Skip to content
Question of 94

Q.Solve the following system of inequalities graphically: x+y≤10x+y \le 10, x+y≥1x+y \ge 1, x−y≤0x-y \le 0, x≥0x \ge 0, y≥0y \ge 0

Assam AhsecAHSEC Higher Secondary (HS) 1st Year Examination 2023Subjective· 4mImportance★★★★★
0% · 0/94 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Plot the boundary lines and axes, shade the common region, and locate its corner points algebraically.

The constraints are: x+y≤10x+y\le10, x+y≥1x+y\ge1, x−y≤0x-y\le0 (i.e. y≥xy\ge x), x≥0x\ge0, y≥0y\ge0.

Boundary lines:

  • L1:x+y=10L_1: x+y=10 — points (10,0)(10,0) and (0,10)(0,10)
  • L2:x+y=1L_2: x+y=1 — points (1,0)(1,0) and (0,1)(0,1)
  • L3:x−y=0L_3: x-y=0, i.e. y=xy=x — the line through the origin at 45°45°

Shading: x+y≤10x+y\le10 means below/left of L1L_1; x+y≥1x+y\ge1 means above/right of L2L_2; y≥xy\ge x means above the line y=xy=x; and we stay in the first quadrant (x,y≥0x,y\ge0).

The feasible region is the set of points satisfying ALL five conditions simultaneously — it is a quadrilateral bounded by parts of L1L_1, L3L_3, L2L_2, and the yy-axis.

Corner points (found by solving pairs of boundary equations):

  • L2∩(x=0)L_2\cap(x=0): 0+y=1⇒(0,1)0+y=1\Rightarrow(0,1)
  • (0,1)(0,1) to (0,10)(0,10) along the yy-axis: corner (0,10)(0,10)
  • L1∩L3L_1\cap L_3: x+x=10⇒x=5,y=5⇒(5,5)x+x=10\Rightarrow x=5,y=5\Rightarrow(5,5)
  • L2∩L3L_2\cap L_3: x+x=1⇒x=0.5,y=0.5⇒(0.5,0.5)x+x=1\Rightarrow x=0.5,y=0.5\Rightarrow(0.5,0.5) …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.