From a point to a region

An equa­tion like a straight line is sat­is­fied by points on a line. An inequal­ity in two vari­ables, such as 2x+3y≤122x + 3y \le 12, is sat­is­fied by a whole region of the plane: every­thing on one side of a line. Such regions are how we pic­ture real lim­its, like a fixed bud­get, as you will see in an exam­ple. We will graph such inequal­i­ties, use a test point to decide which side to shade, and find the region com­mon to a sys­tem of inequal­i­ties. A mixed prac­tice set on the whole chap­ter rounds things off.

One inequal­ity in two vari­ables

The line ax+by=cax + by = c cuts the plane into two half-planes. Every point on one side sat­is­fies ax+by<cax + by < c, and every point on the other side sat­is­fies ax+by>cax + by > c. There is no mix­ing.

Method. First draw the line. Make it dashed for << or >>, since the line itself is not included, and solid for ≤\le or ≥\ge, since it is. Then test any point not on the line; (0,0)(0, 0) is usu­ally eas­i­est. If it sat­is­fies the inequal­ity, shade its side. If not, shade the other side. Because of the no-mix­ing fact above, one test point is enough.

Exam­ple 1. For 2x+3y≤122x + 3y \le 12, draw a solid line through (6,0)(6, 0) and (0,4)(0, 4). Test­ing (0,0)(0, 0) gives 0≤120 \le 12, which is true, so shade the side con­tain­ing the ori­gin.

Graph of the solid line 2x + 3y = 12 through (6, 0) and (0, 4), with the half-plane below it, containing the test point (0, 0), shaded as the solution of 2x + 3y at most 12.
Exam­ple 1: the ori­gin passes the test, so its side is shaded

Exam­ple 2. For y>2x−1y > 2x - 1, the line through (0,−1)(0, -1) and (1,1)(1, 1) is dashed. Test­ing (0,0)(0, 0) gives 0>−10 > -1, true, so shade above the line.

Exam­ple 3. Some lines are par­al­lel to an axis. For x≥3x \ge 3, draw a solid ver­ti­cal line and shade to its right. For y<−2y < -2, draw a dashed hor­i­zon­tal line and shade below it.

Exam­ple 4. A shop­keeper can spend at most ₹600600 on pens at ₹2020 each and note­books at ₹5050 each. So 20x+50y≤60020x + 50y \le 600, with x≥0x \ge 0 and y≥0y \ge 0 because you can­not buy a neg­a­tive num­ber of any­thing. The region is a tri­an­gle in the first quad­rant with ver­tices (0,0),(30,0),(0,12)(0, 0), (30, 0), (0, 12).

Sys­tems

With sev­eral inequal­i­ties, the solu­tion is the region that every one of them shades. A prac­ti­cal tip: shade each one lightly, or bet­ter, mark the side each one rules out, so the com­mon region stands out clearly at the end.

Exam­ple 5. Take x+y≤6x + y \le 6, x−y≥−2x - y \ge -2, x≥0x \ge 0, y≥0y \ge 0. The lines x+y=6x + y = 6 and x−y=−2x - y = -2 meet at (2,4)(2, 4), and the region is the quadri­lat­eral with ver­tices (0,0),(6,0),(2,4),(0,2)(0, 0), (6, 0), (2, 4), (0, 2).

Lines x + y = 6 and x - y = -2 meeting at (2, 4); the shaded region in the first quadrant is the quadrilateral with corners (0, 0), (6, 0), (2, 4) and (0, 2).
Exam­ple 5: the region com­mon to all four inequal­i­ties

Exam­ple 6. A sys­tem can also have no solu­tion. For x+2y≥8x + 2y \ge 8 and x+2y≤4x + 2y \le 4 the two lines are par­al­lel, and the half-planes point away from each other, so they never meet.

Mixed prac­tice

  1. Solve for real xx: 3x−42≥x+14−1\displaystyle \frac{3x - 4}{2} \ge \frac{x + 1}{4} - 1.
  2. Solve: 7≤3x+112≤11\displaystyle 7 \le \frac{3x + 11}{2} \le 11.
  3. Solve the sys­tem 2(x−1)<x+52(x - 1) < x + 5, 3(x+2)>2−x3(x + 2) > 2 - x.
  4. Graph: 3x−2y≥63x - 2y \ge 6; x+y<4x + y < 4; y≤3y \le 3.
  5. Graph the sys­tem 2x+y≥42x + y \ge 4, x+y≤5x + y \le 5, x≥0x \ge 0, y≥0y \ge 0 and list its cor­ner points.
  6. Graph the sys­tem x−2y≤3x - 2y \le 3, 3x+4y≥123x + 4y \ge 12, x≥0x \ge 0, y≥1y \ge 1.
  7. Show that x+y≥7x + y \ge 7 and 2x+2y≤102x + 2y \le 10 have no com­mon solu­tion.
  8. A com­pany makes chairs (44 h of labour each) and tables (66 h each), with at most 240240 labour hours. Write the inequal­ity and graph it for x,y≥0x, y \ge 0.
  9. IQ =MACA×100\displaystyle = \tfrac{MA}{CA} \times 100. For 1212-year-olds the IQ ranges over 80≤IQ≤14080 \le IQ \le 140. Find the range of men­tal age MAMA.
  10. Find all real xx sat­is­fy­ing both ∣x−3∣<4\lvert x - 3 \rvert < 4 and x≥2x \ge 2.

Answers

Show answers
  1. Mul­ti­ply by 44: 6x−8≥x+1−46x - 8 \ge x + 1 - 4; 5x≥55x \ge 5; x≥1x \ge 1.
  2. 14≤3x+11≤2214 \le 3x + 11 \le 22: 1≤x≤113\displaystyle 1 \le x \le \tfrac{11}{3}.
  3. x<7x < 7 and 4x>−44x > -4, x>−1x > -1: (−1,7)(-1, 7).
  4. Solid line through (2,0),(0,−3)(2, 0), (0, -3), shade away from ori­gin; dashed through (4,0),(0,4)(4, 0), (0, 4), shade towards ori­gin; solid y=3y = 3, shade below.
  5. Cor­ners (2,0),(5,0),(0,5),(0,4)(2, 0), (5, 0), (0, 5), (0, 4).
  6. Unbounded region; lines meet x−2y=3x - 2y = 3 and 3x+4y=123x + 4y = 12 at (3.6,0.3)(3.6, 0.3), but y≥1y \ge 1 cuts it: cor­ners (83,1)\displaystyle (\tfrac{8}{3}, 1), (5,1)(5, 1) and (0,3)(0, 3), where 3x+4y=123x + 4y = 12 meets the yy-axis, extend­ing upward.
  7. The sec­ond says x+y≤5x + y \le 5, con­tra­dict­ing x+y≥7x + y \ge 7.
  8. 4x+6y≤2404x + 6y \le 240: tri­an­gle with ver­tices (0,0),(60,0),(0,40)(0, 0), (60, 0), (0, 40).
  9. 80≤MA12×100≤140\displaystyle 80 \le \tfrac{MA}{12} \times 100 \le 140: 9.6≤MA≤16.89.6 \le MA \le 16.8.
  10. −1<x<7-1 < x < 7 and x≥2x \ge 2: [2,7)[2, 7).