A system of linear inequalities is two or more linear inequalities in the same two variables that must hold simultaneously. A point is a solution of the system only if it satisfies every inequality at once. Geometrically, each inequality contributes one half-plane, and the solution of the whole system is the overlap — the region common to all of them.
This common region is called the feasible region (or solution region). It is found by graphing every inequality on the same axes and then keeping only the area where all the shadings agree.
$$\text{Solution of the system} = (\text{half-plane}_1) \cap (\text{half-plane}_2) \cap \dots \cap (\text{half-plane}_n)$$
A reliable procedure for any system:
Many systems include the non-negativity constraints $x \ge 0$ and $y \ge 0$, which confine the solution to the first quadrant. These appear constantly in word problems where the variables count physical things — units produced, hours worked, items bought — that cannot be negative.
Double inequalities such as $3 \le 2x + y \le 8$ are really a compact way of writing two inequalities at once — here $2x + y \ge 3$ and $2x + y \le 8$. They carve out a band (slab) trapped between two parallel boundary lines, and you graph them by shading the region sandwiched in between. A single-variable double inequality like $-2 \le x \le 5$ similarly describes a vertical strip of the plane.
The feasible region may be bounded (enclosed on all sides, forming a polygon) or unbounded (extending to infinity in some direction). It is also possible for a system to have no solution at all: if the half-planes never overlap, the feasible region is empty.
Reading word problems into a system. Mixture, diet, budget and percentage problems all translate the same way. Identify the two quantities and call them $x$ and $y$; turn each limiting phrase ("at least", "no more than", "the mixture must contain") into an inequality; and add $x \ge 0,\ y \ge 0$ whenever the quantities are physical. A percentage condition like "the acid content must be at least $30\%$ of a $x + y$ litre mixture" becomes a linear inequality once you write the acid amount on each side and clear the percentage.
For example, the system $x + y \le 4,\ x \ge 0,\ y \ge 0$ has a feasible region that is the triangle with corners $(0,0)$, $(4,0)$ and $(0,4)$ — a bounded region trapped in the first quadrant beneath the line $x + y = 4$.
Deeper Insight — the feasible region is where linear programming begins: Finding the common region is not an end in itself; it is the first half of one of the most useful techniques in applied mathematics, linear programming. In a typical problem the inequalities are constraints (limited materials, budget, labour) and you want to maximise or minimise some linear quantity such as profit or cost over all points that satisfy them. A beautiful theorem guarantees that the optimum, when it exists, always occurs at a corner (vertex) of the feasible region — never in the smooth interior — so once the region is drawn, the search shrinks to checking a handful of corner points. That is why precision matters here: shade the wrong half-plane and the whole optimisation collapses. Whether the region is bounded also carries meaning: a bounded region guarantees both a maximum and a minimum exist, while an unbounded one may let a quantity grow without limit. Mastering the overlap now makes the optimisation you meet later feel like a short final step rather than a new topic.
Does the point $(1, 1)$ belong to the solution of the system $x + y \le 3$ and $x - y \le 2$?
Solution- Check the first: $1 + 1 = 2 \le 3$. True.
- Check the second: $1 - 1 = 0 \le 2$. True.
- Both hold simultaneously.
Answer: Yes, $(1,1)$ lies in the feasible region of the system.
Find the corner points of the feasible region of $x \ge 0,\ y \ge 0,\ x + y \le 5$.
Solution- $x \ge 0$ and $y \ge 0$ restrict the region to the first quadrant.
- The line $x + y = 5$ meets the axes at $(5,0)$ and $(0,5)$.
- The axes meet at the origin $(0,0)$.
- The bounded region is the triangle with these three vertices.
Answer: Corner points are $(0,0)$, $(5,0)$ and $(0,5)$.
Identify whether the feasible region of $x \ge 1,\ y \ge 1$ is bounded or unbounded.
Solution- $x \ge 1$ is the half-plane to the right of the vertical line $x = 1$.
- $y \ge 1$ is the half-plane above the horizontal line $y = 1$.
- Their overlap is the corner region above-and-right of $(1,1)$, which extends to infinity.
Answer: The feasible region is unbounded.
Show that the system $x + y \le 1$ and $x + y \ge 3$ has no solution.
Solution- The first inequality requires $x + y$ to be at most $1$.
- The second requires $x + y$ to be at least $3$.
- No value of $x + y$ can be both $\le 1$ and $\ge 3$ at once.
- The two half-planes lie on opposite sides of parallel lines and never overlap.
Answer: The system has no solution; the feasible region is empty.
A student has at most $\text{₹}\,120$ to buy pens at $\text{₹}\,10$ each and notebooks at $\text{₹}\,30$ each. With $x$ pens and $y$ notebooks, write the system of inequalities modelling the situation.
Solution- The total cost is $10x + 30y$, which must not exceed $120$: $10x + 30y \le 120$, i.e. $x + 3y \le 12$.
