Systems of Inequalities
- Systems of inequalities let you describe many possible solutions at once.
- Instead of finding a single point (as with a system of equations), you find a region of points that satisfy all given inequalities simultaneously.
- This is a powerful way to build and test models for real decisions.
Inequalities and Solution Sets
A linear inequality looks like a linear equation but uses an inequality sign, for example: $$2 x-3>5, \quad 4 y+7 \leq 19, \quad 3 x-2 y \geq 6$$
Its solution set is the set of all values that make the statement true.
Solution set
The set of all values of the variable(s) that make an inequality (or a system of inequalities) true.
- For one variable, the solution set is usually an interval on a number line (for example, $x < 4$).
- For two variables (like $x$ and $y$), the solution set is typically a shaded region in the coordinate plane.
System of inequalities
A set of two or more inequalities that must all be true at the same time. The solution set is the overlap of the individual solution regions.
Equivalence Transformations
- To solve inequalities algebraically, you use equivalence transformations, similar to solving equations.
- The key idea is that you want to transform the inequality into an equivalent one with the same solution set.
- For real numbers $a$ and $b$ :
- $a>b$ if and only if $a-b>0$.
- $a<b$ if and only if $a-b<0$
- We can transform inequalities, just like equations, as long as we keep the same solution set.
Theorem 1 – Adding/Subtracting the Same Number
Adding or subtracting the same real number on both sides of an inequality gives another equivalent inequality.
- If $a>b$, then for any real $c$:
- $a+c>b+c$
- $a-c>b-c$
- and similarly, if $a<b$, then
- $a+c<b+c$
- $a-c<b-c$
Theorem 2 – Multiplying by a Positive Number
Multiplying both sides of an inequality by the same positive real number leaves the direction the same.
- If $a>b$ and $k>0$, then $a k>b k$.
- If $a<b$ and $k>0$, then $a k<b k$.
Theorem 3 – Multiplying by a Negative Number
Multiplying or dividing both sides of an inequality by a negative real number reverses the direction of the inequality.
- If $a>b$ and $k<0$, then $a k<b k$.
- If $a<b$ and $k<0$, then $a k>b k$.
- A very common mistake is to forget to flip the inequality sign when multiplying or dividing by a negative number.
- For example: since $2 > 1$, multiplying by $-3$ gives $-6 < -3$, so the sign flips.
Solving One-Variable Inequalities
Many inequalities reduce to collecting variable terms on one side and constants on the other.
Solve $5x - 7 \geq 3x + 9$.
Solution
- Subtract $3x$ from both sides: $$2x - 7 \geq 9$$
- Add 7 to both sides: $$2x \geq 16$$
- Divide by 2 (positive, so the sign stays): $$x \geq 8$$
- Check with a value like $x = 9$: $5(9) - 7 = 38$ and $3(9) + 9 = 36$, and $38 \geq 36$ is true.
Double Inequalities
- A double inequality like $-4 < 3x + 2 < 5$ means both inequalities are true: $$-4 < 3x + 2 \quad \text{and} \quad 3x + 2 < 5$$
- You can solve it either by performing the same transformation to all three parts, or by splitting into two inequalities and recombining the result.
If you expect to multiply or divide by a negative number, splitting a double inequality into two separate inequalities can reduce errors about flipping the sign.
Interval Solutions
Some inequalities have infinitely many solutions forming a whole half-line or more.
Solve $6 x-12>0$.
Solution
$$6 x-12 >0 $$
$$6 x >12 $$
$$x >2 $$
So every real number greater than 2 is a solution. We often write the solution set as $(2, \infty)$.
Verify that $3x + 2 < 9x + 6$ has infinitely many solutions.
Solution
- Subtract $3x$ from both sides: $$2 < 6x + 6$$
- Subtract 6: $$-4 < 6x$$
- Divide by 6: $$-\frac{2}{3} < x$$
- So every real number greater than $-\frac{2}{3}$ is a solution, which is infinitely many values.
Graphing One-Variable Inequalities
We can see solutions on a number line.
$$x \geq-3$$
- Draw a solid dot at -3 (included).
- Shade to the right for all numbers greater than -3 .
More generally, you can think of an inequality like $f_1(x) \leq f_2(x)$ as comparing two functions:
- Graph $y=f_1(x)$ and $y=f_2(x)$.
