How to use this set

Here are ten ques­tions on turn­ing every­day sit­u­a­tions into a pair of lin­ear equa­tions, on the three cases, and on the coef­fi­cient test. A few ask you to draw, but most ask you to clas­sify a pair or find a miss­ing con­stant. One habit to keep through­out: put each equa­tion into the form ax+by+c=0ax + by + c = 0 before you read off its coef­fi­cients.

Sec­tion A, Form­ing the equa­tions

  1. A pen and two note­books cost ₹ 70. Three pens and one note­book cost ₹ 60. Write a pair of lin­ear equa­tions, tak­ing a pen to cost ₹ xx and a note­book ₹ yy.
  2. Meera is 66 years older than her brother. In 44 years, the sum of their ages will be 3434. Write the pair of equa­tions for their present ages.
  3. The perime­ter of a rec­tan­gle is 2828 cm and its length is 22 cm more than its width. Form the equa­tions.

Sec­tion B, Clas­si­fy­ing

  1. Say whether each pair is con­sis­tent, incon­sis­tent, or depen­dent: (a) x−4y+2=0x - 4y + 2 = 0 and 3x−12y+6=03x - 12y + 6 = 0; (b) 2x+5y=72x + 5y = 7 and 4x+10y=94x + 10y = 9; (c) x+y=3x + y = 3 and x−y=1x - y = 1.
  2. Draw the lines 2x+y=62x + y = 6 and x+y=4x + y = 4 and read off their point of inter­sec­tion. Then con­firm it alge­braically.
  3. Draw x−y=2x - y = 2 and 2x−2y=82x - 2y = 8. Describe what you see, and state the num­ber of solu­tions.

Sec­tion C, Find­ing a con­stant

  1. Find pp so that 2x+3y=72x + 3y = 7 and px+6y=14px + 6y = 14 have infi­nitely many solu­tions.
  2. Find mm so that mx+3y=4mx + 3y = 4 and 2x+6y=92x + 6y = 9 have no solu­tion.
  3. Find all val­ues of kk for which kx+4y=1kx + 4y = 1 and 3x+2y=53x + 2y = 5 have a unique solu­tion.
  4. A pair of equa­tions is ax+by=cax + by = c and 2ax+2by=2c2ax + 2by = 2c with a,ba, b not both zero. With­out any cal­cu­la­tion, state how many solu­tions it has and why.

Answers

Show answers
  1. x+2y=70x + 2y = 70 and 3x+y=603x + y = 60.
  2. Let the brother be yy years and Meera xx years. Then x−y=6x - y = 6, and (x+4)+(y+4)=34(x + 4) + (y + 4) = 34, i.e. x+y=26x + y = 26.
  3. Width ww, length ll: 2l+2w=282l + 2w = 28, i.e. l+w=14l + w = 14, and l−w=2l - w = 2.
  4. (a) 13=−4−12=26\displaystyle \frac{1}{3} = \frac{-4}{-12} = \frac{2}{6}, all 13\displaystyle \frac{1}{3}: depen­dent. (b) Write as 2x+5y−7=02x + 5y - 7 = 0, 4x+10y−9=04x + 10y - 9 = 0: 12=12≠−7−9\displaystyle \frac{1}{2} = \frac{1}{2} \neq \frac{-7}{-9}: incon­sis­tent. (c) 11≠1−1\displaystyle \frac{1}{1} \neq \frac{1}{-1}: con­sis­tent, unique.
  5. First line through (0,6)(0, 6) and (3,0)(3, 0); sec­ond through (0,4)(0, 4) and (4,0)(4, 0). They meet at (2,2)(2, 2). Check: 4+2=64 + 2 = 6 ✓, 2+2=42 + 2 = 4 ✓.
  6. You see two par­al­lel lines: 12=−1−2≠−2−8\displaystyle \frac{1}{2} = \frac{-1}{-2} \neq \frac{-2}{-8}. There is no solu­tion, because the sec­ond equa­tion is really x−y=4x - y = 4, and x−yx - y can­not be both 22 and 44.
  7. 2p=36=−7−14=12\displaystyle \frac{2}{p} = \frac{3}{6} = \frac{-7}{-14} = \frac{1}{2}, so p=4p = 4.
  8. Need m2=36≠−4−9\displaystyle \frac{m}{2} = \frac{3}{6} \neq \frac{-4}{-9}. So m=1m = 1; and 12≠49\displaystyle \frac{1}{2} \neq \frac{4}{9}, so the lines are par­al­lel.
  9. Unique solu­tion needs k3≠42\displaystyle \frac{k}{3} \neq \frac{4}{2}, i.e. k≠6k \neq 6. Every real kk except 66.
  10. Infi­nitely many. The sec­ond equa­tion is exactly twice the first, so they describe the same line and every point on it is a solu­tion.
Answer 5: line 2x + y = 6 through (0, 6) and (3, 0), and line x + y = 4 through (0, 4) and (4, 0), crossing at the point (2, 2).
Answer 5: the lines meet at (2, 2)
Answer 6: line x - y = 2 through (2, 0) and line 2x - 2y = 8, the same as x - y = 4, through (4, 0). They are parallel and never meet: no solution.
Answer 6: par­al­lel lines, no solu­tion