Solve the system x + y > 4 and 2x - y < 3 graphically

Question

Solve the system of linear inequalities graphically:

x + y > 4 \quad \text{and} \quad 2x - y < 3

Find the solution region and verify with a test point.

Solution — Step by Step

Replace the inequalities with equations to find the boundary lines:

Line 1: $x + y = 4$

Find two points: When $x = 0$ , $y = 4$ → point $(0, 4)$ . When $y = 0$ , $x = 4$ → point $(4, 0)$ .

Line 2: $2x - y = 3$

When $x = 0$ , $y = -3$ → point $(0, -3)$ . When $y = 0$ , $x = 1.5$ → point $(1.5, 0)$ .

Draw both lines. Since inequalities are strict (> and <), draw dashed lines (points on the boundary are NOT included in the solution).

For $x + y > 4$ : Test the origin $(0, 0)$ : $0 + 0 = 0$ , which is NOT > 4. So the origin is NOT in the solution region. The solution half-plane for this inequality is the side away from the origin — above the line $x + y = 4$ .

For $2x - y < 3$ : Test the origin $(0, 0)$ : $2(0) - 0 = 0 < 3$ . Yes, this IS < 3. So the origin IS in the solution region. The solution half-plane is the side containing the origin — below/left of the line $2x - y = 3$ .

The solution to the system is the region that satisfies both inequalities simultaneously — the intersection of the two half-planes.

This is the region that is:

Above (and to the right of) the line $x + y = 4$ , AND
Below (and to the left of) the line $2x - y = 3$

The two dashed lines intersect at a point — find it: $x + y = 4$ and $2x - y = 3$ . Adding: $3x = 7$ , so $x = 7/3$ , $y = 4 - 7/3 = 5/3$ . Intersection: $(7/3, 5/3)$ .

The solution region is the open angular region above line 1 and below line 2, extending to the upper right.

Choose a point in the expected solution region, e.g., $(5, 1)$ :

Check inequality 1: $5 + 1 = 6 > 4$ ✓

Check inequality 2: $2(5) - 1 = 9$ , and $9 < 3$ ? NO ✗

$(5, 1)$ fails inequality 2 — it’s in the wrong region. Try $(2, 3)$ :

Check inequality 1: $2 + 3 = 5 > 4$ ✓

Check inequality 2: $2(2) - 3 = 1 < 3$ ✓

$(2, 3)$ satisfies both — it lies in the solution region.

Why This Works

A linear inequality in two variables divides the plane into two half-planes. The boundary line separates the plane, and one side satisfies the inequality while the other doesn’t. For a system of inequalities, we need the region that satisfies all conditions simultaneously — the intersection of all half-planes.

The test-point method is reliable because any line divides the plane into exactly two sides, and the origin $(0, 0)$ is typically the most convenient test point (unless the line passes through the origin, in which case use $(0, 1)$ or $(1, 0)$ ).

Alternative Method

Algebraic check: From $x + y > 4$ , we get $y > 4 - x$ . From $2x - y < 3$ , we get $y > 2x - 3$ . Combined: $y > \max(4-x, 2x-3)$ . For any specific $x$ , compute both bounds and the solution requires $y$ to exceed the larger one. For $x = 2$ : need $y > \max(2, 1) = 2$ , i.e., $y > 2$ . Point $(2, 3)$ satisfies this ✓.

For CBSE Class 11 (Linear Inequalities chapter), graphical solution questions with 2–3 inequalities are standard 3-mark questions. Always: (1) draw boundary lines, (2) use dashed for strict (<, >) and solid for non-strict (≤, ≥), (3) shade the solution region, (4) verify with a test point. All four steps are required for full marks.

Common Mistake

Students often shade the wrong side. The most reliable method: always test the origin $(0, 0)$ in the inequality. If it satisfies the inequality, shade the side containing the origin. If it doesn’t, shade the other side. Never guess by visual inspection — always test. Also, for strict inequalities, the boundary line is dashed — forgetting the dashed line loses marks in CBSE.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next