- Find where they are equal (their intersection points).
- The solution set is all $x$-values where the graph of $f_1$ lies on or below the graph of $f_2$.
- First find where the two sides are equal.
- Then test one value in each interval to decide where the inequality holds.
Linear Inequalities in Two Variables
- A two-variable linear inequality such as $y \leq x + 1$ divides the coordinate plane into two regions separated by the boundary line $y = x + 1$.
- The boundary line is the related equation (replace the inequality with $=$).
- One side of the line satisfies the inequality, the other side does not.
Boundary line
The line obtained by replacing an inequality sign with an equals sign. It separates the plane into regions that do or do not satisfy the inequality.
Solid vs. Dashed Boundary Lines
- Use a solid line for $\leq$ or $\geq$ (points on the line are included).
- Use a dashed line for $<$ or $>$ (points on the line are not included).
To decide which side to shade, test any point not on the boundary line, often $(0,0)$ if it is not on the line.
Solving Systems Graphically
To solve a system of inequalities in two variables, you:
- Graph each inequality.
- Shade the region that satisfies each one.
- The solution set of the system is the overlap of all shaded regions.
Consider the system
$$
\left\{\begin{array}{l}
y \leq -x+4 \\
y \geq 2 x-1
\end{array}\right.
$$
- Graph $y=-x+4$ as a solid line and shade below it.
- Graph $y=2 x-1$ as a solid line and shade above it.
- The solution set is the region where the two shadings overlap.
Checking a point
Try the point $(1,2)$ :
- Check $y \leq-x+4: 2 \leq-1+4=3$
- Check $y \geq 2 x-1$ : $2 \geq 2(1)-1=1$
So $(1,2)$ is in the solution set.
If the system were
$$
\left\{\begin{array}{l}
y< -x+4 \\
y>2 x-1
\end{array}\right.
$$
then both boundary lines would be dashed, and points on the lines would not be included.
The overlap region is in the same place, but the solution set is slightly smaller.
- Shading correctly but drawing the wrong type of line (solid vs dashed).
- This changes whether boundary points are valid solutions.
Reading Inequalities from Graphs
Sometimes you are given a graph and asked to write inequalities for the shaded (or unshaded) region.
Strategy:
- Identify each boundary line.
- Write its equation.
- Decide which side of the line is included by testing a point from the region.
- Decide whether to use a strict sign ($<$ or $>$) or an inclusive sign ($\leq$ or $\geq$) by checking whether the line is dashed or solid.
If the region is described as "non-shaded," you are still writing inequalities, just for the opposite side of each boundary.
Systems in One Variable
A system of inequalities in one variable is the overlap of their solution intervals.
The pair
- $4x < 4$ (so $x < 1$)
- $3x - 5 \geq 1$ (so $x \geq 2$)
has no common solution, because no number is both less than 1 and at least 2.
Linear Programming
- When a real-life problem is modelled by a system of linear inequalities, we often want the “best” solution according to some goal, such as maximizing profit or minimizing cost or time.
- This process is called linear programming.
- The objective function is the linear function that represents cost, profit, or another quantity to be optimized (maximized or minimized).
- The constraints are the system of linear inequalities that model the practical limits (time, materials, budget, etc.).
- The feasible region is the region of the graph that contains all the solutions to the system of constraints.
- If there is an optimal solution to a linear programming problem, it will occur at one or more vertices (corner points) of the feasible region polygon.
Using Linear Programming to Solve Real-Life Problems
- Identify all variables and parameters.
- Write all constraints as linear inequalities.
- Graph the constraints and identify the polygon that makes up the feasible region.
- Write the objective function in terms of the variables.
- Evaluate the objective function at each vertex of the feasible region.
- Choose the vertex that gives the maximum or minimum value, depending on the problem.
Importance of Graphical Methods
- For two-variable systems, algebra can help you find where boundary lines meet, but it does not easily describe the entire solution region, especially when there are many inequalities.
- Graphing makes the feasible region visible and lets you quickly check whether a proposed solution is possible.
- How do you decide whether to draw a boundary line as dashed or solid?
- What does the solution set of a system of inequalities represent geometrically?
- Why must you reverse the inequality sign when dividing by a negative?