, while the inequality symbol which replaces the question mark is " ". This means the solution zone lies above the parabola's graph.

The best thing to do is to solve the system of the corresponding equations. The number of solutions depends on the sign of the discriminant obtained when the linear equation is substituted into the quadratic one. In the specific case, we have

3x - 2y = 1y = x2 - 2x + 1

We can write 3x - 1 = 2y, multiply the quadratic equation by 2 and then express y in the quadratic equation in terms of x. In this way, we obtain

2y = 2x2 - 4x + 2
3x - 1 = 2x2-4x + 2
2x2 - 4x + 2 - 3x + 1 = 0
2x2 - 7x + 3 = 0

This is a second-order equation with one variable, where its roots represent the intercept of the two original graphs. We have a = 2, b = -7 and c = 3. Thus, the discriminant Δ is

∆ = b2 - 4ac
= (-7)2 - 4 ∙ 2 ∙ 3
= 49 - 24
= 25

Since the discriminant is positive, we have two intercepts for our graphs. Their x-coordinates are found by solving the second-order equation above. Expressing the two intercepts by A and B respectively yields

xA = -b - √∆/2a
= -(-7) - √25/2 ∙ 2
= 7 - 5/4
= 2/4
= 1/2
= 0.5

and

xB = -b + √∆/2a
= -(-7) + √25/2 ∙ 2
= 7 + 5/4
= 12/4
= 3

The corresponding y-values for points A and B are

yA = xA2 - 2xA + 1
= 0.52 - 2 ∙ 0.5 + 1
= 0.25 - 1 + 1
= 2.25
yB = xB2 - 2xB + 1
= 32 - 2 ∙ 3 + 1
= 9 - 6 + 1
= 4

Hence, the two intercepts are A(0.5, 2.25) and B(3, 4).

Another important point of the solution set that helps identify (in this case) the lowest point of the solution set is the parabola vertex. We have briefly discussed this point in previous tutorials, where the formulas used to find the vertex V of a parabola are

xV = -b/2a and yV = -∆/4a

Substituting the known values for the original quadratic equation (a = 1, b = -2 and c = 1) yields

xV = -(-2)/2 ∙ 1
= 2/2
= 1

and

yV = -(b2 - 4ac)/4a
= -(-2)2 - 4 ∙ 1 ∙ 1/4 ∙ 1
= 4 - 4/4
= 0

Hence, point V(1, 0) is the vertex of the parabola, which at the same time acts as the lowest point of the solution set for the original system of inequalities.

The graph solution for the system is shown below.

Math Tutorials: Systems of Inequalities Example

We can draw the following conclusions by looking at the graph:

  1. Vertex V is the minimum point of the system's solution set.
  2. Point A is the leftmost point of the solution set but point A itself is not included in this set.
  3. Point B is both the maximum and the rightmost value of the solution set, where it is also excluded from the solution values.
  4. The two intercepts A and B can be used to plot the linear graph, since plotting the graph of a straight line requires only two known points.

We can use the same approach for systems with more than two inequalities, where not all of them are linear.

More Systems of Inequalities Lessons and Learning Resources

Inequalities Learning Material
Tutorial IDMath Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
10.4Systems of Inequalities
Lesson IDMath Lesson TitleLessonVideo
Lesson
10.4.1Systems of Linear Inequalities
10.4.2The Minimum or Maximum Values of a System of Linear Inequalities
10.4.3Systems of Three Linear Inequalities
10.4.4Systems of Inequalities where one inequality is Quadratic and the other is Linear

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  7. Continuing learning inequalities - read our next math tutorial: Solving Linear Inequalities

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