- Quantities cannot be negative: $x \ge 0$ and $y \ge 0$.
- (If only whole items can be bought, also $x, y \in \mathbb{Z}$.)
Answer: $x + 3y \le 12,\ x \ge 0,\ y \ge 0$.
Find the corner points of the feasible region of $2x + y \le 8,\ x \ge 0,\ y \ge 0$.
Solution- Non-negativity confines the region to the first quadrant.
- The line $2x + y = 8$ meets the $x$-axis where $y = 0$: $2x = 8 \Rightarrow x = 4$, giving $(4,0)$.
- It meets the $y$-axis where $x = 0$: $y = 8$, giving $(0,8)$.
- The third corner is the origin $(0,0)$.
Answer: Corner points are $(0,0)$, $(4,0)$ and $(0,8)$.
Solve the double inequality $-2 \le 3 - 2x \le 7$ and describe the region of the plane it cuts out (taking it as a condition on $x$ alone).
Solution- Subtract $3$ throughout: $-5 \le -2x \le 4$.
- Divide every part by $-2$ and reverse both symbols: $\dfrac{5}{2} \ge x \ge -2$.
- Rewrite in increasing order: $-2 \le x \le \dfrac{5}{2}$.
Answer: $-2 \le x \le \dfrac{5}{2}$ — a vertical strip of the plane between the lines $x = -2$ and $x = \tfrac{5}{2}$.
Find the corner points of the feasible region of $x + y \le 4,\ x + y \ge 1,\ x \ge 0,\ y \ge 0$.
Solution- The two parallel lines $x + y = 1$ and $x + y = 4$ form a slab; with the first quadrant this gives a trapezium.
- $x + y = 1$ meets the axes at $(1,0)$ and $(0,1)$.
- $x + y = 4$ meets the axes at $(4,0)$ and $(0,4)$.
- All four points lie in the first quadrant and bound the region.
Answer: Corner points are $(1,0)$, $(4,0)$, $(0,4)$ and $(0,1)$.
A solution is to be kept between $40^\circ\text{C}$ and $45^\circ\text{C}$. The Celsius–Fahrenheit relation is $C = \dfrac{5}{9}(F - 32)$. Find the range of temperature in degrees Fahrenheit.
Solution- The condition is $40 \le C \le 45$, i.e. $40 \le \dfrac{5}{9}(F - 32) \le 45$.
- Multiply every part by $\dfrac{9}{5}$ (positive — symbols unchanged): $72 \le F - 32 \le 81$.
- Add $32$ throughout: $104 \le F \le 113$.
Answer: The temperature must lie between $104^\circ\text{F}$ and $113^\circ\text{F}$.
How many litres of $30\%$ acid solution must be added to $600$ litres of $12\%$ acid solution so that the resulting mixture is more than $15\%$ but less than $18\%$ acid?
Solution- Let $x$ litres of $30\%$ solution be added. Acid added $= 0.30x$; acid already present $= 0.12 \times 600 = 72$ litres; total volume $= 600 + x$.
- Lower bound: $\dfrac{72 + 0.30x}{600 + x} > 0.15 \Rightarrow 72 + 0.30x > 90 + 0.15x \Rightarrow 0.15x > 18 \Rightarrow x > 120$.
- Upper bound: $\dfrac{72 + 0.30x}{600 + x} < 0.18 \Rightarrow 72 + 0.30x < 108 + 0.18x \Rightarrow 0.12x < 36 \Rightarrow x < 300$.
- Combine the two results.
Answer: Between $120$ and $300$ litres of the $30\%$ solution must be added ($120 < x < 300$).
A manufacturer makes $x$ chairs and $y$ tables. Each chair needs $2$ hours and each table $5$ hours of labour, and at most $40$ labour-hours are available; also at least $4$ chairs must be made. Write the system of constraints.
Solution- Labour: $2x + 5y \le 40$.
- "At least $4$ chairs": $x \ge 4$.
- Tables cannot be negative: $y \ge 0$ (and $x \ge 4$ already forces $x \ge 0$).
Answer: $2x + 5y \le 40,\ x \ge 4,\ y \ge 0$.
Find the corner points of the feasible region of $2x + y \le 10,\ x + 3y \le 15,\ x \ge 0,\ y \ge 0$.
Solution- Non-negativity gives the corners on the axes and the origin: $(0,0)$.
- $2x + y = 10$ meets the $x$-axis at $(5,0)$; $x + 3y = 15$ meets the $y$-axis at $(0,5)$.
- Find where the two slanted lines cross: from $2x + y = 10$, $y = 10 - 2x$; substitute into $x + 3(10 - 2x) = 15 \Rightarrow x + 30 - 6x = 15 \Rightarrow -5x = -15 \Rightarrow x = 3$, then $y = 4$.
- So the intersection is $(3, 4)$, and all four corners bound the region.
Answer: Corner points are $(0,0)$, $(5,0)$, $(3,4)$ and $(0,5)$.